Solving Rational Equations
At times the equation will resemble a proportion. When this is the case, solve the equation by cross multiplication.
Example 1: Cross Multiplication
5(x - 1) = 10(x + 1) Cross multiply.
5x - 5 = 10x + 10 Use the distributive property.
-5x = 15Subtract 10x and add 5 on each side.
x = -3 Divide both sides by -5.
5(x - 1) = 10(x + 1) Cross multiply.
5x - 5 = 10x + 10 Use the distributive property.
-5x = 15Subtract 10x and add 5 on each side.
x = -3 Divide both sides by -5.
When one side of the equation has addition or subtraction with ratios, solve by multiplying every term by the least common denominator (LCD).
Example 2: Multiply by the LCD
The LCD is 4x.
Multiply both sides by the LCD.
32 + 14x = 16x Use the distibutive property.
32 = 2xSubtract both sides by 14x.
16 = xDivide both sides by 2.
The LCD is 4x.
Multiply both sides by the LCD.
32 + 14x = 16x Use the distibutive property.
32 = 2xSubtract both sides by 14x.
16 = xDivide both sides by 2.
Often, when multiplying the LCD, there will be two solutions.
Example 3: More than One Solution
The LCD is (x + 3)(x - 1).
Multiply by the LCD.
(x - 1)9 + (x - 3)(x + 3) = 4Eliminate common terms in the denominators.
9x - 9 + x2 - 9 = 4 Multiply the binomials.
x2 + 9x - 22 = 0 Combine like terms and subtract 4 on both sides.
(x - 2)(x + 11) = 0 Factor.
x = 2, or x = -11 Solve.
The LCD is (x + 3)(x - 1).
Multiply by the LCD.
(x - 1)9 + (x - 3)(x + 3) = 4Eliminate common terms in the denominators.
9x - 9 + x2 - 9 = 4 Multiply the binomials.
x2 + 9x - 22 = 0 Combine like terms and subtract 4 on both sides.
(x - 2)(x + 11) = 0 Factor.
x = 2, or x = -11 Solve.
There are times that solutions are extraneous. When this happens, one of the solutions will return a denominator that is equal to zero.
Example 4: Extraneous Solutions
The LCD is (x - 5)(x + 5).
Multiply by the LCD.
(x + 5)6 = 8x2 - 4x(x - 5)Eliminate common terms in the denominators.
6x + 30 = 8x2 - 4x2 + 20xUse the distributive property.
4x2 + 14x - 30 = 0 Move all terms to the left side.
2(2x - 3)(x + 5) = 0 Factor.
, or x = -5 x = -5 can not be a solution, since the equation is undefined at this coordinate.
The LCD is (x - 5)(x + 5).
Multiply by the LCD.
(x + 5)6 = 8x2 - 4x(x - 5)Eliminate common terms in the denominators.
6x + 30 = 8x2 - 4x2 + 20xUse the distributive property.
4x2 + 14x - 30 = 0 Move all terms to the left side.
2(2x - 3)(x + 5) = 0 Factor.
, or x = -5 x = -5 can not be a solution, since the equation is undefined at this coordinate.
Example 5: Batting Average
In the first 4 months of baseball season, Tommy hit 20 out of 85 at bats. How many consecutive hits does he need to make to have a batting average of 0.350?
Solution: The batting average is the number of hits divided by the number of at bats.
Tommy's current average is 20/85 = 0.235
For Tommy's average to increase to 0.350, solve the equation
Solve by cross multiplication.
Set up the average.
0.35(85 + x) = 20 + x Cross multiply.
29.75 + 0.35x = 20 + xUse the distribute property.
9.75 = 0.65x subtract 20 and 0.35x on both sides
15 = xDivide by 0.65
Tommy needs to hit 15 consecutive bats to raise his batting average to 0.350
In the first 4 months of baseball season, Tommy hit 20 out of 85 at bats. How many consecutive hits does he need to make to have a batting average of 0.350?
Solution: The batting average is the number of hits divided by the number of at bats.
Tommy's current average is 20/85 = 0.235
For Tommy's average to increase to 0.350, solve the equation
Solve by cross multiplication.
Set up the average.
0.35(85 + x) = 20 + x Cross multiply.
29.75 + 0.35x = 20 + xUse the distribute property.
9.75 = 0.65x subtract 20 and 0.35x on both sides
15 = xDivide by 0.65
Tommy needs to hit 15 consecutive bats to raise his batting average to 0.350
Example 6: Working Together
Sue and Ann work together to clean houses. It takes Sue 7 hours to clean one house. With the help of Ann, it only takes them 3 hours to clean. How long does it take Ann to clean the same house by herself?
The best way to answer this scenario, is to add the amount of work being done per hour.
Sue cleans 1/7th of the house per hour.
Ann cleans 1/A of the house per hour.
Sue and Ann Clean 1/3 of a house in one hour.
The LCD is 21A.
Multiply by the LCD.
3A + 21 = 7AUse the distributive property.
21 = 4A Subtract 3A on both sides
5.25 = A
It takes 5 hours and 15 minutes for Ann to clean the house.
Sue and Ann work together to clean houses. It takes Sue 7 hours to clean one house. With the help of Ann, it only takes them 3 hours to clean. How long does it take Ann to clean the same house by herself?
The best way to answer this scenario, is to add the amount of work being done per hour.
Sue cleans 1/7th of the house per hour.
Ann cleans 1/A of the house per hour.
Sue and Ann Clean 1/3 of a house in one hour.
The LCD is 21A.
Multiply by the LCD.
3A + 21 = 7AUse the distributive property.
21 = 4A Subtract 3A on both sides
5.25 = A
It takes 5 hours and 15 minutes for Ann to clean the house.
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Related Links: Math algebra Multiplying Rational Equations Simplifying Rational Expressions Algebra Topics |
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