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10.1.2: Solving Multi-Step Equations


Learning Objectives

  • Use properties of equality together to isolate variables and solve algebraic equations.
  • Use the properties of equality and the distributive property to solve equations containing parentheses, fractions, and/or decimals.

There are some equations that you can solve in your head quickly. For example, what is the value of ( y) in the equation ( 2 y=6)? Chances are you didn’t need to get out a pencil and paper to calculate that ( y=3). You only needed to do one thing to get the answer, divide 6 by 2.

Other equations are more complicated. Solving ( 4left(frac{1}{3} t+frac{1}{2} ight)=6) without writing anything down is difficult! That’s because this equation contains not just a variable but also fractions and terms inside parentheses. This is a multi-step equation, one that takes several steps to solve. Although multi-step equations take more time and more operations, they can still be simplified and solved by applying basic algebraic rules.

Remember that you can think of an equation as a balance scale, with the goal being to rewrite the equation so that it is easier to solve but still balanced. The addition property of equality and the multiplication property of equality explain how you can keep the scale, or the equation, balanced. Whenever you perform an operation to one side of the equation, if you perform the same exact operation to the other side, you’ll keep both sides of the equation equal.

If the equation is in the form, ( a x+b=c), where ( x) is the variable, you can solve the equation as before. First “undo” the addition and subtraction, and then “undo” the multiplication and division.

Example

Solve ( 3 y+2=11).

Solution

( egin{array}{r}
3 y+2=& 11
-2 & -2 \
hline 3 y =&9
end{array})
Subtract 2 from both sides of the equation to get the term with the variable by itself, so ( 3 y=9).

( frac{3 y}{3}=frac{9}{3})

( y=3)

Divide both sides of the equation by 3 to get a coefficient of 1 for the variable.

( y=3)

Example

Solve ( frac{1}{4} x-2=3).

Solution

( egin{array}{r}
frac{1}{4} x-2=&3
+2 & +2 hline
frac{1}{4} x =&5
end{array})
Add 2 to both sides of the equation to get the term with the variable by itself, so ( frac{1}{4} x=5).
( egin{array}{c}
frac{4 x}{4}=5(4)
x=20
end{array})
Multiply both sides of the equation by 4 to get a coefficient of 1 for the variable.

( x=20)

If the equation is not in the form, ( a x+b=c), you will need to perform some additional steps to get the equation in that form.

In the example below, there are several sets of like terms. You must first combine all like terms.

Example

Solve ( 3 x+5 x+4-x+7=88).

Solution

( 3 x+5 x+4-x+7=88)There are three like terms ( 3x), ( 5x) and ( -x) involving a variable.
( egin{array}{r}
7 x+4+7=88
7 x+11=88
end{array})
Combine these like terms. 4 and 7 are also like terms and can be added.
( egin{array}{r}
7 x+11= & 88
-11 & -11 \
hline frac{7 x}{7}= & frac{77}{7}
end{array})

The equation is now in the form ( a x+b=c). So, we can solve ( 7 x+11=88) as before.

Subtract 11 from both sides.

Divide both sides by 7.

( x=11)

Some equations may have the variable on both sides of the equal sign. We need to “move” one of the variable terms in order to solve the equation.

Example

Solve ( 6 x+5=10+5 x). Check your solution.

Solution

( 6 x+5=10+5 x)This equation has ( x) terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign.

( egin{array}{r}
6 x+5=&10+5 x
-5 x & -5 x
hline x+5=&10
end{array})

You can subtract ( 5x) on each side of the equal sign, which gives a new equation: ( x+5=10). This is now a one-step equation!
( egin{array}{rr}
x+5= & 10
-5 & -5 \
hline x=&5
end{array})
Subtract 5 from both sides and you get ( x=5).
Check( egin{aligned}
6 x+5 &=10+5 x
6(5)+5 &=10+5(5) \
30+5 &=10+25 \
35 &=35
end{aligned})

Check your solution by substituting 5 for ( x) in the original equation.

This is a true statement, so the solution is correct.

( x=5)

Here are some steps to follow when you solve multi-step equations.

Solving multi-step equations

  1. If necessary, simplify the expressions on each side of the equation, including combining like terms.
  2. Get all variable terms on one side and all numbers on the other side using the addition property of equality. (( a x+b=c ext { or } c=a x+b))
  3. Isolate the variable term using the inverse operation or additive inverse (opposite) using the addition property of equality.
  4. Isolate the variable using the inverse operation or multiplicative inverse (reciprocal) using the multiplication property of equality to write the variable with a coefficient of 1.
  5. Check your solution by substituting the value of the variable in the original equation.

The examples below illustrate this sequence of steps.

Example

Solve for ( y).

( -20 y+15=2-16 y+11)

Solution

( egin{array}{l}
-20 y+15=2-16 y+11
-20 y+15=-16 y+13
end{array})
Step 1. On the right side, combine like terms: ( 2+11=13).
( egin{array}{rr}
-20 y+15= & -16 y+13
+20 y & +20 y
hline 15= & 4 y+13
-13 & -13 \
hline 2= & 4 y
frac{2}{4}= & frac{4 y}{4}
frac{1}{2}= & y
end{array})

Step 2. Add ( 20y) to both sides to remove the variable term from the left side of the equation.

Step 3. Subtract 13 from both sides.

Step 4. Divide ( 4y) by 4 to solve for ( y).

Check( egin{aligned}
-20 y+15 &=2-16 y+11
-20left(frac{1}{2} ight)+15 &=2-16left(frac{1}{2} ight)+11
-10+15 &=2-8+11 \
5 &=5
end{aligned})
Step 5. To check your answer, substitute ( frac{1}{2}) for ( y) in the original equation. The statement ( 5=5) is true, so ( y=frac{1}{2}) is the solution.

( y=frac{1}{2})

Advanced Example

Solve ( 3 y+10.5=6.5+2.5 y). Check your solution.

Solution

( 3 y+10.5=6.5+2.5 y)This equation has ( y) terms on both the left and the right. To solve an equation like this, you must first get the variables on the same side of the equal sign.
( egin{array}{r}
3 y+10.5=&6.5+2.5 y
-2.5 y & -2.5 y
hline 0.5 y+10.5=&6.5
end{array})
Add ( -2.5 y) to both sides so that the variable remains on one side only.
( egin{array}{rrr}
0.5 y+10.5= & 6.5
-10.5 & -10.5 \
hline 0.5 y =& -4
end{array})
Now isolate the variable by subtracting 10.5 from both sides.
( egin{aligned}
10(0.5 y) &=10(-4)
5 y &=-40
frac{5 y}{5} &=frac{-40}{5}
y &=-8
end{aligned})
Multiply both sides by 10 so that ( 0.5y) becomes ( 5y), then divide by 5.
Check( egin{aligned}
3 y+10.5 &=6.5+2.5 y
3(-8)+10.5 &=6.5+(-20) \
-24+10.5 &=6.5+(-20) \
-13.5 &=-13.5
end{aligned})
Check your solution by substituting -8 in for ( y) in the original equation.

This is a true statement, so the solution is correct.

( y=-8)

Advanced Question

Identify the step that will not lead to a correct solution to the problem.

( 3 a-frac{11}{2}=-frac{3 a}{2}+frac{25}{2})

  1. Multiply both sides of the equation by 2.
  2. Add ( frac{11}{2}) to both sides of the equation.
  3. Add ( frac{3 a}{2}) to the left side, and add ( -3 a) to the right side.
  4. Rewrite ( 3 a) as ( frac{6 a}{2}).
Answer
  1. Multiply both sides of the equation by 2.

    Incorrect. Multiplying both sides by 2 keeps both sides equal; the new equation will be ( 6 a-11=-3 a+25). This will lead to the correct solution. The step that will not lead to a correct solution is: Add ( frac{3 a}{2}) to the left side, and add ( -3 a) to the right side.

  2. Add ( frac{11}{2}) to both sides of the equation.

    Incorrect. Adding ( frac{11}{2}) to both sides of the equation will keep both sides equal; the new equation will be ( 3 a=-frac{3 a}{2}+frac{36}{2}). This will lead to the correct solution. The step that will not lead to a correct solution is: Add ( frac{3 a}{2}) to the left side, and add ( -3 a) to the right side.

  3. Add ( frac{3 a}{2}) to the left side, and add ( -3 a) to the right side.

    Correct. Adding unequal amounts to the left and to the right will unbalance the equation, and you will no longer be able to solve accurately for ( a).

  4. Rewrite ( 3 a) as ( frac{6 a}{2}).

    Incorrect. Rewriting ( 3 a) as ( frac{6 a}{2}) does not change the value of the fraction at all, so this will keep both sides equal. The step that will not lead to a correct solution is: Add ( frac{3 a}{2}) to the left side, and add ( -3 a) to the right side.

More complex multi-step equations may involve additional symbols such as parentheses. The steps above can still be used. If there are parentheses, you use the distributive property of multiplication as part of Step 1 to simplify the expression. Then you solve as before.

The Distributive Property of Multiplication

For all real numbers ( a), ( b), and ( c), ( a(b+c)=a b+a c).

What this means is that when a number multiplies an expression inside parentheses, you can distribute the multiplication to each term of the expression individually. Then, you can follow the routine steps described above to isolate the variable to solve the equation.

Example

Solve for ( a).

( 4(2 a+3)=-3(a-1)+31)

Solution

( egin{aligned}
4(2 a+3) &=-3(a-1)+31
8 a+12 &=-3 a+3+31
end{aligned})
Apply the distributive property to expand ( 4(2 a+3)) to ( 8 a+12) and ( -3(a-1)) to ( -3 a+3).
( egin{array}{r}
8 a+12=&-3 a+34
+3a &+3 a
hline 11 a+12=&+34
-12 &-12 \
hline frac{11 a}{11} =&frac{22}{11}
a=&2
end{array})

Combine like terms.

Add ( 3a) to both sides to move the variable terms to one side.

Subtract 12 to isolate the variable term.

Divide both terms by 11 to get a coefficient of 1.

( a=2)

Example

In which of the following equations is the distributive property properly applied to the equation ( 2(y+3)=7)?

  1. ( y+6=7)
  2. ( 2 y+6=14)
  3. ( 2 y+6=7)
  4. ( 2 y+3=7)

Solution

  1. ( y+6=7)

    Incorrect. All of the terms inside the parentheses must be multiplied by the value outside. The correct answer is ( 2 y+6=7).

  2. ( 2 y+6=14)

    Incorrect. When applying the distributive property, multiplication is spread only to the terms inside the parentheses, not to the other parts of the equation. The correct answer is ( 2 y+6=7).

  3. ( 2 y+6=7)

    Correct. Since the distributive property allows us to distribute the multiplication of an entire expression to each of the terms of the expression separately, ( 2 y+6=7) is correct.

  4. ( 2 y+3=7)

    Incorrect. The correct answer is ( 2 y+6=7).

If you prefer not working with fractions, you can use the multiplication property of equality to multiply both sides of the equation by a common denominator of all of the fractions in the equation. See the example below.

Example

Solve ( frac{1}{2} x-3=2-frac{3}{4} x) by clearing the fractions in the equation first.

Solution

( egin{aligned}
frac{1}{2} x-3 &=2-frac{3}{4} x
4left(frac{1}{2} x-3 ight) &=4left(2-frac{3}{4} x ight)
4left(frac{1}{2} x ight)-4(3) &=4(2)-4left(frac{3}{4} x ight)
frac{4}{2} x-12 &=8-frac{12}{4} x
end{aligned})
Multiply both sides of the equation by 4, the common denominator of the fractional coefficients.

Use the distributive property to expand the expressions on both sides. Multiply.

( egin{array}{r}
2 x-12= & 8-3 x
+3 x & +3 x
hline 5 x-12=&8
+12 & +12 \
hline frac{5 x}{5} =&frac{20}{5}
x=&4
end{array})
Add ( 3x) to both sides to move the variable terms to only one side.

Add 12 to both sides to move the constant terms to the other side.

Divide to isolate the variable.

( x=4)

Of course, if you like to work with fractions, you can just apply your knowledge of operations with fractions and solve.

Example

Solve ( frac{1}{2} x-3=2-frac{3}{4} x).

Solution

( egin{array}{rr}
frac{1}{2} x-3= & 2-frac{3}{4} x
+frac{3}{4} x = & +frac{3}{4} x
hline frac{5}{4} x-3= & 2
+3 & +3 \
hline frac{5}{4} x = & 5
end{array})

Add ( frac{3}{4} x) to both sides to get the variable terms on one side.

( frac{1}{2}+frac{3}{4}=frac{2}{4}+frac{3}{4}=frac{5}{4}-frac{3}{4}+frac{3}{4}=0)

Add 3 to both sides to get the constant terms on the other side.

( egin{aligned}
frac{4}{5} cdot frac{5}{4} x &=frac{4}{5} cdot 5
x &=frac{20}{5}
x &=4
end{aligned})
To get a coefficient of 1, multiply the variable term by its multiplicative inverse.

( x=4)

Advanced Example

Solve ( frac{1}{2}(2+a)=frac{3 a+4}{3^{2}}). Check your solution.

Solution

( frac{1}{2}(2+a)=frac{3 a+4}{3^{2}})Solving this equation will require multiple steps. Begin by evaluating ( 3^{2}=9).
( egin{aligned}
frac{1}{2} cdot 2+frac{1}{2} cdot a &=frac{3 a+4}{9}
1+frac{1}{2} a &=frac{3 a+4}{9}
1+frac{a}{2} &=frac{3 a+4}{9}
end{aligned})
Now distribute the ( frac{1}{2}) on the left side of the equation.
( egin{aligned}
18left(1+frac{a}{2} ight) &=18left(frac{3 a+4}{9} ight)
18 cdot 1+18 cdot frac{a}{2} &=frac{18(3 a+4)}{9}
18+frac{18 a}{2} &=frac{18(3 a+4)}{9}
18+frac{18 a}{2} &=frac{9 cdot 2(3 a+4)}{9}
18+9 a cdot frac{2}{2} &=2(3 a+4) cdot frac{9}{9}
18+9 a &=2(3 a+4)
18+9 a &=2 cdot 3 a+2 cdot 4
18+9 a &=6 a+8
end{aligned})

Multiply both sides of the equation by 18, the common denominator of the fractions in the problem.

Use the distributive property to expand the expression on the left side.

Then remove a factor of 1 from both sides. On the left, you can think of ( frac{18 a}{2}) as ( frac{2}{2}). On the right, you can think of ( frac{18(3 a+4)}{9}) as ( frac{9}{9} cdot frac{2(3 a+4)}{1}).

Continue solving for ( a) using the distributive property.

( egin{array}{rr}
18+9 a= & 6 a+8
-6 a & -6 a
hline 18+3 a= & 8
-18 & -18 \
hline 3 a= & -10
frac{3 a}{3}= & frac{-10}{3}
a= & -frac{10}{3}
end{array})
Then isolate the variable, and solve the remaining one-step problem.
Check( egin{aligned}
frac{1}{2}left[2+left(-frac{10}{3} ight) ight] &=frac{3left(-frac{10}{3} ight)+4}{3^{2}}
frac{1}{2}left[2+left(-frac{10}{3} ight) ight] &=frac{-10+4}{9}
frac{1}{2}left[frac{6}{3}+left(-frac{10}{3} ight) ight] &=frac{-10+4}{9}
frac{1}{2}left(-frac{4}{3} ight) &=frac{-10+4}{9}
frac{-4}{6} &=frac{-10+4}{9}
frac{-4}{6} &=frac{-6}{9}
-frac{2}{3} cdot frac{2}{2} &=-frac{2}{3} cdot frac{3}{3}
-frac{2}{3} &=-frac{2}{3}
end{aligned})

Check your solution by substituting ( -frac{10}{3}) in for ( a) in the original equation.

This is a true statement, so the solution is correct.

( a=-frac{10}{3})

Exercise

To clear the fractions from ( frac{1}{3}-frac{2 y}{9}=19), we can multiply both sides of the equation by which of the following numbers?

( egin{array}{llll}
3 & 6 & 9 & 27
end{array})

  1. 9
  2. 9 and 27
  3. 6
  4. 3 or 9
Answer
  1. 9

    Incorrect. While 9 is a common denominator of ( frac{1}{3}) and ( frac{2}{9}), so is 27. Any denominator will work, not just the least one. The correct answer is 9 and 27.

  2. 9 and 27

    Correct. Both 9 and 27 are common denominators of ( frac{1}{3}) and ( frac{2}{9}).

  3. 6

    Incorrect. You clear fractions by multiplying them by a common denominator. 6 is not a common denominator of ( frac{1}{3}) and ( frac{2}{9}). The correct answer is 9 and 27.

  4. 3 or 9

    Incorrect. While 9 is a common denominator of ( frac{1}{3}) and ( frac{2}{9}), 3 is not. The correct answer is 9 and 27.

Regardless of which method you use to solve equations containing variables, you will get the same answer. You can choose the method you find easier! Remember to check your answer by substituting your solution into the original equation.

Just as you can clear fractions from an equation, you can clear decimals from the equation in the same way. Find a common denominator and use the multiplication property of equality to multiply both sides of the equation.

Example

Solve ( 0.4 x-0.25=1.75) by clearing the decimals first.

Solution

( egin{array}{r}
0.4 x-0.25=1.75
100(0.4 x-0.25)=100(1.75)
end{array})

( 0.4left(frac{4}{10} ight)) and ( 0.25left(frac{25}{100} ight)) and ( 1.75left(frac{175}{100} ight)) have a common denominator of 100.

Multiply both sides by 100.

( egin{array}{r}
40 x-25 =&175
+25 & +25 \
hline frac{40 x}{40} =&frac{200}{40}
x =&5
end{array})

Apply the distributive property to clear the parentheses.

Solve as before. Add 25 to both sides.

Divide both sides by 40.

Check:( egin{array}{r}
0.4 x-0.25=1.75
0.4(5)-0.25=1.75 \
2-1.25=1.75 \
1.75=1.75
end{array})

Substitute ( x=5) into the original equation.

Evaluate.

The solution checks.

( x=5)

Advanced Question

Solve for ( a): ( frac{1}{4}(a+3)=2-a)

  1. ( a=2)
  2. ( a=1)
  3. ( a=0)
  4. ( a=-2)
Answer
  1. ( a=2)

    Incorrect. Try multiplying both sides of the equation by 4, as ( 4 cdot frac{1}{4}=1). The new equation will be ( 4left[frac{1}{4}(a+3) ight]=4(2-a)), which reduces to ( a+3=8-4 a). The correct answer is: ( a=1).

  2. ( a=1)

    Correct. You can solve this equation by multiplying both sides by 4, since ( 4 cdot frac{1}{4}=1). The resulting equation, ( a+3=4(2-a)) can be rewritten as ( a+3=8-4 a), and then ( 5 a=5). You find that ( a=1).

  3. ( a=0)

    Incorrect. Substituting ( a=0) into the equation, you find ( frac{1}{4}(0+3)=2-0), so ( frac{3}{4}=2). This is not accurate. The correct answer is: ( a=1).

  4. ( a=-2)

    Incorrect. Try multiplying both sides of the equation by 4, as ( 4 cdot frac{1}{4}=1). The new equation will be ( 4left[frac{1}{4}(a+3) ight]=4(2-a)), which reduces to ( a+3=8-4 a). The correct answer is: ( a=1).

Complex, multi-step equations often require multi-step solutions. Before you can begin to isolate a variable, you may need to simplify the equation first. This may mean using the distributive property to remove parentheses, or multiplying both sides of an equation by a common denominator to get rid of fractions. Sometimes it requires both techniques.


Math by the Mountain

Unit 3 of Algebra 1 is all about solving equations and their applications. We start off with multi-step equations, because 1-step and 2-step equations were covered in Unit 1: Foundations of Algebra.

Day 1: Multi-Step Equations

In addition to the notes that went into our composition books, students were each given a full-sized flowchart over solving one-variable equations. We did an example as a class, and then I also keep a class set laminated so students can use them with dry-erase markers whenever they like. Students referenced their notes and the laminated flowcharts while working on homework in class.

Day 2: Solving Multi-Step Equations with Special Case Solutions
To start off the lesson, we did a recap warm-up over the prior day’s lesson.

We then went into a foldable that covers what special solutions are and when they arise.

To get even more practice, students did the following Types of Solutions Sort, which emphasized common student errors and misconceptions I’ve noticed in the past.

Day 3: Writing Equations to Solve Multi-Step Equations
We started off the lesson with a recap warm-up that contained special solution types.

From there, we moved into our main set of notes for the day, with an emphasis on marking the text (NOTE: this is the same color-coding we used in Unit 1).

Day 4: Absolute Value Equations
Like usual, we started off the lesson with a recap warm-up of the previous day’s information.

We started off the topic of absolute value equations by really thinking about what an absolute value means/does.

From there, we used the information we’ve gathered to solve absolute value equations a bit more efficiently (without using the modified cover-up question mark method). Students had the even numbered problems as homework that night.

In addition to the notes that went into the composition books, students were given a flowchart for solving absolute value equations to reference whenever they got stuck. Here’s an example of how they could use it! Just like the others, I keep a class set of these laminated so students can use them with dry erase markers whenever they get stuck. I like to color-code each type of flowchart to make it easy to grab the exact one that they need from that unit.

Day 5: Absolute Value Equations Word Problems
To begin the class, we started off by working backwards: writing the absolute value equation that could’ve produced the given solutions.

From there, we went into story problems involving absolute value equations.

Day 6: Ratios and Proportions
We started the day off with a recap warm-up covering the last two days of information (all absolute value equation related).

The first thing that we talked about is what a ratio is and what it means to be proportional.

We then used the definition of proportional to solve equations requiring cross-multiplication.

After these examples, students filled out the other side of the flowchart that they were given on Day 1 with a more difficult example of solving for a variable in a proportion.

Day 7: Percent of Change
Percent of change is a funny topic to cover in Oregon…most of our textbook’s examples are about sales tax, and we have none. If we go to Washington, we just flash our Oregon ID and presto, bingo, bango, no more sales tax (for the little stuff). Anyway, we find other examples to try to make it more meaningful.

After taking notes, we did this Percent of Change Scavenger Hunt. Students worked really hard on it and had a lot of fun. For some of them, it was difficult to remember to put a negative sign on their r-value when it was a percent decrease!

Day 8: Literal Equations, Part 1
We recap percent of change problems and then move into basic solving literal equations problems.

We discuss what a literal equation is, compare and contrast the difference between literal equations and regular equations, and also introduce the flowchart method of solving.

Day 9: Literal Equations, Day 2
We move into more complicated literal equations that require more than one step to solve. After doing a few, students are able to choose which method they wish to solve with (I’m partial to the algebraic method, but some students love the flowchart way).

After notes, we play my favorite Connect 4 game for solving literal equations. We only played until 6 people won, which allowed us to get through about 70% of the problems. From there, students spent the remainder of class working on a festive Carving Pumpkins coloring activity for solving literal equations. This activity was awesome because students were super engaged in the coloring (every last one of them–even the boys! PS: I have 22 boys in this one class…ay, yai, yai), and it was super easy for me to find common trends that I might need to readdress (the eyes for Pumpkin #2 were the most common error). Also, for students, this activity is fairly self-checking, which is a great confidence boost for many of them.

Here’s an example that one student colored! She even named the pumpkins.

Day 10: Stations Review Activity Day
We did a recap warm-up over solving literal equations and then spend the rest of class doing a stations activity with my solving equations unit task cards.

Day 11: Review Day
Day 12: TEST!


Quick Link for All Equations Worksheets

Click the image to be taken to that Equations Worksheets.

One Step EquationsIntegers Worksheets

One Step EquationsDecimals Worksheets

One Step EquationsFractions Worksheets

One Step Equations withIntegers, Decimals, & Fractions

Two Step EquationsIntegers Worksheets

Two Step EquationsDecimals Worksheets

Multiple Step EquationsIntegers Worksheets

Multiple Step EquationsDecimals Worksheets

Absolute ValueEquations Worksheets

Solving ProportionsEquations Worksheets

Percent ProblemsEquations Worksheets

Solving Single VariableEquations Worksheets

One Step EquationWord Problems

Two Step EquationWord Problems

Distance, Rate, and TimeWord Problems

Mixture Word ProblemsEquations Worksheets

Work Word ProblemsEquations Worksheets


Contents

Numerical methods for ordinary differential equations approximate solutions to initial value problems of the form

Consider for an example the problem

y ′ = f ( t , y ) = y , y ( 0 ) = 1.

The exact solution is y ( t ) = e t > .

One-step Euler Edit

A simple numerical method is Euler's method:

Euler's method can be viewed as an explicit multistep method for the degenerate case of one step.

Two-step Adams–Bashforth Edit

Euler's method is a one-step method. A simple multistep method is the two-step Adams–Bashforth method

Three families of linear multistep methods are commonly used: Adams–Bashforth methods, Adams–Moulton methods, and the backward differentiation formulas (BDFs).

Adams–Bashforth methods Edit

The Adams–Bashforth methods with s = 1, 2, 3, 4, 5 are (Hairer, Nørsett & Wanner 1993, §III.1 Butcher 2003, p. 103):

The Lagrange formula for polynomial interpolation yields

The Adams–Bashforth method arises when the formula for p is substituted. The coefficients b j > turn out to be given by

The Adams–Bashforth methods were designed by John Couch Adams to solve a differential equation modelling capillary action due to Francis Bashforth. Bashforth (1883) published his theory and Adams' numerical method (Goldstine 1977).

Adams–Moulton methods Edit

The derivation of the Adams–Moulton methods is similar to that of the Adams–Bashforth method however, the interpolating polynomial uses not only the points t n − 1 , … , t n − s ,dots ,t_> , as above, but also t n > . The coefficients are given by

The Adams–Moulton methods are solely due to John Couch Adams, like the Adams–Bashforth methods. The name of Forest Ray Moulton became associated with these methods because he realized that they could be used in tandem with the Adams–Bashforth methods as a predictor-corrector pair (Moulton 1926) Milne (1926) had the same idea. Adams used Newton's method to solve the implicit equation (Hairer, Nørsett & Wanner 1993, §III.1).

Backward differentiation formulas (BDF) Edit

The central concepts in the analysis of linear multistep methods, and indeed any numerical method for differential equations, are convergence, order, and stability.

Consistency and order Edit

The first question is whether the method is consistent: is the difference equation

a good approximation of the differential equation y ′ = f ( t , y ) ? More precisely, a multistep method is consistent if the local truncation error goes to zero faster than the step size h as h goes to zero, where the local truncation error is defined to be the difference between the result y n + s > of the method, assuming that all the previous values y n + s − 1 , … , y n ,ldots ,y_> are exact, and the exact solution of the equation at time t n + s > . A computation using Taylor series shows that a linear multistep method is consistent if and only if

All the methods mentioned above are consistent (Hairer, Nørsett & Wanner 1993, §III.2).

If the method is consistent, then the next question is how well the difference equation defining the numerical method approximates the differential equation. A multistep method is said to have order p if the local error is of order O ( h p + 1 ) )> as h goes to zero. This is equivalent to the following condition on the coefficients of the methods:

The s-step Adams–Bashforth method has order s, while the s-step Adams–Moulton method has order s + 1 (Hairer, Nørsett & Wanner 1993, §III.2).

These conditions are often formulated using the characteristic polynomials

In terms of these polynomials, the above condition for the method to have order p becomes

ρ ( e h ) − h σ ( e h ) = O ( h p + 1 ) as h → 0. ^)-hsigma (mathrm ^)=O(h^)quad < ext>h o 0.>

In particular, the method is consistent if it has order at least one, which is the case if ρ ( 1 ) = 0 and ρ ′ ( 1 ) = σ ( 1 ) .

Stability and convergence Edit

If the roots of the characteristic polynomial ρ all have modulus less than or equal to 1 and the roots of modulus 1 are of multiplicity 1, we say that the root condition is satisfied. A linear multistep method is zero-stable if and only if the root condition is satisfied (Süli & Mayers 2003, p. 335).

Now suppose that a consistent linear multistep method is applied to a sufficiently smooth differential equation and that the starting values y 1 , … , y s − 1 ,ldots ,y_> all converge to the initial value y 0 > as h → 0 . Then, the numerical solution converges to the exact solution as h → 0 if and only if the method is zero-stable. This result is known as the Dahlquist equivalence theorem, named after Germund Dahlquist this theorem is similar in spirit to the Lax equivalence theorem for finite difference methods. Furthermore, if the method has order p, then the global error (the difference between the numerical solution and the exact solution at a fixed time) is O ( h p ) )> (Süli & Mayers 2003, p. 340).

To assess the performance of linear multistep methods on stiff equations, consider the linear test equation y' = λy. A multistep method applied to this differential equation with step size h yields a linear recurrence relation with characteristic polynomial

This polynomial is called the stability polynomial of the multistep method. If all of its roots have modulus less than one then the numerical solution of the multistep method will converge to zero and the multistep method is said to be absolutely stable for that value of hλ. The method is said to be A-stable if it is absolutely stable for all hλ with negative real part. The region of absolute stability is the set of all hλ for which the multistep method is absolutely stable (Süli & Mayers 2003, pp. 347 & 348). For more details, see the section on stiff equations and multistep methods.

Example Edit

Consider the Adams–Bashforth three-step method

One characteristic polynomial is thus

The other characteristic polynomial is

These two results were proved by Germund Dahlquist and represent an important bound for the order of convergence and for the A-stability of a linear multistep method. The first Dahlquist barrier was proved in Dahlquist (1956) and the second in Dahlquist (1963).

First Dahlquist barrier Edit

The first Dahlquist barrier states that a zero-stable and linear q-step multistep method cannot attain an order of convergence greater than q + 1 if q is odd and greater than q + 2 if q is even. If the method is also explicit, then it cannot attain an order greater than q (Hairer, Nørsett & Wanner 1993, Thm III.3.5).

Second Dahlquist barrier Edit

The second Dahlquist barrier states that no explicit linear multistep methods are A-stable.


#1 I can use the distributive property to expand expressions

The first thing we review when getting started with solving multi-step equations is the distributive property. The distributive property challenges students because they have to remember it over time. They seem to get it in isolation, but when it comes to solving equations it feels like zombies have eaten their brains and all the practice disappeared. What a great opportunity to show kids that some things we learned before show-up in other places. We do a lot of practice with the distributive property and even some number talks about them.

Of course, when negative numbers and variables are thrown into the mix some students struggle. We have references for the rules and patterns with distributive property in our interactive notebook. Students use these references until they don’t need them anymore.

We practice with distributive property throughout the year. We use a lot of whiteboards, Quizizz games, and mazes as practice. I like to make sure that it is fun. When students forget parts of the process, we do quick mini reviews. Once we’ve reviewed expanding expressions with the distributive property, we’re ready for the next step.


Multi-step Equations: Part 3 Variables on Both Sides

Subject Area(s): Mathematics, Mathematics (B.E.S.T. -.

Primary Resource Type: Original Tutorial

Learn alternative methods of solving multi-step equations in this interactive tutorial. This is part five of five in a series on solving multi-step

Subject Area(s): Mathematics, Mathematics (B.E.S.T. -.

Primary Resource Type: Original Tutorial

Explore how to solve multi-step equations using the distributive property in this interactive tutorial. This is part two of five in a series on

Subject Area(s): Mathematics, Mathematics (B.E.S.T. -.

Primary Resource Type: Original Tutorial

Learn how to solve multi-step equations that contain like terms in this interactive tutorial. This is part one of five in a series on

Subject Area(s): Mathematics, Mathematics (B.E.S.T. -.

Primary Resource Type: Original Tutorial

Attachments

  • MAFS.8.EE.3.7: Solve linear equations in one variable.
    1. Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
    2. Solve linear equations with rational number coefficients, including equations whose solutions require expanding expressions using the distributive property and collecting like terms.

Fluency Expectations or Examples of Culminating Standards

Students have been working informally with one-variable linear equations since as early as kindergarten. This important line of development culminates in grade 8 with the solution of general one-variable linear equations, including cases with infinitely many solutions or no solutions as well as cases requiring algebraic manipulation using properties of operations. Coefficients and constants in these equations may be any rational numbers.

Examples of Opportunities for In-Depth Focus

This is a culminating standard for solving one-variable linear equations.

Learn how to solve multi-step equations that contain variables on both sides of the equation in this interactive tutorial.


Step by step guide to solve multi-step equations

  • Combine “like” terms on one side.
  • Bring variables to one side by adding or subtracting.
  • Simplify using the inverse of addition or subtraction.
  • Simplify further by using the inverse of multiplication or division.

Multi–Step Equations – Example 1:

Solve this equation. (-(8 – x)=6)

First use Distributive Property: (−(8 − x)= − 8 + x)
Now solve by adding 8 to both sides of the equation. (−8 + x=6 →−8 + x +8=6 + 8 )
Now simplify: (→x=14)

Multi–Step Equations – Example 2:

Solve this equation. (2x + 5=15 – x)

First bring variables to one side by adding (x) to both sides.
(2x + 5=15 − x→3x + 5=15). Now, subtract (5) from both sides:
(3x + 5 − 5=15 − 5→3x=10)
Now, divide both sides by (3: 3x=10→3 x ÷ 3=frac<10><3>→x=frac<10><3>)

Multi–Step Equations – Example 3:

First use Distributive Property: ( -(2-x)=-2+x)
Now solve by adding (2) to both sides of the equation. (-2+x=5 →-2+x+2=5+2 )
Now simplify: (-2+x+2=5+2 →x=7)

Multi–Step Equations – Example 4:

Solve this equation. (4x+10=25-x)

First bring variables to one side by adding (x) to both sides.
( 4x+10+x=25-x+x→5x+10=25) . Now, subtract (10) from both sides:
(5x+10-10=25-10→5x=15 )
Now, divide both sides by (5: 5x=15 →5x÷5=frac<15><5>→x=3)


Removing Parentheses with the Distributive Property

Just as with simpler equations, solving multi-step equations usually means isolating a variable on one side of the equals sign. Step-by-step, we must get the variable out of parentheses, away from other terms, and with a coefficient of `1` . If a variable is inside parentheses, they can be cleared by applying the Distributive Property of Multiplication.

The Distributive Property states that for all real numbers `a` , `b` , and `c` , `a(b+c)=ab+ac` . What that means is that when a number multiplies an expression inside parentheses, we can distribute the multiplication to each term of the expression individually. Let&rsquos go back to the equation `6(1/4t+3/8)=2` . We can apply the distributive property and clear the parentheses by multiplying each term inside of them by `6` . The expression `6(1/4t+3/8)` then becomes `6/4t+18/8` . Now there are no parentheses.

In which of the following equations is the distributive property properly applied to the equation `x(y+3)=7` ?


Multi-Step Equations

To solve a multi-step equation, we would start by trying to simplify the equation by combining like terms and using the distributive property whenever possible.

Consider the equation 2( x + 1) &ndash x = 5. First, we will use the distributive property to remove the parenthesis and then we can combine like terms and the isolate the variable.

Look at the lesson on Solving Multi-Step Equations if you need help on how to solve multi-step equations.

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

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Solving Multi-Step Equations (12-2)

Solving Multi-Step Equations

Like Terms - terms taht have the same value and the same corresponding exponents.