# HOW TO SOLVE WORD PROBLEMS WITH RATIONAL EQUATIONS

The following steps would be useful to solve word problems with rational equations.

Step 1 :

Assign a variable to the quantity that has to be found.

Step 2 :

Translate the verbal phrase to a rational equation.

Step 3 :

Solve for the variable

Note :

Rational equations can be solved easily by getting rid of the denominators. To get rid of the denominators one by one, multiply both sides of the equation by the expression in the denominator.

Problem 1 :

3 less than reciprocal of a number is equal to ⁻¹⁴⁄₅. Find the number.

Solution :

Let x be the number.

¹⁄ₓ - 3 = ⁻¹⁴⁄₅

Multiply both sides by x to get rid of the denominator x on the left side.

x(¹⁄ₓ - 3) = x(⁻¹⁴⁄₅)

x(¹⁄ₓ) - x(3) = ⁻¹⁴ˣ⁄₅

1 - 3x = ⁻¹⁴ˣ⁄₅

Multiply both sides by 5 to get rid of the denominator 5 on the right side.

5(1 - 3x) = 5(⁻¹⁴ˣ⁄₅)

5 - 15x = -14x

5 = x

The number is 5.

Problem 2 :

5 more than two times of reciprocal of a number is equal to ¹⁷⁄₃. Find the number.

Solution :

Let x be the number.

2(¹⁄ₓ) + 5 = ¹⁷⁄₃

²⁄ₓ + 5 = ¹⁷⁄₃

Multiply both sides by x to get rid of the denominator x on the left side.

x(²⁄ₓ + 5) = x(¹⁷⁄₃)

x(²⁄ₓ) + x(5) = ¹⁷ˣ⁄₃

2 + 5x = ¹⁷ˣ⁄₃

Multiply both sides by 3 to get rid of the denominator 5 on the right side.

3(2 + 5x) = 3(¹⁷ˣ⁄₃)

6 + 15x = 17x

Subtract 15x from both sides.

6 = 2x

Divide both sides by 2.

3 = x

The number is 3.

Problem 3 :

Peterson's car can travel 300 miles with 15 gallons of gas. If Peterson currently has 3 gallons of gas in his car, how many more gallons of gas are needed to travel 200 miles?

Solution :

Let x be the additional gallons of gas needed to cover 200 miles distance.

15 gallons : 300 miles = (3 + x) gallons : 200 miles

Write each ratio in the equation above as a fraction.

¹⁵⁄₃₀₀ = ⁽³ ⁺ ˣ⁾⁄₂₀₀

¹⁄₂₀ = ⁽ˣ ⁺ ³⁾⁄₂₀₀

Multiply both sides by 20 to get rid of the denominator 5 on the right side.

20(¹⁄₂₀) = 20(⁽ˣ ⁺ ³⁾⁄₂₀₀)

1 = ⁽ˣ ⁺ ³⁾⁄₁₀

Multiply both sides by 10 to get rid of the denominator 5 on the right side.

10(1) = 10(⁽ˣ ⁺ ³⁾⁄₁₀)

10 = x + 3

Subtract 3 from both sides.

7 = x

Peterson needs 7 more gallons of gas to cover 200 miles.

Problem 4 :

Kent can paint a certain room in 6 hours, but Kendra needs 4 hours to paint the same room. How long does it take them to paint the room if they work together?

Solution :

Givren : Kent can paint a certain room in 6 hours, but Kendra needs 4 hours to paint the same room.

Part of the room painted by Kent in 1 hour :

=

Part of the room painted by Kendra in 1 hour :

¼

Let x be the number of hours required to paint the room, if both Kent and Kendra work together.

Part of the room painted by both Kent and Kendra in 1 hour :

¹⁄ₓ

¹⁄ₓ ⅙ ¼

Multiply both sides by x to get rid of the denominator on the left side.

x(¹⁄ₓ) = x(⅙ ¼)

1 = x() + x(¼)

1 = ˣ⁄₆ + ˣ⁄₄

The least common multiple of (6, 4) = 12.

Multiply both sides by 12 to get rid of the denominators 6 and 4 on the right side.

24(1) = 24(ˣ⁄₆ + ˣ⁄₄)

24 = 24(ˣ⁄₆) + 24(ˣ⁄₄)

24 = 4x + 6x

24 = 10x

Divide both sides by 10.

2.4 = x

If both Kent and Kendra work together, they can paint the room in 2.4 hours.

Problem 5 :

The numerator of a fraction is 5 less than the denominator. If 2 be added to both numerator and denominator, the fraction becomes ½. Find the original fraction.

Solution :

Let x be the denominator of the fraction.

Then the numerator is (x - 5).

Fraction = ⁽ˣ ⁻ ⁵⁾⁄ₓ

Givren : When 2 is added to both numerator and denominator, the fraction becomes ½.

⁽ˣ ⁻ ⁵ ⁺ ²⁾⁄₍ₓ ₊ ₂₎ = ½

⁽ˣ ⁻ ³⁾⁄₍ₓ ₊ ₂₎ = ½

By cross multiplying,

2(x - 3) = 1(x + 2)

2x - 6 = x + 2

Subtract x from both sides.

x - 6 = 2

x = 8

x - 5 = 8 - 5 = 3

⁽ˣ ⁻ ⁵⁾⁄ₓ

The required fraction is .

Problem 6 :

The sum of two numbers is 10. If the sum of their reciprocals is ⁵⁄₁₂, find the numbers.

Solution :

Let x be one of the numbers.

Then the other number is (10 - x).

Given : Sum of the reciprocals of the numbers is ⁵⁄₁₂.

¹⁄ₓ + ¹⁄₍₁₀ ₋ ₓ₎ = ⁵⁄₁₂

Multiply both sides by x to get rid of the denominator x on the left side.

x[¹⁄ₓ + ¹⁄₍₁₀ ₋ ₓ₎] = x(⁵⁄₁₂)

x(¹⁄ₓ) + x[¹⁄₍₁₀ ₋ ₓ₎] = ⁵ˣ⁄₁₂

1 + ˣ⁄₍₁₀ ₋ ₓ₎ = ⁵ˣ⁄₁₂

Multiply both sides by (10 - x) to get rid of the denominator (10 - x) on the left side.

(10 - x)[1 + ˣ⁄₍₁₀ ₋ ₓ₎] = (10 - x)(⁵ˣ⁄₁₂)

(10 - x)(1) + (10 - x)[ˣ⁄₍₁₀ ₋ ₓ₎] = (10 - x)(⁵ˣ⁄₁₂)

10 - x + x = (10 - x)(⁵ˣ⁄₁₂)

10 (10 - x)(⁵ˣ⁄₁₂)

Multiply both sides by 12 to get rid of the denominator 12 on the right side.

12(10) = 12(10 - x)(⁵ˣ⁄₁₂)

120 = (10 - x)(5x)

120 = 50x - 5x2

5x+ 120 = 50x

Subtract 50x from both sides.

5x- 50x + 120 = 0

Divide both sides by 5.

x- 10x + 24 = 0

Factor and solve.

x- 4x - 6x + 24 = 0

x(x - 4) - 6(x - 4) = 0

(x - 4)(x - 6) = 0

x - 4 = 0  or x - 6 = 0

x = 4  or  x = 6

 If x = 4,10 - x = 10 - 410 - x = 6 If x = 6,10 - x = 10 - 610 - x = 4

The numbers are 4 and 6.

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