SOLVING EQUATIONS WITH VARIABLE IN DENOMINATOR

In an equation, if there is a variable like x orin denominator of a fraction, you have to eliminate the fraction by multiplying both sides of the equation by the denominator (just as you do when numbers are in the denominators).

For example, if you want to solve the equation ⁵⁄ₓ = 40, multiply both sides of the equation by x to get 5 = 40x, and then divide both sides of the equation by 40 to solve for x :

x = ⁵⁄₄₀ = 

Note that sometimes, having a variable in denominator can lead to extraneous solutions. Extraneous solution is the value of the variable which makes the denominator zero and the fraction becomes undefined. Such solution always has to be excluded.

Examples 1-10 : Solve for x.

Example 1 :

¹⁄ₓ + ²⁄₍ₓ ₋ ₂₎ = 0

Solution :

¹⁄ₓ + ²⁄₍ₓ ₋ ₂₎ = 0

In the above equation, there are  two denominators,

x and (x - 2)

To get rid of the first denominator x, multiply both sides of the equation by x.

x[¹⁄ₓ + ²⁄₍ₓ ₋ ₂₎] = x(0)

x(¹⁄ₓ) + x[²⁄₍ₓ ₋ ₂₎] = 0

1 + ²ˣ⁄₍ₓ ₋ ₂₎ = 0

Multiply both sides of the equation by (x - 2) to get rid of the denominator (x - 2).

(x - 2)[1 + ²ˣ⁄₍ₓ ₋ ₂₎)] = (x - 2)(0)

(x - 2)(1) + (x - 2)[²ˣ⁄₍ₓ ₋ ₂₎] = 0

x - 2 + 2x = 0

3x - 2 = 0

Add 2 to both sides.

3x = 2

Divide both sides by 3.

x = ⅔.

Example 2 :

ˣ⁄₄ = ³⁄₂

Solution :

ˣ⁄₄ = ³⁄₂

In the equation above, there are two denominators 4 and 2.

Least common multiple of (4, 2) = 4.

Multiply both sides of the equation by 4 to get rid of the denominators.

4(ˣ⁄₄) = 4(³⁄₂)

x = 2(3)

x = 6

Example 3 :

³⁄₂ₓ -  = ²⁄₃ₓ

Solution :

³⁄₂ₓ -  = ²⁄₃ₓ

In the equation above, there are three different denominators 2x, 6 and 3x.

Least common multiple of (2, 6, 3) = 6.

Since x is multiplied in two of the denominators, multiply both sides of the equation by 6x to get rid of all the denominators.

6x(³⁄₂ₓ - ) = 6x(²⁄₃ₓ)

6x(³⁄₂ₓ) + 6x(-) = 2(2)

3(3) + x(-5) = 4

9 - 5x = 4

Subtract 9 from both sides.

-5x = -5

Divide both sides by -5.

x = 1

Example 4 :

¹⁄₃ₓ + ¼ = ⁴⁄ₓ

Solution :

¹⁄₃ₓ + ¼ = ⁴⁄ₓ

In the equation above, there are three different denominators 3x, 4 and x.

Least common multiple of (3, 4) = 12.

Multiply both sides of the equation by 12x to get rid of all the denominators.

12x(¹⁄₃ₓ + ¼) = 12x(⁴⁄ₓ)

12x(¹⁄₃ₓ) + 12x(¼) = 12(4)

4(1) + 3x(1) = 48

4 + 3x = 48

Subtract 4 from both sides.

3x = 44

Divide both sides by 3.

x = ⁴⁴⁄₃

Example 5 :

¹⁄₂ₓ + ¹⁄₂ = ²⁄ₓ

Solution :

¹⁄₂ₓ + ¹⁄₂ = ²⁄ₓ

Multiply both sides of the equation by 2x to get rid of all the denominators.

2x(¹⁄₂ₓ + ¹⁄₂) = 2x(²⁄ₓ)

2x(¹⁄₂ₓ) + 2x(¹⁄₂) = 2(2)

1(1) + x(1) = 4

1 + x = 4

Subtract 1 from both sides.

x = 3

Example 6 :

⁽⁷ ⁻ ˣ⁾⁄₍₅ ₋ ₓ₎ = ³⁄₂

Solution :

⁽⁷ ⁻ ˣ⁾⁄₍₅ ₋ ₓ₎ = ³⁄₂

That is, numerator on the left side to be multiplied by denominator on the right side and numerator on the right side has to be multiplied by denominator on the left side.

2(7 - x) = 3(5 - x)

Using Distributive Property,

14 - 2x = 15 - 3x

Add 3x to both sides.

14 + x = 15

Subtract 15 from both sides.

x = 1

Example 7 :

½ = ⁽³ ⁻ ²ˣ⁾⁄₍₃ₓ ₋ ₈₎

Solution :

½ = ⁽³ ⁻ ²ˣ⁾⁄₍₃ₓ ₋ ₈₎

By corss multiplying,

1(3x - 8) = 2(3 - 2x)

Using Distributive Property,

3x - 8 = 6 - 4x

Add 4x to both sides.

7x - 8 = 6

Add 8 to both sides.

7x = 14

Divide both sides by 7.

x = 2

Example 8 :

⁽³ˣ ⁻ ⁶⁾⁄₍ₓ ₋ ₄₎ = 2

Solution :

⁽³ˣ ⁻ ⁶⁾⁄₍ₓ ₋ ₄₎ = 2

On the right of the equation, write 2 as a fraction by taking denominator 1.

⁽³ˣ ⁻ ⁶⁾⁄₍ₓ ₋ ₄₎ = ²⁄₁

By corss multiplying,

1(3x - 6) = 2(x - 4)

Using Distributive Property,

3x - 6 = 2x - 8

Subtract 2x from both sides.

x - 6 = -8

Add 6 to both sides.

x = -2

Example 9 :

⁽ˣ ⁻ ²⁾⁄₍₃ₓ ₋ ₁₂₎ = ⁻¹⁄₃

Solution :

⁽ˣ ⁻ ²⁾⁄₍₃ₓ ₋ ₁₂₎ = ⁻¹⁄₃

By corss multiplying,

3(x - 2) = -1(3x - 12)

Using Distributive Property,

3x - 6 = -3x + 12

Add 3x to both sides.

6x - 6 = 12

Add 6 to both sides.

6x = 18

Divide both sides by 6.

x = 3

Example 10 :

⁽ˣ ⁻ ⁵⁾⁄₍₂ₓ ₋ ₃₎ = ⁻²⁄₃

Solution :

⁽ˣ ⁻ ⁵⁾⁄₍₂ₓ ₋ ₃₎ = ⁻²⁄₃

By corss multiplying,

3(x - 5) = -2(2x - 3)

Using Distributive Property,

3x - 15 = -4x + 6

Add 4x to both sides.

7x - 15 = 6

Add 15 to both sides.

7x = 21

Divide both sides by 7.

x = 3

Example 11 :

3a/b = -1

Find the value of b when a = 7.

Solution :

3a/b = -1

Multiply both sides by b to get rid of the denominator b.

3a = -b

Multiply both sides by -1.

-3a = b

Substitute a = 7.

-3(7) = b

-21 = b

Example 12 :

Subtracting 2 from 5 times the reciprocal of a number results ½. Find the number.

Solution :

Let x be the number.

It is given that subtracting 2 from 5 times the reciprocal of the number results 1/2.

5(¹⁄ₓ) - 2 = ½

⁵⁄ₓ - 2 = ½

Multiply both sides of the equation by 2x to get rid of the denominators x and 2.

2x(⁵⁄ₓ - 2) = 2x(½)

2x(⁵⁄ₓ) - 2x(2) = x

2(5) - 4x = x

10 - 4x = x

Add 4x to both sides.

10 = 5x

Divide both sides by 5.

2 = x

The number is 2.

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