1: Linear Equations - Mathematics

Learning Objectives

In this chapter, you will learn to:

  1. Graph a linear equation.
  2. Find the slope of a line.
  3. Determine an equation of a line.
  4. Solve linear systems.
  5. Do application problems using linear equations.

Thumbnail: The red and blue lines on this graph have the same slope (gradient); the red and green lines have the same y-intercept (cross the y-axis at the same place). (CC BY-SA 1.0; ElectroKid via Wikipedia)

Mathematics PreCalculus Mathematics at Nebraska

Up to this point, we have learned how to create graphs that reflect linear equations and how to construct linear equations given graphs. In this section, we will introduce two characteristics that pairs of lines can have so that we can continue to construct linear equations with various properties.

In this section, you will.

learn what it means for two lines to be

learn what it means for two lines to be

determine whether lines are parallel, perpendicular, or neither from their equations

construct linear equations for lines with various properties

Subsection Parallel and Perpendicular Lines

are lines in the same plane that never intersect. Two different lines in the same plane are parallel if their slopes are the same in symbols, if the slope of the first line is (m_1) and the slope of a different line is (m_2 ext<,>) then the slopes are parallel if (m_1=m_2 ext<.>)

Example 90

Find an equation of the line passing through ((4,1)) and parallel to (x-2y=-2 ext<.>)

To find the slope of the given line, solve for (y ext<.>)

Here the given line has slope (m_1=frac<1><2> ext<.>) The line we are constructing is parallel to this line and will therefore have the same slope, so (m_2=frac<1><2> ext<.>) Since we are given a point and we now have the slope, we will choose to use point-slope form of a linear equations to determine the slope-intercept form of the equation.

Our point is ((4,1)) and our slope is (m=frac<1><2> ext<.>)

The equation of the line is given by (y-1 = dfrac<1><2>(x-4) ext<.>)

It is important to have a geometric understanding of this question. The line we were given, (x-2y=-2 ext<,>) is shown in blue. The line we constructed, (y-1=frac<1><2>(x-4) ext<,>) is shown in magenta. notice that the slope is the same at the given line, but they are distinct lines (that is, they do not have the same (y)-intercept).

are lines in the same plane that intersect at right angles (90 degrees). Two nonvertical lines, in the same plane with slopes (m_1) and (m_2 ext<,>) are perpendicular if the product of their slopes is (-1 ext<,>) (m_1cdot m_2=-1 ext<.>) We can solve for (m_1) and obtain (m_1=-frac<1> ext<.>) In this form, we see that perpendicular lines have slopes that are . In general, given non-zero real numbers (a) and (b ext<,>) if the slope of the first line is given by (m_1=frac ext<,>) then the slope of the perpendicular line is (m_2=-frac ext<.>)

For example, the opposite reciprocal of (m_1=-frac<3><5>) is (m_2=frac<5><3> ext<.>) We can verify that two slopes produce perpendicular lines if their product is (-1 ext<.>)

Example 91

Find an equation of the line passing through ((-5,-2)) and perpendicular to the graph of (x+4y=4 ext<.>)

To find the slope of the given line, solve for (y ext<.>)

The given line has slope (m_1=-frac<1><4> ext<.>) Since we are constructing a line that is perpendicular to this line, their slopes should be opposite reciprocals, and thus the line we are constructing should have a slope of (m_2=+frac<4><1>=4 ext<.>) Now we can substitute the slope, (m_2=4 ext<,>) and the given point, ((-5,-2) ext<,>) into point-slope form:

The equation of the perpendicular line is given by (y+2=4(x+5) ext<.>)

Geometrically, we see that the graph of (y=4x+18 ext<,>) shown as the dashed line in the graph, passes through ((-5,-2)) and is perpendicular to the graph of (y=-frac<1><4>x+1 ext<.>)

Example 92

Find an equation of the line passing through ((-5,-1)) and perpendicular to (frac<1><3>x-frac<1><2>y=-2 ext<.>)

To find the slope of the given line, we solve for (y ext<:>)

The given line has slope (m_1=frac<2><3> ext<,>) therefore the line we are constructing must have a slope of (m_2=-frac<3><2> ext<.>) Using this and the point ((-5,-1) ext<,>) we can use point-slope form to write the following equation:

Maths Formulas For Class 10 (Chapterwise)

Before getting into the list of the formulas, let’s check out the major chapters of Class 10 Maths for which formulas are needed:

Class 10 Maths Formulas For Arithmetic Progression (AP)

If a1, a2, a3, a4….. be the terms of an AP and d be the common difference between each term, then the sequence can be written as: a, a + d, a + 2d, a + 3d, a + 4d…… a + nd. where a is the first term and (a + nd) is the (n – 1) th term. So, the formula to calculate the nth term of AP is given as:

n th term = a + (n-1) d

The sum for the nth term of AP where a is the 1st term, d is the common difference, and l is the last term is given as:

Sn = n/2 [2a + (n-1) d] or Sn = n/2 [a + l]

Class 10 Maths Formulas For Linear Equations

Linear equations in one, two, and three variables have the following forms:

Linear Equation in one Variableax + b=0Where a ≠ 0 and a & b are real numbers
Linear Equation in Two Variables ax + by + c = 0Where a ≠ 0 & b ≠ 0 and a, b & c are real numbers
Linear Equation in Three Variablesax + by + cz + d = 0Where a ≠ 0, b ≠ 0, c ≠ 0 and a, b, c, d are real numbers

The pair of linear equations in two variables are given as:

Quick Note: Linear equations can also be represented in graphical form.

Trigonometry Formulas For Class 10 Maths

The Trigonometric Formulas for Class 10 covers the basic trigonometric functions for a right-angled triangle i.e. Sine (sin), Cosine (cos), and Tangent (tan) which can be used to derive Cosecant (cos), Secant (sec), and Cotangent (cot).

Let a right-angled triangle ABC is right-angled at point B and have (angle heta) is one of the other two angles.

The Trigonometric Table comprising the values of these trigonometric functions for standard angles is as under:


Some other trigonometric formulas are given below:

  1. sin (90° – θ) = cos θ
  2. cos (90° – θ) = sin θ
  3. tan (90° – θ) = cot θ
  4. cot (90° – θ) = tan θ
  5. sec (90° – θ) = cosecθ
  6. cosec (90° – θ) = secθ
  7. sin 2 θ + cos 2 θ = 1
  8. sec 2 θ = 1 + tan 2 θ for 0° ≤ θ < 90°
  9. Cosec 2 θ = 1 + cot 2 θ for 0° ≤ θ ≤ 90°

Class 10 Maths Formulas For Algebra & Quadratic Equations

To know the algebra formulas for Class 10, first, you need to get familiar with Quadratic Equations.

The Quadratic Formula: For a quadratic equation px 2 + qx + r = 0, the values of x which are the solutions of the equation are given by:

Now you know the basic quadratic equation.

Let us now go through the list of algebra formulas for Class 10:

  1. (a+b) 2 = a 2 + b 2 + 2ab
  2. (a-b) 2 = a 2 + b 2 – 2ab
  3. (a+b) (a-b) = a 2 – b 2
  4. (x + a)(x + b) = x 2 + (a + b)x + ab
  5. (x + a)(x – b) = x 2 + (a – b)x – ab
  6. (a + b) 3 = a 3 + b 3 + 3ab(a + b)
  7. (a – b) 3 = a 3 – b 3 – 3ab(a – b)
  8. (x – a)(x + b) = x 2 + (b – a)x – ab
  9. (x – a)(x – b) = x 2 – (a + b)x + ab
  10. (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2xz
  11. (x + y – z) 2 = x 2 + y 2 + z 2 + 2xy – 2yz – 2xz
  12. (x – y + z) 2 = x 2 + y 2 + z 2 – 2xy – 2yz + 2xz
  13. (x – y – z) 2 = x 2 + y 2 + z 2 – 2xy + 2yz – 2xz
  14. x 3 + y 3 + z 3 – 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz -xz)
  15. x 2 + y 2 =½ [(x + y) 2 + (x – y) 2 ]
  16. (x + a) (x + b) (x + c) = x 3 + (a + b +c)x 2 + (ab + bc + ca)x + abc
  17. x 3 + y 3 = (x + y) (x 2 – xy + y 2 )
  18. x 3 – y 3 = (x – y) (x 2 + xy + y 2 )
  19. x 2 + y 2 + z 2 -xy – yz – zx = ½ [(x-y) 2 + (y-z) 2 + (z-x) 2 ]

Quick Note: These formulas will be important in higher classes and various comeptitive examninations. So, memorize them and understand them well.

Class 10 Maths Formulas For Circle

Circle formulas act as a base for Mensuration. The Class 10 Maths Circle formulas for a circle of radius r are given below:

  • 1. Circumference of the circle = 2 π r
  • 2. Area of the circle = π r 2
  • 3. Area of the sector of angle θ = (θ/360) × π r 2
  • 4. Length of an arc of a sector of angle θ = (θ/360) × 2 π r

Class 10 Maths Formulas For Surface Area & Volume

These formulas are very important for successfully solving mensuration questions. Find below the formulas in a tabulated form for your convenience.

Here, LSA = Lateral Surface Area,

Class 10 Maths Formulas For Statistics

Statistics in Class 10 is majorly about finding the Mean, Median, and Mode of the given data. The statistic formulas are given below:

(I) The Mean of Grouped Data can be found by 3 methods.

  1. Direct Method: x̅ = (frac^f_x_>^f_>), where fi xi is the sum of observations for i = 1 to n And fi is the number of observations for i = 1 to n
  2. Assumed Mean Method : = a+(frac^f_d_>^f_>)
  3. Step Deviation Method : x̅ = a+(frac^f_u_>^f_> imes h)

(II) The Mode of Grouped Data: Mode = l +(frac<>-f_<0>><2f_<1>-f_<0>-f_<2>> imes h)

(III) The median for a grouped data: Median = l+(frac<2>-cf> imes h)

Some FAQs On Maths Formulas Class 10

Here are some frequently asked questions that students ask regarding Class 10 Maths Formulas.

Ans: Learning or memorizing maths formulas requires a lot of practice. First, familiarize yourself with the chapter and concepts, then try to understand how a formula is derived and then memorize it.

Ans: A formal is a set of instructions that produces the desired result whereas an equation contains numerical operators. For and equation the LHS should be equal to RHS.

Ans: To learn maths formulas easily you can take the help of formulas provided in this article. You can learn them directly from the article or you can take a printout.

Ans: Students looking for Surface Areas and Volume formulas can view them through this article.

These are some of the important formulas for Class 10 Maths. These Maths formulas for Class 10 will prove to be helpful in your learning process. You will find them useful while revising the CBSE Class 10 Maths syllabus.

Solve the free Class 10 Maths questions and refer to these formulas to score better in Class 10 board exams:

Maths Formulas For Class 8Mensuration Formulas
Trigonometry TableTrigonometric Ratios

If you have any queries regarding this article on Maths formulas for class 10 , feel free to ask in the comment section below. We will get back to you at the earliest.


A N EQUATION is an algebraic statement in which the verb is "equals" = . An equation involves an unknown number, typically called x . Here is a simple example:

"Some number, plus 4, equals 10."

We say that an equation has two sides : the left side, x + 4, and the right side, 10.

Because x appears to the first power, we call that a linear equation. A linear equation is also called an equation of the first degree .

The degree of any equation is the highest exponent that appears on the unknown number. An equation of the first degree is called linear because, as we will see much later, its graph is a straight line .

The equation -- that statement -- will become true only when the unknown has a certain value, which we call the solution to the equation.

The solution to that equation is obviously 6:

6 is the only value of x for which the statement " x + 4 = 10" will be true. We say that x = 6 satisfies the equation.

Now, algebra depends on how things look. As far as how things look, then, we will know that we have solved an equation when we have isolated x on the left .

Why the left? Because that's how we read, from left to right. " x equals . . ."

In the standard form of a linear equation -- ax + b = 0 -- x appears on the left.

In fact, we have seen that, for any equation that looks like this:

x + a = b ,
the solution will always look like this:
x = b &minus a .
x + 4 = 10,
x = 10 &minus 4
= 6.

There are two pairs of inverse operations . Addition and subtraction, multiplication and division.

Formally, to solve an equation we must isolate the unknown on one side of the equation.

We must get a, b , c over to the other side, so that x is alone.

How do we shift a number from one side of an equation
to the other?

By writing it on the other side with the inverse operation.

That is the law of inverses . It follows from the two Rules of Lesson 5.

Example 1. Solve this equation:

a x &minus b + c = d .
Solution. Since b is subtracted on the left, we will add it on the right:
a x + c = d + b .
Since c is added on the left, we will subtract it on the right:
ax = d + b &minus c .
And finally, since a multiplies on the left, we will divide it on the right:
x = d + b &minus c

We have solved the equation.

Solving any linear equation, then, will fall into four forms, corresponding to the four operations of arithmetic. The following are the basic rules for solving any linear equation. In each case, we will shift a to the other side.

1. If x + a = b , then x = b &minus a .

"If a number is added on one side of an equation,
we may subtract it on the other side."

2. If x &minus a = b , then x = b + a .

"If a number is subtracted on one side of an equation,
we may add it on the other side."

"If a number multiplies one side of an equation,
we may divide it on the other side."

"If a number divides one side of an equation,
we may multiply it on the other side."

In every case, a was shifted to the other side by means of the inverse operation. It will be possible to solve any linear equation by applying one or more of those rules..

When the operations are addition or subtraction (Forms 1 and 2), we call that transposing .

We may shift a term to the other side of an equation
by changing its sign .

+ a goes to the other side as &minus a .

&minus a goes to the other side as + a .

Transposing is one of the most characteristic operations of algebra, and it is thought to be the meaning of the word algebra , which is of Arabic origin. (Arabic mathematicians learned algebra in India, from where they introduced it into Europe.) Transposing is the technique of those who actually use algebra in science and mathematics -- because it is skillful. And as we are about to see, it maintains the clear, logical sequence of statements . Moreover, it emphasizes that you do algebra with your eyes. When you see

x + a = b ,
then you immediately see that + a goes to the other side as &minus a :
x = b &minus a .

What is often taught, though, is to actually write &minus a on both sides, draw a line and add.

First, you will never see that in any calculus text. What you will see is a logical sequence of statements , which we are about to come to.

What is more, we proved that we may simply transpose. It is not necessary to prove it again every time you solve an equation.

(Do you have to prove the Pythagorean theorem every time you apply it? No, you do not.)

If you want to imagine that you have subtracted a from both sides, fine. But to have to write it is not skillful.

Here is what you will see in your calculus text.

Let us consider again the equation of Example 1.

That algebraic sentence -- that statement -- will logically imply other statements. We will now see the logical sequence that leads to the final statement, which is the solution.

The original equation (1) is "transformed" by first transposing the terms . Statement (1) implies statement (2).

That statement is then transformed by dividing by a . Statement (2) implies statement (3), which is the solution.

Thus we solve an equation by transforming it -- changing how it looks -- statement by statement, line by line according to the rules of algebra, until x finally is isolated on the left. That is how books on mathematics are written (but unfortunately not books that teach algebra!). Each line is its own readable statement that follows from the line above -- with no crossings out.

In other words, What is a calculation? It is a discrete transformation of symbols. In arithmetic we transform "19 + 5" into "24". In algebra we transform " x + a = b " into " x = b &minus a ."

Problem 1. Write the logical sequence of statements that will solve this equation for x :

To see the answer, pass your mouse from left to right
over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!

First, transpose the terms . Line (2).

It is not necessary to write the term 0 on the right.

Then divide by the coefficient of x .

Problem 2. Write the logical sequence of statements that will solve this equation for x :

In Problems 3, 4, and 5, only the solution is given. The student should write the logical sequence of statements that leads to it.

Problem 3. Solve for x : ( p &minus q ) x + r = s

Problem 5. Solve for x : 2 x + 1= 0

Each of the equations above are in the standard form , namely:

a does not mean a . It means the coefficient of x . And b does not mean b . It means whatever the terms are.

That is why it is called a form. Whatever looks like that.

Problem 6 . Solve: ax + b = 0.
x = &minus b

This simple equation illustrates doing algebra with your eyes. The student should see the solution immediately. You should see that b will go to the other side as &minus b , and that a will divide.

Problem 7. Solve for x : ax = 0 ( a 0).

Now, when the product of two numbers is 0, then at least one of them must be 0. (Lesson 6.) Therefore, any equation with that form has the solution,

We could solve that formally, of course, by dividing by a .

Problem 9. Write the sequence of statements that will solve this equation:

When we go from line (1) to line (2), &minus x remains on the left. For, the terms in line (1) are 6 and &minus x .

We have "solved" the equation when we have isolated x -- not &minus x -- on the left. Therefore we go from line (3) to line (4) by changing the signs on both sides. (Lesson 5.)

Alternatively, we could have eliminated &minus x on the left by changing all the signs immediately:

We can easily solve this -- in one line -- simply by transposing x to the left, and what is on the left to the right:

In this Example, + x is on the right. Since we want + x on the left, we can achieve that by exchanging sides:

Note: When we exchange sides, no signs change.

Upon transposing c , the solution easily follows:

In summary, when &minus x is on the right, it is skillful simply to transpose it. But when + x is on the right, we may exchange the sides.

Problem 18. Solve for cos &theta ("cosine thay -ta").

The thing to see is that this equation has exactly the same form as Problem 17. cos &theta is the unknown. You will solve it exactly the same as Problem 17.

Algebra consists in recognizing the form. And there are only a finite number.

Praxis Core Math: Linear Equation Practice Questions

Linear equations are classified as a type of algebra problem in the official Core Math Study Companion, but they behave like function questions in many ways. A linear equation is an equation that states the way that two algebraic variables will always behave in relation to each other, even when the values change.

In other words, linear equations show variables that are dependent on each other—when the value of one variable changes, the other variable’s value will also change. And the changes will be predictable. To give a very simple example, suppose the values of x and y are connected, so that whenever the value of x goes up by 1, the value of y also increases by 1. This is pretty predictable, right?

But even then, there still must be a point at which x and y intercept. This is the point at which x is 0. In the equation described in the above paragraph, if y is 0 when x is also 0, than the linear equation would be x = y. This is because x and y will always have the same value if they intersect at 0 and each number goes up by 1 when the other number increases by 1. You can see how the linear equation of x and y’s exact equivalency goes up in the following table:

table for linear equation x = y

This same relationship could be expressed by drawing a line (hence the name linear equation) on a coordinate plane:

graph for linear equation x = y
The linear equations you’ll come across in Praxis Core Math questions will be a good deal more complicated, of course. A more typical linear relationship might involve y increasing by 8 every time x increases by 2. 8 is four times as much as 2, so this means that y increases 4 times as quickly as x.

So in terms of the inter-related increase of x and y, y = 4x. Now if y and x both meet each other at 0, then y = 4x would be the correct linear equation. However, on the Praxis Core, the intercept will probably also be more complex than the example above. Let’s say that y is 3 when x is 0. To add this intercept to y = 4x, you can rewrite the linear equation as y = 4x + 3. On the Core Math test, the table for this equation would look like this:

table for linear equation y =4x + 3

And the graph for the equation might look something like this:

graph for linear equation x = y

Equations reducible to linear form

Consider the following examples to see how we can reduce equations involving ratios into linear form.

    Transpose the variable term &lsquox&rsquo to the left side and the numerical terms on the right side of the equation by changing their sign.

Divide both sides by 4 to find the value of x.

Cramer's Rule and Gaussian Elimination are two good, general-purpose algorithms (also see Simultaneous Linear Equations). If you're looking for code, check out GiNaC, Maxima, and SymbolicC++ (depending on your licensing requirements, of course).

EDIT: I know you're working in C land, but I also have to put in a good word for SymPy (a computer algebra system in Python). You can learn a lot from its algorithms (if you can read a bit of python). Also, it's under the new BSD license, while most of the free math packages are GPL.

You can solve this with a program exactly the same way you solve it by hand (with multiplication and subtraction, then feeding results back into the equations). This is pretty standard secondary-school-level mathematics.

If you plug these values back into A, B and C, you'll find they're correct.

The trick is to use a simple 4x3 matrix which reduces in turn to a 3x2 matrix, then a 2x1 which is "a = n", n being an actual number. Once you have that, you feed it into the next matrix up to get another value, then those two values into the next matrix up until you've solved all variables.

Provided you have N distinct equations, you can always solve for N variables. I say distinct because these two are not:

They are the same equation multiplied by two so you cannot get a solution from them - multiplying the first by two then subtracting leaves you with the true but useless statement:

By way of example, here's some C code that works out the simultaneous equations that you're placed in your question. First some necessary types, variables, a support function for printing out an equation, and the start of main :

Next, the reduction of the three equations with three unknowns to two equations with two unknowns:

Next, the reduction of the two equations with two unknowns to one equation with one unknown:

Now that we have a formula of the type number1 = unknown * number2 , we can simply work out the unknown value with unknown <- number1 / number2 . Then, once you've figured that value out, substitute it into one of the equations with two unknowns and work out the second value. Then substitute both those (now-known) unknowns into one of the original equations and you now have the values for all three unknowns:

Linear Equations

Linear equations are the simplest kind of equations you come across in maths. You may be asked to solve a linear equation:

or to draw the graph of a linear equation such as y =2x+1, which is a straight line, or to solve simultaneous linear equations:

We all solve linear equations in our heads all the time without even noticing it. If, for example, you have bought two CDs, for the same price, and a book, and know that you spent £20 pounds in total and that the book was £6, then you solve the linear equation 2x+6=20 to find out that the price of each CD was £7.

But the use of linear equations goes far beyond every-day problems like this one.

Computer graphics

Computer graphics programmers use what is called “linear algebra”. As a simple example take the figure below. We can describe the location of each point on the image by a coordinate system. A point has co-ordinates (x,y) if you can get to it from the point where the two axes meet by taking x steps in the horizontal direction and y steps in the vertical direction. Now suppose you would like to rotate that image clockwise through a sixth of a turn. Then you can calculate that every point that previously had the co-ordinates (x,y) now moves to the position with co-ordinates

The two new coordinates have a similar form to the linear equations above and this is why graphics programmers end up having to solve a lot of linear equations.

This is a simple example – you can imagine that if the images involved are a lot more complicated, for example three-dimensional, and if the object is to move in a much more complicated way, then the programmer must be able to handle a great number of simultaneous linear equations, involving many variables.


Economics is another area that makes use of linear equations a lot. Think, for example, of supply and demand. Your demand for a product – a chocolate bar or a car – will depend on the price of the product and how much you earn, among other things. If the product is very expensive, and you’re not that rich, you probably don’t want too much of it. If it’s cheap, like the chocolate bar, you may want a lot (unless you’re worried about your teeth). The demand is often expressed as linear equation:
D = a – bP +cI, where D is the demand, P the price and I your income. You can see that D goes down if P goes up (that is the reason for the minus sign), and it goes up as your income goes up. In a similar way, the supply can be expressed by a linear equation. Economists and other people in the financial world work with these kind of equations all the time.

The study of genes

Systems of linear equations also come up a lot in the study of genes. If you want to find out what a particular gene actually does, you have to see how it influences all the chemical processes in our body. There are hundreds of those going on in our bodies all the time, for example we produce sugars and proteins. The way these processes work, and how they influence each other, can be expressed by large systems of linear equations.

These are just three examples of how these equations are used in the real world. But there is another very important thing about them: they are the simplest equations there are. Equations are used in many areas to model the world around us. A biologist will use them to get an idea of how a population of animals might change over time. An economist or financial adviser will use them to predict the economy or the future profits of a company. An engineer will use them to work out the exact proportions of a building, like a bridge or a sky scraper, and how much and what kind of materials to use. In short, equations are a fact of life for many people, and to be able to work with them you need to start with the simplest ones – the linear equations.

Wolfram Web Resources

The #1 tool for creating Demonstrations and anything technical.

Explore anything with the first computational knowledge engine.

Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more.

Join the initiative for modernizing math education.

Solve integrals with Wolfram|Alpha.

Walk through homework problems step-by-step from beginning to end. Hints help you try the next step on your own.

Unlimited random practice problems and answers with built-in Step-by-step solutions. Practice online or make a printable study sheet.

Collection of teaching and learning tools built by Wolfram education experts: dynamic textbook, lesson plans, widgets, interactive Demonstrations, and more.

About the Book

This text covers the standard material for a US undergraduate first course: linear systems and Gauss's Method, vector spaces, linear maps and matrices, determinants, and eigenvectors and eigenvalues, as well as additional topics such as introductions to various applications. It has extensive exercise sets with worked answers to all exercises, including proofs, beamer slides for classroom use, and a lab manual for computer work. The approach is developmental. Although everything is proved, it introduces the material with a great deal of motivation, many computational examples, and exercises that range from routine verifications to a few challenges. Ancillary materials are available at the publisher link.