SOLVING EQUATIONS WITH SPECIAL SOLUTIONS

When we solve linear equations in one variable, we may get three types of solution.

1) One solution

2) No solution

3) Infinitely many solution

Tell whether the following equation has one solution, no solution, or infinitely many solution.

Example 1 :

2(x + 4) + 6x = 12x + 8 - 3x

Solution :

2(x + 4) + 6x = 12x + 8 - 3x

Distributing 2, we get

2x + 8 + 6x = 12x + 8 - 3x

Combining the like terms, we get

8x + 8 = 9x + 8

Subtract 9x on both sides.

8x - 9x + 8 = 8

-x + 8 = 8

Subtract 8 on both sides

-x + 8 - 8 = 8 - 8

-x = 0

x = 0

It has one solution or unique solution.

Example 2 :

3 + 8x - 12 = 5x + 3(x - 4)

Solution :

3 + 8x - 12 = 5x + 3(x - 4)

Distributing 3, we get

3 + 8x - 12 = 5x + 3x - 12

Combining the constants and like terms, we get

-9 + 8x = 8x - 12

No real values will satisfy the equation above. So, it has no solution.

Example 3 :

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

Solution :

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

Distributing 5, we get

3 + 5x = 5x - 10 - 7

3 + 5x = 5x - 17

No value of x will not satisfy the equation. So, it has no solution.

Example 4 :

4(x + 3) - 4 = 8 (x/2 + 1)

Solution :

4(x + 3) - 4 = 8(x/2 + 1)

Distributing 4 and 8, we get

4x + 12 - 4 = 4x + 8

4x + 8 = 4x + 8

All real values will make the equation true. So, it has infinitely many solution.

Example 5 :

3 + 3x/2 + 4 = 4x- 5x/2

Solution :

3 + 3x/2 + 4 = 4x- 5x/2

Combining the like terms, we get

7 + 3x/2 = (8x - 5x)/2

7 + 3x/2 = 3x/2

No value of x will satisfy the equation. So, it has no solution.

Example 6 :

(3/2)(2x + 6) = 3x + 9

Solution :

(3/2)(2x + 6) = 3x + 9

Distributing 3/2, we get

3x + 18/2 = 3x + 9

3x + 9 = 3x + 9

All real values will satisfy the solution. So, it has infinitely many solution.

Example 7 :

-14 - 8x = -2(-3x + 7)

Solution :

-14 - 8x = -2(-3x + 7)

Distributing -2, we get

-14 - 8x = 6x - 14

Subtract 6x on both sides.

-8x - 6x - 14 = -14

-14x - 14 = -14

Add 14 on both sides.

-14x = -14 + 14

-14x = 0

Divide by 14 on both sides.

x = 0

Example 8 :

3x+ 7x + 1 = 2(5x + 1)

Solution :

3x+ 7x + 1 = 2(5x + 1)

Combining the like terms, we get

10x + 1 = 10x + 2

No real values of x will satisfy the solution. So, it has no solution.