ORCCA Solving Equations in General

Section 7.4 Solving Equations in General

Objectives: PCC Course Content and Outcome Guide

In your algebra studies, you have learned how to solve linear equations, quadratic equations, and radical equations. In this section, we examine some similarities among the processes for solving these equations. Understanding these similarities can improve your general equation solving ability, even into the future with new equations that are not of these three types.

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Figure 7.4.1. Alternative Video Lesson

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Subsection 7.4.1 Equations Where the Variable Appears Once

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Here are some examples of equations that all have something in common: the variable only appears once.

\begin{aligned} 2 \overset{↓}{x} + 1 & = 7 & (\overset{↓}{x} + 4)^{2} & = 36 & \sqrt{2 \overset{↓}{x} - 3} = 3 \end{aligned}

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For equations like this, there is a strategy for solving them that will keep you from overcomplicating things. In each case, according to the order of operations, the variable is having some things “done” to it in a specific order.

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With

2 x + 1 = 7,

$x$ is multiplied by $2$
then that result is added to $1$
and this result is a number, $7$

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With

(x + 4)^{2} = 36,

$x$ is added to $4$
then that result is squared
and this result is a number, $36$

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With

\sqrt{2 x - 3} = 3,

$x$ is multiplied by $2$
then that result has $3$ subtracted from it
then that result has a square root applied
and this result is a number, $3$

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Because there is just one instance of the variable, and then things happen to that value in a specific order according to the order of operations, then there is a good strategy to solve these equations. We can just undo each step in the opposite order.

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Example 7.4.2.

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Solve the equation

2 x + 1 = 7 .

Explanation.

The actions that happen to

x

are multiply by

2,

and then add

1 .

So we will do the opposite actions in the opposite order to each side of the equation. We will subtract

1

and then divide by

2 .

\begin{aligned} 2 x + 1 & = 7 & now subtract 1 from each side \\ 2 x + 1 - 1 & = 7 - 1 \\ 2 x & = 6 & now divide by 2 on each side \\ \frac{2 x}{2} & = \frac{6}{2} \\ x & = 3 \end{aligned}

You should check this solution by substituting it into the original equation.

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Example 7.4.3.

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Solve the equation

(x + 4)^{2} = 36 .

Explanation.

The actions that happen to

x

are add

4,

and then square. So we will do the opposite actions in the opposite order to each side of the equation. We will apply the Square Root Property and then subtract

4 .

\begin{aligned} (x + 4)^{2} & = 36 & now apply the Square Root Property \\ x + 4 & = \pm \sqrt{36} \\ x + 4 & = \pm 6 & now subtract 4 on each side \\ x + 4 - 4 & = \pm 6 - 4 \\ x & = \pm 6 - 4 \end{aligned}

\begin{aligned} x & = - 6 - 4 & or & x & = 6 - 4 \\ x & = - 10 & or & x & = 2 \end{aligned}

You should check these solutions by substituting them into the original equation.

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Example 7.4.4.

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Solve the equation

\sqrt{2 x - 3} = 3 .

Explanation.

The actions that happen to

x

are multiply by

2,

and then subtract

3,

and then apply the square root. So we will do the opposite actions in the opposite order to each side of the equation. We will square both sides, add

3

and then divide by

2 .

\begin{aligned} \sqrt{2 x - 3} & = 3 & now square both sides \\ 2 x - 3 & = 9 & now add 3 to each side \\ 2 x - 3 + 3 & = 9 + 3 \\ 2 x & = 12 & now divide by 2 on each side \\ \frac{2 x}{2} & = \frac{12}{2} \\ x & = 6 \end{aligned}

You should check this solution by substituting it into the original equation.

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Subsection 7.4.2 Equations With More Than One Instance of the Variable

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Now consider equations like

\begin{aligned} 5 \overset{↓}{x} + 1 & = 3 \overset{↓}{x} + 2 & \overset{↓}{x^{2}} + 6 \overset{↓}{x} & = - 8 & \sqrt{\overset{↓}{x} - 3} = \sqrt{\overset{↓}{x}} - 1 \end{aligned}

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In these examples, the variable appears more than once. We can’t exactly dive in to the strategy of undoing each step in the opposite order. For each of these equations, remind yourself that you can apply any operation you want, as long as you apply it to both sides of the equation. In many cases, you will find that there is some basic algebra move you can take that will turn the equation into something more “standard” that you know how to work with.

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With

5 x + 1 = 3 x + 2,

we have a linear equation. If we can simply reorganize the terms to combine like terms, a solution will be apparent.

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With

x^{2} + 6 x = - 8,

adding

8

to both sides would give us a quadratic equation in standard form. And then the quadratic formula can be used.

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With

\sqrt{x - 3} = \sqrt{x} - 1,

the complication is those two radicals. We can take any action we like as long as we apply it to both sides, and squaring both sides would remove at least one radical. Maybe after that we will have a simpler equation.

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Example 7.4.5.

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Solve the equation

5 x + 1 = 3 x + 2 .

Explanation.

We’ll use basic algebra to rearrange the terms.

\begin{aligned} 5 x + 1 & = 3 x + 2 & now subtract 1 from each side \\ 5 x & = 3 x + 1 & now subtract 3 x from each side \\ 2 x & = 1 & now divide by 2 on each side \\ x & = \frac{1}{2} \end{aligned}

You should check this solution by substituting it into the original equation.

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Example 7.4.6.

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Solve the equation

x^{2} + 6 x = - 8 .

Explanation.

Adding

8

to each side will give us a quadratic equation in standard form, and then we may apply The Quadratic Formula.

\begin{aligned} x^{2} + 6 x & = - 8 & now add 8 to each side \\ x^{2} + 6 x + 8 & = 0 & now apply The Quadratic Formula \end{aligned}

\begin{aligned} x & = \frac{- 6 \pm \sqrt{6^{2} - 4 (1) (8)}}{2 (1)} \\ = \frac{- 6 \pm \sqrt{36 - 32}}{2} \\ = \frac{- 6 \pm \sqrt{4}}{2} \\ = \frac{- 6 \pm 2}{2} \end{aligned}

\begin{aligned} x & = \frac{- 6 - 2}{2} & or & x & = \frac{- 6 + 2}{2} \\ x & = \frac{- 8}{2} & or & x & = \frac{- 4}{2} \\ x & = - 4 & or & x & = - 2 \end{aligned}

You should check these solutions by substituting them into the original equation.

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Example 7.4.7.

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Solve the equation

\sqrt{x - 3} = \sqrt{x} - 1 .

Explanation.

Hoping to obtain a simpler equation, we will square each side. This will eliminate at least one radical, which may help.

\begin{aligned} \sqrt{x - 3} & = \sqrt{x} - 1 & now square both sides \\ {(\sqrt{x - 3})}^{2} & = {(\sqrt{x} - 1)}^{2} \\ x - 3 & = {(\sqrt{x})}^{2} - 2 \sqrt{x} + 1 \\ x - 3 & = x - 2 \sqrt{x} + 1 & now note that there are some like terms \\ - 3 & = - 2 \sqrt{x} + 1 & now we have an equation with only one instance of the variable \\ - 4 & = - 2 \sqrt{x} \\ 2 & = \sqrt{x} \\ 2^{2} & = x \\ x & = 4 \end{aligned}

You should check this solution by substituting it into the original equation. It is especially important to do this when the equation was a radical equation. At one point, we squared both sides, and this can introduce extraneous solutions (see Remark 6.4.4).

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Subsection 7.4.3 Solving For a Variable in Terms of Other Variables

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In the examples so far in this section, there has been one variable (but possibly more than one instance of that variable). This leaves out important situations in science applications where you have a formula with multiple variables, and you need to isolate one of them. Fortunately these situations are not more difficult than what we have explored so far, as long as you can keep track of which variable you are trying to solve for.

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Example 7.4.8.

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In physics, there is a formula for converting a Celsius temperature to Fahrenheit:

F = \frac{9}{5} C + 32

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Solve this equation for

C

in terms of

F .

Explanation.

The variable we are after is

C,

and that variable only appears once. So we will apply the strategy of undoing the things that are happening to

C .

First

C

is multiplied by

\frac{9}{5},

and then it is added to

32 .

So we will undo these actions in the opposite order: subtract

32

and then multiply by

\frac{5}{9}

(or divide by

\frac{9}{5}

if you prefer).

\begin{aligned} F & = \frac{9}{5} \overset{↓}{C} + 32 \\ F - 32 & = \frac{9}{5} \overset{↓}{C} + 32 - 32 \\ F - 32 & = \frac{9}{5} \overset{↓}{C} \\ \frac{5}{9} \cdot (F - 32) & = \frac{5}{9} \cdot \frac{9}{5} \overset{↓}{C} \\ \frac{5}{9} (F - 32) & = \overset{↓}{C} \\ C & = \frac{5}{9} (F - 32) \end{aligned}

We are satisfied, because we have isolated

C

in terms of

F .

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Example 7.4.9.

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In physics, when an object of mass

m

is moving with a speed

v,

its “kinetic energy”

E

is given by:

E = \frac{1}{2} m v^{2}

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Solve this equation for

v

in terms of the other variables.

Explanation.

The variable we are after is

v,

and that variable only appears once. So we will apply the strategy of undoing the things that are happening to

v .

First

v

is squared, then it is multiplied by

m

and by

\frac{1}{2} .

So we will undo these actions in the opposite order: multiply by

2,

divide by

m,

and apply the square root.

\begin{aligned} E & = \frac{1}{2} m \overset{↓}{v^{2}} \\ 2 \cdot E & = 2 \cdot \frac{1}{2} m \overset{↓}{v^{2}} \\ 2 E & = m \overset{↓}{v^{2}} \\ \frac{2 E}{m} & = \frac{m \overset{↓}{v^{2}}}{m} \\ \frac{2 E}{m} & = \overset{↓}{v^{2}} \\ \pm \sqrt{\frac{2 E}{m}} & = \overset{↓}{v} \\ v & = \sqrt{\frac{2 E}{m}} \end{aligned}

At the very end, we chose the positive square root, since a speed

v

cannot be negative. We are satisfied, because we have isolated

v

in terms of

E

and

m .

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Solve the equation

A = s^{2}

for

s .

Assume

s > 0 .

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52.

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Solve the equation

A = π r^{2}

for

r .

Assume

r > 0 .

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53.

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Solve the equation

V = π r^{2} h

for

r .

Assume

r > 0 .

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54.

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Solve the equation

V = \frac{1}{3} s^{2} h

60.

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Solve the equation

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Solve the equation

F = m a

for

a .

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67.

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Solve the equation

a = \frac{v^{2}}{r}

for

v .

Assume

v > 0 .

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68.

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Solve the equation

K = \frac{1}{2} m v^{2}

for

v .

Assume

v > 0 .

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69.

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Solve the equation

T = 2 π \sqrt{\frac{ℓ}{g}}

for

l .

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70.

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Solve the equation

T = 2 π \sqrt{\frac{ℓ}{g}}

for

g .

You have attempted 1 of 2 activities on this page.

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Section 7.4 Solving Equations in General

Objectives: PCC Course Content and Outcome Guide

Subsection 7.4.1 Equations Where the Variable Appears Once

Example 7.4.2.

Example 7.4.3.

Example 7.4.4.

Subsection 7.4.2 Equations With More Than One Instance of the Variable

Example 7.4.5.

Example 7.4.6.

Example 7.4.7.

Subsection 7.4.3 Solving For a Variable in Terms of Other Variables

Example 7.4.8.

Example 7.4.9.

Reading Questions 7.4.4 Reading Questions

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