Using the Distance Formula to Solve Word Problems
Solve a Formula for a Specific Variable
You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be familiar with formulas and be able to manipulate them easily.
In Example 2.58 and Example 2.59, we used the formula \(d=rt\). This formula gives the value of \(d\), distance, when you substitute in the values of \(r \text{and }t\), the rate and time. But in Example 2.59, we had to find the value of \(t\). We substituted in values of \(d \text{and }r\) and then used algebra to solve for \(t\). If you had to do this often, you might wonder why there is not a formula that gives the value of \(t\) when you substitute in the values of \(d\text{and }r\). We can make a formula like this by solving the formula \(d=rt\) for \(t\).
To solve a formula for a specific variable means to isolate that variable on one side of the equals sign with a coefficient of 1. All other variables and constants are on the other side of the equals sign. To see how to solve a formula for a specific variable, we will start with the distance, rate and time formula.
Example 2.60
Solve the formula \(d=rt\) for \(t\):
- when \(d=520\) and \(r=65\)
- in general
Solution
We will write the solutions side-by-side to demonstrate that solving a formula in general uses the same steps as when we have numbers to substitute.
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| Write the formula. | \(d=rt\) |
Write the formula. | \(d=rt\) |
| Substitute. | \(520=65t\) |
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| Divide, to isolate \(t\). |
\(\frac{520}{65}= \frac{65t}{65}\) |
Divide, to isolate \(t\). |
\(\frac{d}{r} = \frac{rt}{t}\) |
| Simplify. |
\(8=t\) |
Simplify. | \(\frac{d}{r} = t\) |
We say the formula \(t=dr\) is solved for \(t\).
Try It 2.119
Solve the formula \(d=rt\) for \(r\):
- when \(d=180\) and \(t=4\)
- in general
Try It 2.120
Solve the formula \(d=rt\) for \(r\)
- when \(d=780\) and \(t=12 \)
- in general
Example 2.61
Solve the formula \(A=\frac{1}{2}bh\) for \(h\):
- when \(A=90\) and \(b=15 \)
- in general
Solution
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| Write the formula. | \(A=\frac{1}{2}bh\) |
Write the formula. | \(A=\frac{1}{2}bh\) |
| Substitute. | \(90=\frac{1}{2} \cdot 15 \cdot h\) |
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| Clear the fractions. |
\(2 \cdot 90 = 2 \cdot \frac{1}{2}15h\) |
Clear the fractions. |
\(2 \cdot A = 2 \cdot \frac{1}{2}bh\) |
| Simplify. |
\(180=h\) |
Simplify. | \(2A = bh\) |
| Solve for \(h\). |
\(12=h\) | Solve for \(h\). |
\(\frac{2A}{b}=h\) |
We can now find the height of a triangle, if we know the area and the base, by using the formula \(h=\frac{2A}{b}\).
Try It 2.121
Use the formula \(A=\frac{1}{2}bh\) to solve for \(h\):
- when \(A=170\) and \(b=17\)
- in general
Try It 2.122
Use the formula \(A=\frac{1}{2}bh\) to solve for \(b\):
- when \(A=62\) and \(h=31 \)
- in general
The formula \(I=Prt\) is used to calculate simple interest, \(I\), for a principal, \(P\), invested at rate, \(r\), for \(t\) years.
Example 2.62
Solve the formula \(I=Prt\) to find the principal, \(P\):
- when \(I=$5,600\), \(r=4%\), \(t=7\) years
- in general
Solution
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| Write the formula. | \(I=Prt\) |
Write the formula. | \(I=Prt\) |
| Substitute. | \(5600=P(0.04)(7)\) |
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| Simplify. | \(5600=P(28)\) | Simplify. | \(I=P(rt)\) |
| Divide, to isolate P. | \(\frac{5600}{0.28}=\frac{P(0.28)}{0.28}\) |
Divide, to isolate P. |
\(\frac{I}{rt}=\frac{P(rt)}{rt}\) |
| Simplify |
\(20,000=P\) |
Simplify. | \(\frac{I}{rt}=P\) |
| The principal is |
\( \ $20,000\) | \(P=\frac{I}{rt}\) | |
Try It 2.123
Use the formula \(I=Prt\) to find the principal, \(P\):
- when \(I=$2,160\), \(r=6%\), \(t=3\) years
- in general
Try It 2.124
Use the formula \(I=Prt\) to find the principal, \(P\):
- when \(I=$5,400\), \(r=12%\), \(t=5\) years
- in general
Later in this class, and in future algebra classes, you'll encounter equations that relate two variables, usually \(x\) and \(y\). You might be given an equation that is solved for \(y\) and need to solve it for \(x\), or vice versa. In the following example, we're given an equation with both \(x\) and \(y\) on the same side and we'll solve it for \(y\).
Example 2.63
Solve the formula \(3x+2y=18\) for \(y\):
- when \(x=4\)
- in general
Solution
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| \(3x+2y=18\) |
\(3x+2y=18\) |
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| Substitute. | \(3(4)+2y=18\) |
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| Subtract to isolate the y-term. |
\(12-12+2y=18-12\) |
Subtract to isolate the y-term. |
\(3x-3x+2y=18-3x\) |
| Divide. |
\(\frac{2y}{2}=\frac{6}{2}\) |
Divide. | \(\frac{2y}{2}=\frac{18}{2}-\frac{3x}{2}\) |
| Simplify. |
\(y=3\) | Simplify. |
\(y=-\frac{3x}{2}+9\) |
Try It 2.125
Solve the formula \(3x+4y=10\) for \(y\):
- when \(x=\frac{14}{3}\)
- in general
Try It 2.126
Solve the formula \(5x+2y=18\) for \(y\):
- when \(x=4\)
- in general
In Examples 1.60 through 1.64 we used the numbers in part ⓐ as a guide to solving in general in part ⓑ. Now we will solve a formula in general without using numbers as a guide.
Example 2.64
Solve the formula \(P=a+b+c\) for \(a\).
Solution
| We will isolate \(a\) on one side of the equation. | \(P=a+b+c\) |
| Both \(b\) and \(c\) are added to \(a\), so we subtract them from both sides of the equation. | \(P - b - c =a+b+c - b - c\) |
| Simplify. |
\(P -b - c =a\) \(a=P -b -c\) |
Try It 2.127
Solve the formula \(P=a+b+c\) for \(b\).
Try It 2.128
Solve the formula \(P=a+b+c\) for \(c\).
Example 2.65
Solve the formula \(6x+5y=13\) for \(y\).
Solution
| \(6x+5y=13\) | |
| Subtract \(6x\) from both sides to isolate the term with \(y\). |
\(6x+5y-6x=13-6x\) |
| Simplify. | \(5y=13-6x\) |
| Divide by 5 to make the coefficient 1. | \(\frac{5y}{5}=\frac{13-6x}{5}\) |
| Simplify. | \(y=\frac{13-6x}{5}\) |
The fraction is simplified. We cannot divide \(13−6x\) by 5.
Try It 2.129
Solve the formula \(4x+7y=9\) for \(y\).
Try It 2.130
Solve the formula \(5x+8y=1\) for \(y\).