Linear Equations and Modeling

Contents: This page corresponds to § 2.1 (p. 168) of the text.

Suggested Problems from Text

p. 176 #2, 3, 5, 17, 19, 22, 26, 28, 29, 31, 34, 35, 39, 42, 44, 47, 65, 66, 67, 73

Solving Equations

Checking Solutions

Revenue Problem

Solving Equations

We have seen examples of equation solving already in this course. For example, to decide whether or not an equation in x and y determines y as a function of x, we solve the equation for y.

In this section we start looking at the details of solving equations in one variable.

A solution for an equation in x is a number such that when we substitute that number for x in the equation we have a true statement.

To solve an equation is to find the set of all solutions of the equation.

Example 1.

x2 - 9 = 0.

This equation has two solutions, x = 3 and x = -3.

For some equations the set of solutions is obvious.

Example 2.

x = 3.

It is clear that the number 3 is the only solution for this equation.

The general approach to solving equations is to reduce the problem to one as simple as that in Example 2 above. You do this by replacing the given equation by simpler equivalent equations. Equations are equivalent if they have the same set of solutions, and there is a collection of operations on the equations that produce equivalent equations.

Generating Equivalent Equations

(1) Simplify either side by combining like terms, reducing fractions, etc.

3(4 + x) + 2x = 7x + 5

is equivalent to

12 + 3x + 2x = 7x + 5

is equivalent to

12 + 5x = 7x + 5.

(2) Add or subtract the same quantity to or from both sides of the equation.

12 + 5x = 7x + 5

is equivalent to

12 + 5x -5 = 7x + 5 - 5

is equivalent to

7 + 5x = 7x

is equivalent to

7 + 5x - 5x = 7x - 5x

is equivalent to

7 = 2x.

(3) Multiply or divide both sides of the equation by the same non zero quantity.

7 = 2x

is equivalent to

7 ÷ 2 = 2x ÷ 2

7/2 = 2x/2

7/2 = x

(4) Interchange the two sides of the equation.

7/2 = x

is equivalent to

x = 7/2

Putting together the chain of equivalent equations used as examples above, we see that 3(4 + x) + 2x = 7x + 5 is equivalent to x = 7/2. Therefore, the number 7/2 is the only solution for 3(4 + x) + 2x = 7x + 5.

Linear Equations

A linear equation in x is one that can be written in the form ax + b = 0 for some numbers a and b with a not equal to 0.

A linear equation has exactly one solution. Furthermore, the approach used in the last example of finding an equivalent equation of the form x = c always works with linear equations.

Notes on Common Mistakes

(1) When we multiplied 3 through the parentheses above, we were using the distributive property which says 3(4 + x) is equal to 3*4 + 3*x. One common error is to multiply 3 times 4, then forget to multiply 3 times x.

(2) Adding or subtracting the same quantity from both sides of an equation is such a common operation that we usually start to skip steps. For example

x - 6 = 0.

x - 6 + 6 = 0 + 6.

x = 6.

We would usually go directly from the first equation to the third, and students often think of this as "moving the 6 to the other side." This is likely to cause problems unless you remember exactly what we are really doing to get from the first equation to the third. We are not saying that you have to write out the second equation. In fact, we will usually not include that step in these course notes, but it is important to understand the step we are leaving out.

(3) We also start to skip steps when multiplying or dividing both sides of an equation by the same non zero quantity.

5x = 2.

5x ÷ 5 = 2 ÷ 5.

x = 2/5.

When we go from the first to the third equation, students start to think of this as "moving the 5 to the other side." When you think in terms of moving numbers from side to side, it starts to get complicated. Sometimes the sign changes, sometimes it does not. Sometimes you multiply, sometimes you divide. This confusion can be eliminated if you keep in mind exactly what steps are being carried out.

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Checking Solutions

In general, solving an equation requires several steps and provides numerous opportunities for making mistakes. You should always check your solutions.

Checking Algebraically

The most obvious way to check a solution for an equation is to substitute the number for the variable in the equation and see if the resulting statement is true.

For example: We found the solution x = 7/2 for the equation 3(4 + x) + 2x = 7x + 5.

We will use the decimal form 7/2 = 3.5, and substitute for x on both sides of the equation.

Left Side: 3(4 + 3.5) +2*3.5 = 3*7.5 + 2*3.5 = 22.5 + 7 = 29.5.

Right Side: 7*3.5 + 5 = 24.5 + 5 = 29.5.

Since the left side equals the right side when we substitute x = 3.5, the solution checks.

Checking Graphically

Consider the two equations we get by setting y equal to the two sides of equation 3(4 + x) + 2x = 7x + 5.

Equation 1:  y = 3(4 + x) + 2x


Equation 2: y = 7x + 5

We think that x = 3.5 is a solution for the original equation. If that is true, then substituting x = 3.5 into equations 1 and 2 above will yield the same y value. This means that the graphs of equations 1 and 2 will intersect at a point with first coordinate x = 3.5. We can see if this is true by using a graphing utility to graph equations 1 and 2 and then using a trace utility to check the intersection point.

Exercise 1:

(a) Open the Java Grapher and enter 3*(4 + x) + 2*x in the f box and 7*x + 5 in the g box and graph.

Click the Trace button to start tracing the graph of f. Notice that you cannot see the point of intersection because it is not in the viewing rectangle. Drag the graph (down) until the intersection point is visible. Now click the Trace button and move the trace point to the intersection of the two graphs. The coordinates of the trace point should be close to (3.5,29.5).

Note that the two lines are steep and close to parallel, and this makes it difficult to see just when the trace point is on the intersection point. If you set the window coordinates to Xmin = 3, Xmax = 4, Ymin = 28, Ymax = 32 (and click Reset), then it is easier to see the location of the intersection point. We also could have improved our picture by dragging a "tall and narrow" Zoom box. We will get more practice with adjusting the viewing rectangle to find intersection points in later sections.

In this exercise we are checking a solution, rather than finding one from scratch, so we can use the Table feature to easily see that f(3.5) is equal to g(3.5).

(b) Click Table and enter 3.5 (or 7/2) in the x box at the bottom. The top row of the table shows the x value of 3.5 and the values of the two functions f and g at x = 3.5. This amounts to using the grapher to do the algebraic check that we did by hand earlier.

Note: You can usually check a solution more quickly by hand than you can with a calculator. However, if you made a simplification mistake while solving the equation, then you may repeat that mistake when checking, so it is a good to have another way to verify solutions.

Exercise 2:

(a) Solve the equation 2(x - 3) + 1 = 1 - x.

(b) Check by hand.

(c) Check graphically.


Exercise 3:

(a) Solve .

(b) Check by hand.

(c) Check graphically.

One method of generating equivalent equations is to multiply or divide both sides of the equation by a non zero quantity. It is obvious that multiplying both sides of an equation by the number 0 is not a good idea! But you must be careful when you multiply or divide both sides of an equation by an expression that contains a variable. That expression should not be zero.

Example 3.

The LCD is x2 -1 = (x + 1)(x - 1). (Recall that the LCD is the Least Common Denominator.) When we multiply both sides of the equation by x2 -1 and cancel, we get

2x + (x - 1) = x + 1.

2x = 2.

x = 1.

However, x = 1 is not a solution for the original equation. The expressions in the original equation are not defined for x = 1.

Multiplying by x2 -1 has introduced an extraneous solution x = 1. Since x = 1 was our only candidate for a solution, the original equation has no solutions.

Whenever you multiply, or divide, both sides of an equation by an expression with a variable, you need to check the solutions.

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Revenue Problem

The Acme Manufacturing Company produces a product called a widget. In this problem we will be given information about the cost of producing widgets and how their price is related to the number produced. The goal is find out how much AMC should charge for a widget in order to maximize profit.


The cost of producing one widget is $200.

The demand for widgets is determined by the price. In other words, the lower the price, the more they can sell.

When AMC charges $375 for a widget, they sell 25 per week. Management estimates that each $10 decrease in price will result in selling 2 more widgets per week.


Find the price per widget that maximizes profit.


The first step is to determine what quantities are important and assign variables to them. The Goal statement mentions price per widget and profit, so these quantities clearly need variable names. Because the number sold is closely related to price, it will also have a variable name.

n = number of widgets sold per week.

p = price for which AMC sells a widget.

f = weekly profit.

Demand Equation

In the Data section we are told (in words) how n and p are related. We need to find an equation, called the demand equation, that describes this relationship.

When we discussed equations in two variables, we used x and y; but, here the variable names are n and p. We will let the horizontal axis be the n-axis and the vertical axis will be the p-axis.

The fact that every $10 decrease in p results in an increase of 2 in n implies that the graph of the demand equation is a line, and we can find its slope.

We use the Point-Slope form of the equation of a line to find the demand equation.

p - 375 = -5(n - 25).

p - 375 = -5n + 125.

p = -5n + 500.

Profit Function

The weekly profit, f, is the number produced, n, times the price per widget, p, minus the cost of producing n widgets.

f = np - 200n.

Substituting the expression for p from the demand equation, yields

f = n(-5n + 500) - 200n.

f = -5n2 + 500n -200n.

f = -5n2 + 300n.

Maximizing Profit

Exercise 4:

Use a graphing utility to verify that the weekly profit of f = $4500 corresponding to selling n = 30 widgets is the largest possible. See Approximating Relative Extrema for help.

Note that in general, a graphing utility will only approximate a relative extreme point, but in this case the maximum occurs at a whole number value of n, so we actually get the true location of the relative maximum.

We are almost done! The goal was to find the price that AMC should charge to maximize profit, and we have found the weekly sales that will maximize profit. However, the demand equation determines price p as a function of sales n. So we see that when n = 30,

p = -5*30 + 500 = 350

Therefore, charging $350 per widget leads to weekly sales of 30 units which yields the maximum profit of $4500, and we have completed the problem.

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