HOW TO SOLVE WORD PROBLEMS WITH RATIONAL EQUATIONS

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

Add 15x to both sides.

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 :

= ⅙

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(⅙) + 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

Add 6 to both sides.

x = 8

x - 5 = 8 - 5 = 3

 ⁽ˣ ⁻ ⁵⁾⁄ₓ = ⅜

The required fraction is ⅜.

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

Add 5x² to both sides.

5x² + 120 = 50x

Subtract 50x from both sides.

5x² - 50x + 120 = 0

Divide both sides by 5.

x² - 10x + 24 = 0

Factor and solve.

x² - 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

The numbers are 4 and 6.

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