SOLVING EQUATIONS BY ISOLATING THE VARIABLE

When we want to solve an equation for a variable like x, we have to isolate the variable x whose value we want to find. To isolate a variable in an equation, we have to get rid of all the values around it. To do so, we have to undo the mathematical operations that are being performed on the variable we want to isolate.

To undo the mathematical operations, we have to perform their the inverse operations.

Addition and subtraction are inverse operations :

To undo addition, we have to perform its inverse operation subtraction.

For example. consider the equation 'x + 3 = 0'.

In the equation 'x + 3 = 0', 3 is added to the variable x. To isolate the variable x in the equation, we have to undo the addition of 3. To do so, we have to perform its inverse operation. That is, we have to subtract 3 from both sides of the equation.

x + 3 = 0

x + 3 - 3 = 0 - 3

x + 0 = -3

x = -3

In the similar manner, we can undo subtraction by performing its inverse operation addition.

y - 7 = 0

y - 7 + 7 = 0 + 7

y - 0 = 7

y = 7

Multiplication and division are inverse operations :

To undo multiplication, we have to perform its inverse operation division.

For example. consider the equation '7x = 14'.

In the equation '7x = 14', x is multiplied by 7. To isolate the variable x in the equation, we have to undo the multiplication of 7. To do so, we have to perform its inverse operation. That is, we have to divide  both sides of the equation by 7.

⁷ˣ⁄₇¹⁴⁄

x = 2

In the similar manner, we can undo division by performing its inverse operation multiplication.

ˣ⁄₅ = 2

5(ˣ⁄₅) = 5(2)

x = 10

Exponent and radical are inverse operations :

To undo exponent, we have to perform its inverse operation root.

For example. consider the equation 'x2 = 4'.

In the equation 'x2 = 4', x is having the exponent 2. To isolate the variable x in the equation, we have to undo the exponent 2. To do so, we have to perform its inverse operation. That is, we have to take square root on both sides of the equation.

x2 = 4
x24

x = ±2

In the similar manner, we can undo radical by performing its inverse operation exponent. on both sides of the equation.

x = 3

Take exponent 2 on both sides.

(√x)2 = 32

x = 9

Example 1 :

Solve for x :

3x + 4 = 31

Solution :

3x + 4 = 31

Subtract 4 from both sides.

3x + 4 - 4 = 31 - 4

3x = 27

Divide both sides by 3.

³ˣ⁄₃²⁷⁄₃

x = 9

Example 2 :

Solve for x :

ˣ⁄₅ - 1 = 2

Solution :

ˣ⁄₅ - 1 = 2

Add 1 to both sides.

ˣ⁄₅ - 1 + 1 = 2 + 1

ˣ⁄₅ = 3

Multiply both sides by 3.

5 ⋅ ˣ⁄₅ = 3

x = 15

Example 3 :

Solve for x :

²ˣ⁄₃ = ⁵⁄₂

Solution :

²ˣ⁄₃ = ⁵⁄₂

Multiply both sides by ³⁄₂.

³⁄₂  ²ˣ⁄₃ = ⁵⁄₂ ⋅ ³⁄₂

⁶ˣ⁄₆ ¹⁵⁄₄

x = ¹⁵⁄₄

Example 4 :

If ⁽ˣ ⁻ ³⁾⁄₅ = k and k = 3, what is the value of x?

Solution :

⁽ˣ ⁻ ³⁾⁄₅ = k

Substitute k = 3.

⁽ˣ ⁻ ³⁾⁄₅ = 3

Multiply both sides by 3.

 x - 3 = 15

Add 3 to both sides.

 x = 18

Example 5 :

If 5x + 6 = 7, what is the value of 10x + 3?

Solution :

5x + 6 = 7

Subtract 6 from both sides.

5x = 1

Multiply both sides by 2.

10x = 2

Add 3 to both sides.

10x + 3 = 5

Example 6 :

Solve for y :

5(y + 1) + 4(y - 1) = 3y

Solution :

5(y + 1) + 4(y - 1) = 3y

Using Distributive Property,

5y + 5 + 4y - 4 = 3y

Combine the like terms.

9y + 1 = 3y

Subtract 3y from both sides.

6y + 1 = 0

Subtract 1 from both sides.

6y = -1

Divide both sides by 6.

y = ⁻¹⁄₆

Example 7 :

Subtracting 12 from 5 times of a number results -2. Find the number.

Solution :

Let x be the number.

Given : Subtracting 12 from 5 times of a number results -2.

5x - 12 = -2

Add 12 to both sides.

5x = 10

Divide both sides by 5.

x = 2

The number is 2.

Example 8 :

When 10 times the number x is added to 22, the result is 2. What number results when 5 times x is added to 4?

Solution :

Given : 10 times the number x is added to 22 results is 2.

10x + 22 = 2

Subtract 22 from both sides.

10x = -20

Divide both sides by 2.

5x = -10

Add 4 to both sides.

5x + 4 = -6

Example 9 :

If 3a + 2b + 7 = 7(b + 1), what is the value of ᵃ⁄b?

Solution :

3a + 2b + 7 = 7(b + 1)

Using Distributive Property,

3a + 2b + 7 = 7b + 7

Subtract 7 from both sides.

3a + 2b = 7b

Subtract 2b from both sides.

3a = 5b

Divide both sides by b.

³ᵃ⁄b = 5

Multiply both sides by .

ᵃ⁄b = ⁵⁄₃

Example 10 :

Twice the age of a man 7 years ago is equal to 8 more than his age 5 years hence. Find his present age.

Solution :

Let x be the present of the man.

Age of the man 7 years ago = x - 8

Age of the man 5 years hence = x + 5 

Given : Twice the age of a man 7 years ago is equal to 8 more than his age 5 years hence.

2(x - 7) = (x + 5) + 8

Using Distributive Property,

2x - 14 = x + 5 + 8

2x - 14 = x + 13

Subtract x from both sides.

x - 14 = 13

Add 14 to both sides.

x = 27

The present age of the man is 27 years.

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