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2.5: Linear Equations - Manipulating and Solving (Solving the Puzzle)


You are shopping at Old Navy for seven new outfits. How do you spend $110 to acquire all the needed outfits without exceeding your budget while getting as many $30 items as possible?

This is a problem of linear equations, and it illustrates how you can use them to make an optimal decision. Let (L) represent the quantity of clothing at the low price point of $10, and (H) represent the quantity of clothing at the high price point of $30. This results in the following algebraic equations:

[L+H=7 ext { (the total number of outfits you need) } onumber]

[$ 10 L+$ 30 H=$ 110 ext { (your total budget) } onumber]

By simultaneously solving these equations you can determine how many outfits at each price point you can purchase.

You will encounter many situations like this in your business career, for example, in making the best use of a manufacturer’s production capacity. Assume your company makes two products on the same production line and sells all its output. Each product contributes differently to your profitability, and each product takes a different amount of time to manufacture. What combination of each of these products should you make such that you operate your production line at capacity while also maximizing the profits earned? This section explores how to solve linear equations for unknown variables.

Understanding Equations

To manipulate algebraic equations and solve for unknown variables, you must first become familiar with some important language, including linear versus nonlinear equations and sides of the equation.

The goal in manipulating and solving a linear equation is to find a value for the unknown variable that makes the equation true. If you substitute a value of (x = −1) into the above example, the left-hand side of the equation equals the right-hand side of the equation (see Figure below). The value of (x = −1) is known as the root, or solution, to the linear equation.

Solving One Linear Equation with One Unknown Variable

In your study of solving linear equations, you need to start by manipulating a single equation to solve for a single unknown variable. Later in this section you will extend from this foundation to the solution of two linear equations with two unknowns.

How It Works

To determine the root of a linear equation with only one unknown variable, apply the following steps:

Step 1: Your first goal is to separate the terms containing the literal coefficient from the terms that only have numerical coefficients. Collect all of the terms with literal coefficients on only one side of the equation and collect all of the terms with only numerical coefficients on the other side of the equation. It does not matter which terms go on which side of the equation, so long as you separate them.

To move a term from one side of an equation to another, take the mathematical opposite of the term being moved and add it to both sides. For example, if you want to move the +3 in (4x + 3 = −2x − 3) from the left-hand side to the right-hand side, the mathematical opposite of +3 is −3. When you move a term, remember the cardinal rule: What you do to one side of an equation you must also do to the other side of the equation. Breaking this rule breaks the equality in the equation.

Step 2: Combine all like terms on each side and simplify the equation according to the rules of algebra.

Step 3: In the term containing the literal coefficient, reduce the numerical coefficient to a 1 by dividing both sides of the equation by the numerical coefficient.

Important Notes

When you are unsure whether your calculated root is accurate, an easy way to verify your answer is to take the original equation and substitute your root in place of the variable. If you have the correct root, the left-hand side of the equation equals the right-hand side of the equation. If you have an incorrect root, the two sides will be unequal. The inequality typically results from one of the three most common errors in algebraic manipulation:

  1. The rules of BEDMAS have been broken.
  2. The rules of algebra have been violated.
  3. What was done to one side of the equation was not done to the other side of the equation.

Things To Watch Out For

When you move a term from one side of the equation to another using multiplication or division, remember that this affects each and every term on both sides of the equation. To remove the (x) from the denominator in the following equation, multiply both sides of the equation by (x):

(dfrac{5}{x}+dfrac{1}{x}=dfrac{2}{x}+2) becomes (xleft(dfrac{5}{x}+dfrac{1}{x} ight)=left(dfrac{2}{x}+2 ight) x) which then becomes (5+1=2+2 x)

Multiplying every term on both sides by (x) maintains the equality.

Paths To Success

Negative numbers can cause some people a lot of grief. In moving terms from a particular side of the equation, many people prefer to avoid negative numerical coefficients in front of literal coefficients. Revisiting (4x + 3 = −2x − 3), you could move the (4x) from the left side to the right side by subtracting (4x) from both sides. However, on the right side this results in (−6x). The negative is easily overlooked or accidentally dropped in future steps. Instead, move the variable to the left side of the equation, yielding a positive coefficient of (6x).

Example (PageIndex{1}): How to Solve the Opening Example

Take the ongoing example in this section and solve it for (x): (4x + 3 = −2x − 3)

Solution

This is a linear equation since the exponent on the variable is 1. You are to solve the equation and find the root for (x).

What You Already Know

The equation has already been provided.

How You Will Get There

Apply the three steps for solving linear equations. To arrive at the root, you must follow the rules of algebra, BEDMAS, and equality.

Perform

Step 1: Move terms with literal coefficients to one side and terms with only numerical coefficients to the other side. Let’s collect the literal coefficient on the left-hand side of the equation. Move (−2x) to the left-hand side by placing (+2x) on both sides.

[4x + 3 = −2x – 3 onumber ]

On the right-hand side, the (−2x) and (+2x) cancel out to zero.

[4x + 3 (f{+ 2x}) = −2x − 3 (f{+ 2x}) onumber ]

Step 1 (continued): All of the terms with the literal coefficient are now on the left. Let’s move all of the terms containing only numerical coefficients to the right-hand side. Move the +3 to the right-hand side by placing −3 on both sides.

[4x + 3 + 2x = −3 onumber ]

On the left-hand side, the +3 and −3 cancel out to zero.

[4x + 3 + 2x (f{– 3}) = −3 (f{− 3}) onumber ]

Step 2: The terms are now separated. Combine like terms according to the rules of algebra.

[4x + 2x = −3 – 3 onumber ]

Step 3: The term with the literal coefficient is being multiplied by the numerical coefficient of 6. Therefore, divide both sides by 6.

[f{6x = −6} onumber ]

The left-hand side numerical coefficients will divide to 1. Resolve the numerical coefficients on the right-hand side.

[dfrac{6 x}{f{6}}=dfrac{-6}{f{6}} onumber ]

This is the root of the equation.

[x = −1 onumber]

The root of the equation is (x = −1). To verify the accuracy of your manipulation, take the root of (x = −1) and substitute it into the original equation:

[4(−1) + 3 = −2(−1) − 3 onumber]

[−4 + 3 = 2 − 3 onumber]

[−1 = −1 onumber]

The left-hand side equals the right-hand side, so the root is correct.

Example (PageIndex{2}): Solving a Linear Equation with One Unknown Variable

Solve the following equation for (m): (dfrac{3 m}{4}+2 m=4 m-15)

Solution

This is a linear equation since the exponent on the variable is a 1. You are to solve the equation and find the root for (m).

What You Already Know

The equation has already been provided.

How You Will Get There

Simplify the equations first and then apply the three steps for solving linear equations. To arrive at the root you must follow the rules of algebra, BEDMAS, and equality. You can use an approach that avoids negatives.

Perform

First, simplify all fractions to make the equation easier to work with.

[dfrac{3 m}{4}+2 m=4 m-15 onumber ]

Still simplifying, collect like terms where possible.

[(f{0.75m}) + 2m = 4m − 15 onumber ]

Step 1: Collect all terms with the literal coefficient on one side of the equation. Move all terms with literal coefficients to the right-hand side.

[(f{2.75m })= 4m − 15 onumber ]

Step 1 (continued): Combine like terms and move all terms with only numerical coefficients to the left-hand side.

[2.75m (f{− 2.75m}) = 4m − 15 (f{− 2.75m}) onumber ]

On the left-hand side, the (+2.75m) and (−2.75m) cancel each other out. Now move the numerical coefficients to the left-hand side.

[(f{0}) = 4m − 15 (f{− 2.75m}) onumber ]

On the right-hand side, the −15 and +15 cancel each other out.

[0 (f{+ 15 })= 4m − 15 − 2.75m (f{+ 15}) onumber ]

Step 2: Combine like terms on each side.

[0 (f{+ 15}) = 4m − 2.75m onumber ]

Step 3: Divide both sides by the numerical coefficient that accompanies the literal coefficient.

[f{15 = 1.25m} onumber ]

Simplify.

[dfrac{15}{f{1.25}}=dfrac{1.25 m}{f{1.25}} onumber ]

This is the root of the equation.

[12=m onumber ]

The root of the equation is (m = 12).

This makes both sides of the equation,

(dfrac{3 m}{4}+2 m) and (4 m-15), equal 33.

Example (PageIndex{3}): Solving a Linear Equation with One Unknown Variable Containing Fractions

Solve the following equation for (b) and round your answer to four decimals: (dfrac{5}{8} b+dfrac{2}{5}=dfrac{17}{20}-dfrac{b}{4})

Solution

This is a linear equation since the exponent on the variable is a 1. You are to solve the equation and find the root for (b).

What You Already Know

The equation has already been provided. Although you could attempt to clear each and every fraction or try to find a common denominator, recall that you can eliminate fractions by converting them to decimals.

How You Will Get There

Simplify the fractions into decimal form. Then apply the three steps for solving linear equations. To arrive at the root, you must follow the rules of algebra, BEDMAS, and equality.

Perform

Simplify the fractions and convert to decimals.

[dfrac{5}{8} b+dfrac{2}{5}=dfrac{17}{20}-dfrac{b}{4} onumber ]

Step 1: Move the literal coefficient terms to the left-hand side.

[(f{0.625})b (f{+ 0.4}) = (f{0.85 − 0.25})b onumber ]

The literal coefficients on the right-hand side cancel each other out.

[0.625b + 0.4 + (f{0.25b}) = 0.85 − 0.25b + (f{0.25b}) onumber ]

Move the numerical coefficient terms to the right-hand side.

[0.625b + 0.4 + 0.25b = 0.85 onumber ]

The numerical coefficients on the left-hand side cancel each other out.

[0.625b + 0.4 +0.25b (f{− 0.4}) = 0.85 (f{− 0.4}) onumber ]

Step 2: Combine like terms on each side.

[0.625b + 0.25b = 0.85 − 0.4 onumber ]

Step 3: Divide both sides by the numerical coefficient that accompanies the literal coefficient.

[f{0.875b = 0.45} onumber ]

Simplify.

[dfrac{0.875 b}{f{0.875}}=dfrac{0.45}{f{0.875}} onumber ]

Round to four decimals as instructed.

[b = 0.514285 onumber ]

This is the root.

[b = 0.5143 onumber ]

The root of the equation, rounded to four decimals, is (b = 0.5143).

Solving Two Linear Equations with Two Unknown Variables

The manipulation process you have just practiced works well for solving one linear equation with one variable. But what happens if you need to solve two linear equations with two variables simultaneously? Remember when you were at Old Navy purchasing seven outfits earlier in this chapter (equation 1)? You needed to stay within a pricing budget (equation 2). Each equation had two unknown variables representing the number of lower-priced and higher-priced outfits.

The goal is to reduce two equations with two unknowns into a single linear equation with one unknown. Once this transformation is complete, you then identify the unknown variable by applying the three-step procedure for solving one linear equation, as just discussed.

When you work with two linear equations with two unknowns, the rules of algebra permit the following two manipulations:

  1. What you do to one side of the equation must be done to the other side of the equation to maintain the equality. Therefore, you can multiply or divide any equation by any number without changing the root of the equation. For example, if you multiply all terms of (x + y = 2) by 2 on both sides, resulting in (2x + 2y = 4), the equality of the equation remains unchanged and the same roots exist.
  2. Terms that are on the same side of an equation can be added and subtracted between equations by combining like terms. Each of the two equations has a left side and right side. This rule permits taking the left side of the first equation and either adding or subtracting like terms on the left side of the second equation. When you perform this action, remember the first rule above. If you add the left sides of the equations together, you then must add the right side of both equations together to maintain equality.

How It Works

Follow these steps to solve two linear equations with two unknown variables:

Step 1: Write the two equations one above the other, vertically lining up terms that have the same literal coefficients and terms that have only the numerical coefficient. If necessary, the equations may need to be manipulated such that all of the literal coefficients are on one side with the numerical coefficients on the other side.

Step 2: Examine your two equations. Through multiplication or division, make the numerical coefficient on one of the terms containing a literal coefficient exactly equal to its counterpart in the other equation.

Step 3: Add or subtract the two equations as needed so as to eliminate the identical term from both equations.

Step 4: In the new equation, solve for the last literal coefficient.

Step 5: Substitute the root of the known literal coefficient into either of the two original equations. If one of the equations takes on a simpler structure, pick that equation.

Step 6: Solve your chosen equation for the other literal coefficient.

Paths To Success

Sometimes it is unclear exactly how you need to multiply or divide the equations to make two of the terms identical. For example, assume the following two equations:

[4.9x + 1.5y = 38.3 onumber]

[2.7x − 8.6y = 17.8 onumber]

If the goal is to make the terms containing the literal coefficient (x) identical, there are two alternative solutions:

  1. Take the larger numerical coefficient for (x) and divide it by the smaller numerical coefficient. The resulting number is the factor for multiplying the equation containing the smaller numerical coefficient. In this case, (4.9 div 2.7 =1 . overline{814}). Multiply all terms in the second equation by (1 . overline{814}) to make the numerical coefficients for (x) equal to each other, resulting in this pair of equations:

[4.9x + 1.5y = 38.3 onumber]

[4.9 x-15.6 overline{074} y=32.3 overline{037} ext { (every term multiplied by } 1 . overline{814}) onumber]

  1. Take the first equation and multiply it by the numerical coefficient in the second equation. Then take the second equation and multiply it by the numerical coefficient in the first equation. In this case, multiply all terms in the first equation by 2.7. Then multiply all terms in the second equation by 4.9.

[13.23 x+4.05 y=103.41 ext { (every term multiplied by } 2.7) onumber]

[13.23 x-42.14 y=87.22 ext { (every term multiplied by 4.9) } onumber]

Note that both approaches successfully result in both equations having the same numerical coefficient in front of the literal coefficient (x).

Paths To Success

Ultimately, every pairing of linear equations with two unknowns can be converted into a single equation through substitution. To make the conversion, do the following:

  1. Solve either equation for one of the unknown variables.
  2. Take the resulting algebraic expression and substitute it into the other equation. This new equation is solvable for one of the unknown variables.
  3. Substitute your newfound variable into one of the original equations to determine the value for the other unknown variable.

Take the following two equations:

[a + b = 4 quad quad 2a + b = 6 onumber]

  1. Solving the first equation for a results in (a = 4 - b).
  2. Substituting the expression for a into the second equation and solving for b results in (2(4 - b) + b = 6), which solves as (b = 2).
  3. Finally, substituting the root of b into the first equation to calculate a gives (a + 2 = 4) resulting in (a = 2). Therefore, the roots of these two equations are (a = 2) and (b = 2).

Example (PageIndex{4}): Buying Those Outfits

Recall from the section opener that in shopping for outfits there are two price points of $10 and $30, your budget is $110, and that you need seven articles of clothing. The equations below represent these conditions. Identify how many low-priced outfits ((L)) and high-priced outfits ((H)) you can purchase.

[L + H = 7 ext{ } $10L + $30H = $110 onumber]

Solution

You need to determine the quantity of low-price-point items, or (L), and high-price-point items, or (H), that are within your limited budget. Note that the exponents on the variables are 1 and that there are two unknowns. So there are two linear equations with two unknowns.

What You Already Know

You require seven articles of clothing and only have a budget of $110. The equations express the relationships of quantity and budget.

How You Will Get There

Apply the six-step procedure for solving two linear equations with two unknowns.

Step 1:

Write the equations one above the other and line them up.

[egin{array} {lllll} {L} & + &{H}& = &{7} {$10L} & + &{$30H}& = &{$110} end{array} onumber ]

Step 2:

Multiply all terms in the first equation by 10 so that (L) has the same numerical coefficient in both equations.

[egin{array} {lllll} {10L} & + &{10H}& = &{70} {$10L} & + &{$30H}& = &{$110} end{array} onumber ]

Step 3:

Subtract the equations by subtracting all terms on both sides.

[egin{array} {llllll} { } &{10L} & + &{10H}& = &{70} { ext{subtract}} &{$10L} & + &{$30H}& = &{$110}{ } &{ } & - &{$20H}& = &{−$40} end{array} onumber ]

Step 4:

Solve for (H) by dividing both sides by −20.

[dfrac{-$ 20 H}{-$ 20}=dfrac{-$ 40}{-$ 20} quad H=2 onumber ]

Step 5:

Substitute the known value for (H) into one of the original equations. The first equation is simple, so choose that one.

[egin{array} {lllll} {L} & + &{H}& = &{7} {L} & + &{2}& = &{7} end{array} onumber ]

Step 6:

Solve for (L) by subtracting 2 from both sides. You now have the roots for (L) and (H).

[egin{array} {lllllllll} {L}&+&{2} & - &{2}& = &{7}&-&{2} { } & { } &{ } & { } &{L}& = &{5} & { } & { } end{array} onumber ]

You can purchase five articles of clothing at the low price point and two articles of clothing at the high price point. This allows you to purchase seven articles of clothing and stay within your budget of $110.

Paths To Success

One of the most difficult areas of mathematics involves translating words into mathematical symbols and operations. To assist in this translation, the table below lists some common language and the mathematical symbol that is typically associated with the word or phrase.

LanguageMath Symbol

Sum

Addition

In addition to

In excess

Increased by

Plus

+

Subtract

Decreased by

Diminished by

Less Minus

Difference

Reduced by

-

Multiplied by

Times

Percentage of

Product of

Of

×

Divide

Division

Divisible

Quotient

Per÷

Becomes

Is/Was/Were

Will be

Results in

Totals

=
More thanGreater than>
Less thanLower than<
Greater than or equal to
Less than or equal to
Not equal to

Example (PageIndex{5}): Solving Two Linear Equations with Two Unknowns for an Amusement Park

Tinkertown Family Fun Park charges $15 for a child wrist band and $10.50 for an adult wrist band. On a warm summer day, the amusement park had total wrist band revenue of $15,783 from sales of 1,279 wrist bands. How many adult and child wrist bands did the park sell that day?

Solution

You need the number of both adult and child wrist bands sold on the given day. Therefore, you must identify two unknowns.

What You Already Know

The price of the wrist bands,total quantity, and sales are known:

Child wrist band price = $15

Adult wrist band price = $10.50

Total revenue = $15,783

Total unit sales = 1,279

The quantity of adult wrist bands sold and the quantity of child wrist bands sold are unknown:

Adult wrist bands quantity = (a)

Child wrist bands quantity = (c)

How You Will Get There

  1. Work with the quantities first. Calculate the total unit sales by adding the number of adult wrist bands to the number of child wrist bands:

[# ext { of adult wrist bands }+# ext { of child wrist bands }= ext { total unit sales } onumber]

[a + c = 1,279 onumber]

  1. Now consider the dollar figures. Total revenue for any company is calculated as unit price multiplied by units sold. In this case, you must sum the revenue from two products to get the total revenue.

[ ext { Total adult revenue }+ ext { Total child revenue }= ext { Total revenue } onumber]

[ ext { (Adult price } imes ext { Adult guantity })+ ext { (Child price } imes ext { Child quantity) }= ext { Total revenue } onumber]

[$ 10.50 a+$ 15 c=$ 15,783 onumber]

  1. Apply the six-step procedure for solving two linear equations with two unknowns.

Perform

Step 1:

Write the equations one above the other and line them up.

[egin{array} {lllll} {a} & + &{c}& = &{1,279} {$10.50a} & + &{$15c}& = &{$15,783} end{array} onumber ]

Step 2:

Multiply all terms in the first equation by 10.5, resulting in a having the same numerical coefficient in both equations.

[egin{array} {lllll} {f{10.50} a} & + &{f{10.50} c} & = & {f{13,429.50}} {$10.50a} & + &{$15c}& = &{$15,783} end{array} onumber ]

Step 3:

Subtract the equations by subtracting all terms on both sides.

[egin{array} {llllll} { } & {f{10.50} a} & + &{f{10.50} c} & = & {f{13,429.50}} { ext{Subtract}} & {underline{$10.50a}} & {underline{+}} &{underline{$15c}} & {underline{=}} &{underline{$15,783}} { } & { } & { } & {f{-4.5c}} & {f{=}} & {f{-2,353.50}} end{array} onumber ]

Step 4:

Solve for (c) by dividing both sides by −4.5.

[dfrac{-4.5 c}{-4.5}=dfrac{-2,353.50}{-4.5} quad c=523 onumber ]

Step 5:

Substitute the known value for (c) into one of the original equations. The first equation is simple, so choose that one.

[egin{array} {lllll} {a} & + & {c} & = &{1,279} {a} & + & {f{523}} & = &{1,279} end{array} onumber ]

Step 6:

Solve for a by subtracting 523 from both sides. You now have the roots for (a) and (c).

[egin{aligned} a+523 f{-523} &=1,279 f{-523} a &=756 end{aligned} onumber ]

Tinkertown Family Fun Park sold 523 child wrist bands and 756 adult wrist bands.


2.5: Linear Equations - Manipulating and Solving (Solving the Puzzle)

An equation Statement indicating that two algebraic expressions are equal. is a statement indicating that two algebraic expressions are equal. A linear equation with one variable An equation that can be written in the standard form a x + b = 0 , where a and b are real numbers and a ≠ 0 . , x, is an equation that can be written in the standard form a x + b = 0 where a and b are real numbers and a ≠ 0 . For example,

A solution Any value that can replace the variable in an equation to produce a true statement. to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation 3 x − 12 = 0 is x and the solution is x = 4 . To verify this, substitute the value 4 in for x and check that you obtain a true statement.

3 x − 12 = 0 3 ( 4 ) − 12 = 0 12 − 12 = 0 0 = 0 ✓

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example 1

Is a = − 1 2 a solution to − 10 a + 5 = 25 ?

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

− 10 a + 5 = − 10 ( − 1 2 ) + 5 = 5 + 5 = 10 ≠ 25 ✗

Answer: No, a = − 1 2 does not satisfy the equation.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations Equations with the same solution set. as equations with the same solution set.

3 x − 5 = 16 3 x = 21 x = 7 > E q u i v a l e n t e q u a t i o n s

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, <7>. To obtain equivalent equations, use the following properties of equality Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers. . Given algebraic expressions A and B, where c is a nonzero number:

Addition property of equality:

Subtraction property of equality:

Multiplication property of equality:

Division property of equality:

Note: Multiplying or dividing both sides of an equation by 0 is carefully avoided. Dividing by 0 is undefined and multiplying both sides by 0 results in the equation 0 = 0.

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form a x + b = c , then we can solve it in two steps. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking the solution in the following examples is left to the reader.

Example 2

7 x − 2 = 19 7 x − 2 + 2 = 19 + 2 A d d 2 t o b o t h s i d e s . 7 x = 21 7 x 7 = 21 7 D i v i d e b o t h s i d e s b y 7 . x = 3

Example 3

When no sign precedes the term, it is understood to be positive. In other words, think of this as 56 = + 8 + 12 y . Therefore, we begin by subtracting 8 on both sides of the equal sign.

56 − 8 = 8 + 12 y − 8 48 = 12 y 48 12 = 12 y 12 4 = y

It does not matter on which side we choose to isolate the variable because the symmetric property Allows you to solve for the variable on either side of the equal sign, because x = 5 is equivalent to 5 = x . states that 4 = y is equivalent to y = 4 .

Example 4

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient 5 3 .

5 3 x + 2 = − 8 5 3 x + 2 − 2 = − 8 − 2 S u b t r a c t 2 o n b o t h s i d e s . 5 3 x = − 10 3 5 ⋅ 5 3 x = 3 5 ⋅ ( − 10 ) − 2 M u l t i p l y b o t h s i d e s b y 3 5 . 1 x = 3 ⋅ ( − 2 ) x = − 6

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Try this! Solve: 2 3 x + 1 2 = − 5 6 .


Solving by Combination

The basic principle of solving by combination is to manipulate two equations so that, when the equations are added together, one of the variables cancels out. Since one of the variables cancels, this method is sometimes called the elimination method.

Let’s use combination to solve this system of two equations:

This system of equations is suited for combination, because there is already a 2x in both equations. Therefore, if we subtract equation (1) from equation (2) – or, equivalently, multiply equation (1) by -1 and add the two equations – we have a single equation with y:

Dividing both sides, we find that y = -4/3. We can then plug y back into either original equation to get the value of x, as we did when solving by substitution.

We can still solve by combination even if the variables aren’t lined up so nicely. For example, we can start over and solve the system of equations by making the y‘s cancel, rather than the x‘s. To do that, we can multiply the first equation (1) by the number 2 on both sides:

Now subtracting (2) from that result gives us:

Solving, we find x = 7. To finish the job, we substitute x = 7 into either of the original equations. If we plug x = 7 into (1), we get:

Subtracting 14 from both sides, we get

And dividing by 3, we find that

So the solution to the two equations (1) and (2) is:

Most people prefer the substitution method to the combination method. However, the combination method will prove much faster on certain questions, so if you don’t consider using it, you are likely to lose time or a correct answer on the Quant section. Furthermore, you want to be comfortable with the concept that equations can be added, since a given equation is after all equal on both sides, since that fact can be useful even when you are not solving a system of linear equations by the combination method.


As I mentioned before, our variables are formatted as below:

To ensure that each box has only value, we can keep the row, col constant and vary the value from 1 to 9. The sum of the binary values should be equal to 1 since only one variable will be equal to 1 and others must be 0.

(value = 1, row =1, col =1) + (value = 2, row =1, col =1) + (value = 3, row =1, col =1) + (value = 4, row =1, col =1) + (value = 5, row =1, col =1) + (value = 6, row =1, col =1) + (value = 7, row =1, col =1) +(value = 8, row =1, col =1) +(value = 9, row =1, col =1) == 1

We will need to perform this check for all different combinations of row, col


How to Find Solving Linear Equations Calculator?

A linear equation is defined as an equation that is written for two different variables. This equation will be a linear combination of these two variables and a constant.

An equation of the form Ax + By = C. Here, x and y are variables, and A, B, and C are constants.

Solved Example:

Solve 2x + y = 7 and x + y = 5

Solution:

Similarly, you can try the calculator to find the algebra value for a given equation


ALGEBRAIC MANIPULATION PROBLEMS

If x > 0 and  x 2 - 2x - 35  =  0, then find the value of :

If t < 0 and (t - 1) 2  = 16, what is the value of t 2  ?

Let "x" be a real number which satisfies the relations (2x-5)  > 2 and (3x+3) < 18. Which of the values can "x" take ?

Calculate the fifth term of the sequence defined as

Find the domain of the function : 

We hope that the students will be able to solve the a lgebraic manipulation problems1 on SAT above. 

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Maze Solving Equations Activities

Solving One-Step Equations

  • 2-1 Solving One-Step Equations - Answers - Maze Activity (PDF - Member Only)
  • 2-1 Solving One-Step Equations - Maze Activity (Editable - Member Only)
  • ⭐ Solving One-Step Equations - Maze Activity(PDF - FREEBIE)

Solving Two-Step Equations

  • 2-2 Solving Two-Step Equations - Answers - Maze Activity (PDF - Member Only)
  • 2-2 Solving Two-Step Equations - Maze Activity (Editable - Member Only)
  • ⭐ Solving Two-Step Equations - Maze Activity(PDF - FREEBIE)

Solving Multi-Step Equations

  • 2-3 Solving Multi-Step Equations - Answers - Maze Activity (PDF - Member Only)
  • 2-3 Solving Multi-Step Equations - Maze Activity (Editable - Member Only)
  • ⭐ Solving Multi-Step Equations - Maze Activity(PDF - FREEBIE)

Solving Equations with Variables on Both Sides

  • 2-4 Solving Equations with Variables on Both Sides - Answers - Maze Activity (PDF - Member Only)
  • 2-4 Solving Equations with Variables on Both Sides - Maze Activity (Editable - Member Only)
  • ⭐ Solving Equations with Variables on Both Sides - Maze Activity(PDF - FREEBIE)

Literal Equations and Formulas

  • 2-5 Literal Equations and Formulas - Answers - Maze Activity (PDF - Member Only)
  • 2-5 Literal Equations and Formulas - Maze Activity (Editable - Member Only)
  • ⭐ Literal Equations and Formulas - Maze Activity(PDF - FREEBIE)

Ratios, Rates, and Conversions

  • 2-6 Ratios, Rates, and Conversions - Answers - Maze Activity (PDF - Member Only)
  • 2-6 Ratios, Rates, and Conversions - Maze Activity (Editable - Member Only)
  • ⭐ Ratios, Rates, and Conversions - Maze Activity(PDF - FREEBIE)

Solving Proportions

  • 2-7 Solving Proportions - Answers - Maze Activity (PDF - Member Only)
  • 2-7 Solving Proportions - Maze Activity (Editable - Member Only)
  • ⭐ Solving Proportions - Maze Activity(PDF - FREEBIE)

Proportions and Similar Figures

  • 2-8 Proportions and Similar Figures - Answers - Maze Activity (PDF - Member Only)
  • 2-8 Proportions and Similar Figures - Maze Activity (Editable - Member Only)
  • ⭐ Proportions and Similar Figures - Maze Activity(PDF - FREEBIE)

Percentages

  • 2-9 Percentages - Answers - Maze Activity (PDF - Member Only)
  • 2-9 Percentages - Maze Activity (Editable - Member Only)
  • ⭐ Percentages - Maze Activity(PDF - FREEBIE)

Change Expressed as a Percent

  • 2-10 Change Expressed as a Percent - Answers - Maze Activity (PDF - Member Only)
  • 2-10 Change Expressed as a Percent - Maze Activity (Editable - Member Only)
  • ⭐ Change Expressed as a Percent - Maze Activity(PDF - FREEBIE)

Solving Linear Equations

The simplest equation to solve is a linear equation. A linear equation is an equation where the highest exponent of the variable is ( ext<1>) . The following are examples of linear equations:

Solving an equation means finding the value of the variable that makes the equation true. For example, to solve the simple equation (x + 1 = 1) , we need to determine the value of (x) that will make the left hand side equal to the right hand side. The solution is (x = 0) .

The solution, also called the root of an equation, is the value of the variable that satisfies the equation. For linear equations, there is at most one solution for the equation.

To solve equations we use algebraic methods that include expanding expressions, grouping terms, and factorising.

Check the answer by substituting (x=-cfrac<1><2>) .

egin ext & = 2x + 2 & = 2(-cfrac<1><2>) + 2 & = -1 + 2 & = 1 ext & =1 end

The following video gives an introduction to solving linear equations.

[Attributions and Licenses]

This article is licensed under a CC BY-NC-SA 4.0 license.

Note that the video(s) in this lesson are provided under a Standard YouTube License.


Let&rsquos Start &hellip Coding!

The Games::LMSolve::Base class tries to solve a game by iterating through its various positions, recording every one it passes through, and trying to reach the solution. However, it does not know in advance what the games rules are, and what the meaning of the positions and moves are. In order for it to know that, we need to inherit it and code several methods that are abstract in the base class.

We will code a derived class that will implement the logic specific to the Jumping Cards game. It will implement the following methods, which, together with the methods of the base class, enable the solver to solve the game:

  1. input_board
  2. pack_state
  3. unpack_state
  4. display_state
  5. check_if_final_state
  6. enumerate_moves
  7. perform_move
  8. render_move

Here&rsquos the beginning of the file where we put the script:

As can be seen, we declared a new package, Jumping::Cards , imported the Games::LMSolve::Base namespace, and inherited from it. Now let&rsquos start declaring the methods. First, a method to input the board in question.

Since our board is constant, we just return an array reference that contains the initial sequence.

When Games::LMSolve::Base iterates over the states, it stores data about each state in a hash. This means we&rsquore going to have to provide a way to convert each state from its expanded form into a uniquely identifying string. The pack_state method does this, and in our case, it will look like this:

It is a good idea to use functions like pack , join or any other serialization mechanism here. In our case, we simply used join .

It is not very convenient to manipulate a packed state, and so we need another function to expand it. unpack_state does the opposite of pack_state and expands a packed state.

display_state() converts a packed state to a user-readable string. This is so that it can be displayed to the user. In our case, the comma-delimited notation is already readable, so we leave it as that.

We need to determine when we have reached our goal and can terminate the search with a success. The check_if_final_state function accepts an expanded state and checks if it qualifies as a final state. In our case, it is final if it&rsquos the 8-to-1 sequence.

Now we need a function that will tell the solver what subsequent states are available from each state. This is done by enumerating a set of moves that can be performed on the state. The enumerate_moves function does exactly that.

What enumerate_moves does is iterate over the indices of the locations twice, and checks every move for the validity of the resultant board. If it&rsquos OK, it pushes the exchanged indices to the array @moves , which is returned at the end.

We also need a function that will translate an origin state and a move to a resultant state. The perform_move function performs a move on a state and returns the new state. In our case, it simply swaps the cards in the two indices specified by the move.

Finally, we need a function that will render a move into a user-readable string, so it can be displayed to the user.


Solving linear equations

This unit teaches students to identify linear relationships and solve linear equations in context.

  • Identify and find values for variables in context.
  • Identify linear relationships in context.
  • Represent linear relationships using tables, graphs and simple linear equations.
  • Draw strip diagrams to represent linear equations.
  • Solve simple linear equations and interpret the answers in context.

Algebra started with the need to solve problems. Al Khwarizmi, a Persian mathematician, was arguably the first person to represent linear and quadratic problems in symbolic form and solved the problems by processes of ‘restoration’, i.e. equivalent operations that conserved equality. In fact, the word for algebra comes from the Arabic word for restoration.

It is fitting then that modern approaches to algebra focus on the thinking that underpins the symbolic systems. Algebraic thinking is concerned with generalisation. Letters, words, tables, graphs, networks, etc. are cultural tools that enable us to represent, then think with, those generalisations. With representational tools we are capable of ‘amplified cognition’ in that we can anticipate results that would never be possible if we relied solely on the physical environment, and on our limited capacity to process ideas just mentally.

Generalisation begins with noticing patterns and structures. A pattern is a consistency, that is something that occurs in a predictable way. It is the ‘what’ of algebraic thinking. Structure is about the organisation of patterns. It is the ‘how’ and sometimes the ‘why’ of generalisation. From noticing pattern and structure, we develop properties. For example, early counting involves pattern and structure. The ‘fourness’ of a collection comes from noticing sameness among collections of four, irrespective of the size, colour, texture, etc. of the objects. Structure of counting involves ideas like the order of counting the objects doesn’t matter.

Specific Teaching Points

In upper primary school, learning experiences for algebraic thinking typically begin with patterns. Usually these patterns are spatial and may be connected to some meaningful life context, though number patterns are also rich in opportunity. Patterns involve variables, that is features, some of which can be quantified. For example, consider this simple spatial pattern.

Among the variables we might discern that the ‘tower’ has height and each ‘tower’ is made of some number of squares. Height and number of cubes may not be the only variables, just those we notice. Variables change, that is height varies and so does the number of squares in the ‘tower’. We might try to find a relation between the variables, describe and represent that relation, and use it to predict how the pattern grows beyond what we can see. Then we are thinking with the properties and representations in a sophisticated way.

On the way it is likely we will need to organise the data from the pattern systematically. A table of values is a productive generic strategy, so we represent the pattern like this:

The danger in moving to an organised numeric strategy like a table too early is that it may negate what we can ‘see’ in the pattern visually. Noticing and reasoning may be inductive, that is tied to the incremental change of the figures. For example:

Noticing and reasoning can also be abductive, that is based on the structure of one example.

Noticing and reasoning can be deductive, that is based on making assumptions about structure and reasoning with the assumptions. For example, we might assume that the tower is composed of an array of something multiplied by three plus two.

From the assumptions we might deduce the appearance of towers much further on in the sequence, e.g. A tower 100 high will contain 2 + 99 x 3 squares. Ways of ‘seeing’ the pattern are manifest in relations within the table of values. For example, inductive thinking leads to seeing the values in the bottom row increasing by three each time. Abductive reasoning might support seeing this relation in the table:

Representing the relation as an algebraic equation involves two important and connected types of knowledge, related to the language conventions (semiotics), and to the nature of variables. We might write s = 3h – 1, or s = 3(h - 1) + 2, or s = 2h + (h – 1), depending on what we notice. The equations are meaningless to anyone else unless we clearly define what the variables, s and h, represent. Note that both and s refer to quantities that vary and are not fixed objects, such as houses or towers. Quantities are a combination of count and measurement unit. In this case h expresses unit lengths in height, and s refers to an area of squares. 3h means h multiplied by three, not thirty-something, and 3(h - 1) means that one is subtracted from h before the multiplication by three occurs. Working with variables requires acceptance of lack of closure, that is thinking with an object (h in this case) without specifically knowing what it is. For example, knowing that 3(h – 1) = 3h – 3 is true, irrespective of whatever the value of h, is itself a generalisation. The equals sign represents a statement of ‘transitive balance’ meaning that the balance is conserved if equivalent operations are performed on both sides of the equation. Knowledge of which operations conserve equality and those which disrupt it are important generalisations about the properties of numbers under those operations, e.g. distributive property of multiplication.

This unit specifically deals with relations that are linear. The first sign of linearity is that there is constant difference in the increase or decrease of one variable, as the value of the other increases by one. In the table above the number of squares increases by three as height increases by one.

Note that this graph shows a relation, not a function, since the values of variables are discrete, not continuous. There are some important connections between features of the algebraic equation, the table and the graph of a linear relation. Constant difference is represented by the co-efficient of the independent variable (s = 3h -1 in this case), differences of three in the bottom values of the table, and a slope of three (change in s for every unit change in h). The constant in the equation (- 1) is reflected in the table by a need to adjust the value of 3h by subtracting one to get the value of s, and reflected in the graph as a downward translation (shift) of the graph for s = 3h by one unit. This results in the intercept of the graph with the s axis being (0, -1), not the origin (0, 0).

Simple linear equations occur when the value of one variable in a relation or function is set and the other must be found. For example, with the tower problem this problem might be posed “A tower in the pattern has 98 squares. How high is the tower?” Depending on the equation used to represent the relation, this problem can be expressed as 3h – 1 = 98, 3(h – 1) + 2 = 98 or 2h + (h – 1) = 98. Linear equations with the variable on both sides occur when two conditions are equalised. An example might be, “Both Lilly and Todd look at the same tower. Lilly notices that the number of squares in the tower is three times the height less one. Todd notices that the number of squares is two times the height plus 18. How tall is the tower?” This problem can be written as 3h – 1 = 2h + 18.

  • Attachments as listed at the bottom of the unit
  • Access to the two digital learning objects:

Prior Experience

It is anticipated that students at Level 4 understand, and are proficient with, multiplicative thinking. However, the tasks in this unit are also accessible for students whose preference is additive thinking. In fact, the experiences may prompt a move towards multiplicative thinking.

Session One: Maia the Moa

In this session students are shown a spatial growth pattern for a moa made from square tiles. As Maia the moa ages she grows in her legs, body and neck while her feet and head remain constant. Session One is driven using PowerPoint One. The approach is to structure one example of the pattern then transfer that structure to other members of the pattern.

  1. Show the students Slide One. Aim to identify features of the pattern that might become variables. Ask: What do you notice about this figure?
    Students might notice different features such as colour, height, width, age, total number of squares, etc.
  2. Ask: Is there an easy way to count the number of squares that Maia is made of?
  3. Give students a while to structure their counting then ask them to share their method with others. Building a model of Maia at age three years with connecting cubes allows students to experiment with ways to partition the model. Encourage them to express their counting method as an expression. Use these videos to show examples of how to do this, but only if needed:
  4. Ask students to apply their counting structures to Maia at age two years (Slide Two). Ask them to record expressions for their counting strategy and compare them to what they recorded for year four.
  5. Ask: What changes and what stays the same in your expressions?
    For example, from Casey’s method these two expressions emerge:
    4 + 2 x 3 + 2 (Age two) 6 + 2 x 5 + 2 (Age four)
    The ‘+ 2’ is constant and ‘2 x’ is present in both expressions. The other numbers vary.
  6. Ask: What will your expression for Maia at age three years look like? Write the expression then check it by drawing a picture of Maia at age three (See Slide 3).
  7. Ask students to show where the parts of their expressions come from in the picture. For Casey’s method the expression is 5 + 2 x 4 + 2. Slide 4 shows how parts of the diagram can be linked to parts of the expression. Look at the strategies of the students.
    Are their strategies based on induction? That is sequential processing. For example, 4 , ? , 6, so ? = 5, and 2 x 3, 2 x ?, 2 x 5, so ? = 4.
  8. Are their strategies based on deduction? That is reasoning about the structure of any term. For example, the first number is two more than the age, and the multiplier of two is one more than the age. So, for y = 3 Casey’s expression is 5 + 2 x 4 + 2.
  9. Pose this problem for students to explore individually or in small co-operative groups:
    Imagine that Maia celebrates her twentieth birthday.
    How many squares will she be made of?
    Find a way to predict the number of squares that Maia is made of for any age in years?
  10. Allow students plenty of time to explore the problem. Look for the following:
    • Do the students record the data systematically? For example, if they draw Maia at age five years. Are their structural counting methods consistent? Is their recording in sequence?
    • Do students use inductive methods? For example, Maia increases by three squares each year.
    • Do students use deductive methods? For example, applying Casey’s method Maia should be (20 + 2) + 2 x (20 + 1) + 2 on her twentieth birthday.
    • How do students express their general rules? Do they use words?, e.g. “I take the age and add two to it to get the first number..” or do they attempt to symbolise their rules, e.g. Next number = number before + 3.
  11. Bring the class together to discuss their methods with emphasis on the points above. Acknowledge the legitimacy of inductive methods but also highlight the power of deductive methods. Use questions like, “Which strategy would be better for finding out about Maia at 100 years of age?”

Session Two

This session builds on the Maia, the moa, pattern to represent the relation between age and number of squares using a table, a graph and an equation. Features of these representations are connected through looking at the effect of changing the original spatial pattern with focussed variation.

  1. Open Excel or a similar spreadsheet program and create a blank workbook. You may need to have Slide 3 of PowerPoint One available for source data. Ask one of the students to set up a table like this:
  2. Ask students what they notice in the table.
    Some may notice missing values in the Age column, particularly the ages 0, and 1. Others may notice that the number of squares are all multiples of three. They may express this idea inductively, “The number of squares goes up by three.”
    How can we continue the table to get more values?
  3. Induction can be used to ‘fill down’ the values in both columns but deductive rules across the columns are more sophisticated. Video 2A shows how to create values by filling down. Video 2B is about using formulae across the columns. The videos can be stopped at any point for discussion. Video 2B goes straight to the most efficient rule but students could enter the rules they developed in Lesson One.
  4. Ask: Can you use Excel to show that your rule from yesterday works?
  5. Next a graph is created from the table of values. Video 2C shows how to do this. Ask the students to create their own graph of Maia’s growth patterns and record some features that they notice.
    Why are the points in a line? (This tell us that the relation is linear)
    How steep is the line?
    Note (0, 6) represents Maia’s situation upon hatching.
    Where does it cross the s axis?
    Why does it cross there? offers three scenarios in which Maia’s shape is changed in some way. The reason for doing this is to connect features of the table and graph with the spatial pattern. For each scenario students may need to draw the progression of each pattern back until Maia hatches. That will lead a table of values that can be graphed. Video 2D shows what happens when the original Maia growth pattern is altered by a constant, - 1 for losing her foot and + 2 for gaining a backpack. Video 2E and Video 2F show the effect of changing the co-efficient (multiplier) of a, in that the slope of the graph alters from three to four. Copymaster 1 provides printable versions and the start of a table for each.

Session Three

  1. Remind the students of the rule that was entered into Excel to create the pattern in the Number of squares column for Maia’s original growth pattern (e.g. =(A2+2)*3).
    What does A# represent? (Maia’s age in years, a). So instead of A# we could write = (a + 2) x 3 or = 3(a + 2).
    What does this expression tell us? (The number of squares Maia is made up of). So we could write s = 3(a + 2).
  2. Show the students PowerPoint Three which shows how linear equations can be represented using a length model. Work through the slides.
    Do the students observe that a is free to take up different values? a is a variable. The twos remain equal in length as the value of a changes. So, +2 is a constant.
    Pose this problem to the students.
  3. Maia is made up of 144 squares. How old is she, in years?
    This situation constrains s to 144 so a linear equation is created which might be expressed as 3(a + 2) = 144 or in other forms, dependent on the structure of the rule. For example, Katia’s method would yield 3a + 6 = 144. Student may need access to a picture of Maia’s growth pattern, e.g. Slide 3 of PowerPoint One.
  4. Look to see whether the students use deductive reasoning or whether they are reliant on inductive methods.
    For example, inductive methods might involve creating a table of values and extending it until the matching value of a is found. Spreadsheets make inductive methods easy to implement. A sign of reliance on additive methods would be repeated adding of three to find next values of s.
    Deductive methods involve applying inverse operations to rules. For example, “I divided 144 by three to get 48, so the age plus two must equal 48.”
  5. After a suitable time gather the class to discuss their strategies. Highlight the efficiency of deductive rules, which are sometimes referred to as function or direct rules, compared to lengthy inductive rules, which are sometimes referred to as recursive. Slides 5 and 6 show one way to solve the problem of Maia’s age when she is made of 144 squares.
  6. The 144 squares problem shows how solving linear equations can lead to solutions efficiently. Play this video which introduces how to use the simplest version of the Visual Linear Algebra learning object. Allow students plenty of time to explore the object.

Session Four

In this session students investigate linear equations where the variable is present on both sides.

  1. Begin with a reminder of how to solve linear equations in their simplest form by looking at the structural similarity of possible rules for Maia’s growth pattern. PowerPoint Four gives two possible rules attributed to hypothetical students. The rules may be alike some that the students created in Session One and Two. Slide Four shows the lengths rearranged end on end.
  2. Ask: Why do these rules give the same total for any value of a?
    Do students recognise that both rules can be rearranged to give 3a + 6 which is Katia’s rule?
  3. Possibly link the algebraic manipulation that matches the lengths in the diagram is students show interest. For example:
    (Leah’s rule) 3 (a + 1) + 3 = 3a + 3 + 3
    = 3a + 6 (Katia’s rule)
  4. Ask the students to use Katia’s rule to solve this problem:
    Maia the moa is made of 222 squares. How old is Maia?
    Do students apply inverse operations to both sides of the equation, 3a + 6 = 222, to find the solution?
  5. Pose this problem:
    Ken and Katia are looking at the same picture of Maia.
    Katia says that the number of squares equals three times Maia’s age plus six.
    Ken says that the number of squares equals four times Maia’s age minus 18.
    They are both correct. How old is Maia?
  6. Let the students work in small groups to solve the problem. Look for the following:
    • Do they build up a table of value inductively to find a value for a that meets both conditions?
    • Do they try values of a and ‘close in’ on the solution?
    • Do they use their knowledge of equations to solve the problem?
  7. Bring the class together to share their solution methods. Trial and improvement strategies can be very efficient in solving these types of problems, especially if the initial attempts are based on reasonable estimation. For example, setting a = 30 gives Ken’s number of squares at 102 and Katia’s at 96. So, is 30 too big or too small?
    An equation based solution looks like:
    3a + 6 = 4a – 18
    3a + 24 = 4a (adding 18)
    24 = a (subtracting 3a)
    Note that there are many possible first moves.
  8. Introduce the second learning object in the Visual Linear Algebra collection using this video. Allow students plenty of time to explore the tool.

Session Five

This session is intended as an opportunity for students to practice applying their understanding of linear relations and their techniques for solving linear equations.

Provide the students with copies of Copymaster 2 and encourage them to solve the problems in co-operative groups.

Dear parents and caregivers,

This week we are learning about linear relationships. Real life is full of situations where things grow at a constant rate, such as the money we earn for the hours we work, or the total cost related to the quantity we buy.

In the unit we will learn to represent linear relationships using tables of values, graphs and equations. We will use spreadsheets to solve problems with linear relations, and use a learning object to solve linear equations.