Solving a formula for a specified variable
A formula is solved for a variable when that variable is totally isolated on one side of the equation.
Example. Solve for C: V = C - ![]()
Solution:
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V = C - |
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L · V = L (C - |
Multiply both sides by LCD, L |
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LV = LC L · |
Remove parentheses |
| LV = LC (C S)N | Simplify |
| LV = LC (CN SN) | Distributive Property |
| LV = LC CN + SN | Distributive Property |
| LV SN = LC CN + SN SN | Subtract SN from both sides to isolate terms containing C on one side |
| LV SN = LC CN | Simplify |
| LV SN = C(L N) | Factor out C to isolate it |
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Divide both sides by L N |
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Simplify |
Strategy for Solving Verbal Problems
1. Read the problem carefully.
2. Use diagrams or charts if you think it will make the information clearer.
3. Find the relationship or formula relevant to the problem.
4. Identify the unknown quantity (or quantities), and label them, using one variable.
5. Write an equation involving the unknown quantity, using the relationship or formula from step 3.
6. Solve the equation.
7. Answer the question.
8. Check the answer in the original words of the problem.
Example. A total of $20,000 was invested in two bond mutual funds, a junk bond fund and a government bond fund. The junk bond fund is risky and yields 11% interest. The safer government bond fund yields only 5%. The total income for the year from the two investments was $1300. How much was invested in each fund?
Solution. Let x = the amount of money invested in the junk bond fund
20,000 x = the amount of money invested in the government bond fund
To calculate interest, we use the simple interest formula: interest = principle · rate · time
Since the time is one year, in this problem, interest = principle · rate
|
Amount invested (Principal) |
Interest Rate |
Interest Earned | |
| Junk Bonds | x | 0.11 | 0.11x |
| Government Bonds | 20,000 x | 0.05 | 0.05(20,000 x) |
| Total | 20,000 | 1300 |
The total interest earned is the sum of the interest earned from each fund. We obtain the following equation from the "Interest Earned" column of the chart:
| 0.11x + 0.05(20,000 x) = 1300 | |
| 0.11x + 1000 0.05x = 1300 | Distributive Property |
| 0.06x + 1000 = 1300 | Combine like terms |
| 0.06x = 300 | Subtract 1000 from each side |
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x = 5000 |
Divide both sides by 0.06 |
The amount invested in the junk bond fund, x, is $5000.
The amount invested in the government bond fund, 20000 x, is $15,000.
Example. Two travelers left a restaurant in Oklahoma City and traveled in opposite directions on Interstate 40. If one driver averaged 65 mph and the other averaged 60 mph, how long was it before they were 400 miles apart?
Solution. This problem involves distance, rate, and time; these variables are related by the distance formula: distance = rate · time
Let t = the time the two travelers drive until they are 400 miles apart
| Rate | Time | Distance | |
| One driver | 65 | t | 65t |
| Other driver | 60 | t | 60t |
| Total | 400 |
The sum of the distances of each driver is the total, 400. From the "Distance" column in the chart, we obtain the equation:
| 65t + 60t = 400 | |
| 125t = 400 | Combine like terms |
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Divide both sides by 125 |
| t = 3.2 | Simplify |
The travelers will be 400 miles apart in 3.2 hours.
Example. Braums Dairy mixed two grades of milk, one containing 3% butterfat and the other containing 4.5% butterfat, to obtain 150 gallons of milk that contained 4% butterfat. How many gallons of each were used in the mixture?
Solution. Let x = number of gallons of the 3% butterfat milk
150 x = number of gallons of the 4.5% butterfat milk
We multiply the number of gallons of solution by the percent of butterfat to obtain the number of gallons of butterfat in each solution.
|
Gallons of Solution |
Percent of Butterfat |
Gallons of Butterfat |
|
| 3% Milk | x | 0.03 | 0.03x |
| 4.5% Milk | 150 x | 0.045 | 0.045(150 x) |
| Mixture | 150 | 0.04 | 0.04(150) |
The total amount of butterfat in the mixture is the sum of the amounts of butterfat in each of the two original milks. Thus, from the "Gallons of Butterfat" column in the chart, we obtain the equation:
| 0.03x + 0.045(150 x) = 0.04(150) | |
| 0.03x + 6.75 0.045x = 6 | Distributive Property |
| -0.015x + 6.75 = 6 | Combine like terms |
| -0.015x = -0.75 | Subtract 6.75 from both sides |
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x = 50 |
Divide both sides by 0.015 |
The number of gallons of 3% butterfat milk in the mixture, x, is 50 gallons. The number of gallons of 4.5% butterfat milk in the mixture, 150 x, is 100 gallons.
Example. A winery has a vat to hold Chardonnay. An inlet pipe can fill the vat in 9 hours, while an outlet pipe can empty the vat in 12 hours. How long will it take to fill the vat if both the outlet and the inlet pipes are open?
Solution. In solving a work problem, begin by using the following
fact to express each rate of work: If a job can be done in t units
of time, then the rate of work is
job per unit of
time.
Let x = number of hours it will take to fill the vat with both pipes open
= rate of the inlet pipe
-
= rate of the outlet pipe (This rate is negative
because the outlet pipe is working
against the inlet pipe rather than with it)
| Rate | Time | Part of Job Done | |
| Inlet Pipe |
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x |
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| Outlet Pipe |
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x |
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Part of the job done Part of the job done
by one pipe + by other pipe = 1 (Whole job done)
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Multiply both sides by the LCD, 36 |
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Distributive Property |
| 4x 3x = 36 | Simplify |
| x = 36 | Combine like terms |
It would take 36 hours to fill the vat with both pipes open.