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.

Example 9 :

0.125x = 0.025(5x + 1)

Solution :

0.125x = 0.025(5x + 1)

0.125x = 0.025(5x) + 0.025(1)

0.125x = 0.125x + 0.025

On both sides, we have same quantities of x, subtracting 0.125x on both sides.

0.125x - 0.125x = 0.025

0x = 0.025

So, there is no solution.

Example 10 :

0.5(3q + 87) = 1.5q + 43.5

Solution :

0.5(3q + 87) = 1.5q + 43.5

0.5(3q) + 0.5(87) = 1.5q + 43.5

1.5q + 43.5 = 1.5q + 43.5

1.5q - 1.5q = 43.5 - 43.5

0q = 0

All real numbers are solutions.

Example 11 :

On dividing $200 between A and B such that twice of A’s share is less than 3 times B’s share by 200, B’s share is?

Solution :

Let x be A's share

B's share = 200 - x

2x = 3(200 - x) - 200

2x = 600 - 3x - 200

2x + 3x = -200 + 600

5x = 400

x = 400/5

x = 20

A's share = $20

B's share = 200 - 20 ==> $180

Example 12 :

Madhulika thought of a number, doubled it and added 20 to it. On dividing the resulting number by 25, she gets 4. What is the number

Solution :

Let x be the number.

(2x + 20)/25 = 4

Multiplying by 25 on both sides, we get

2x + 20 = 4(25)

2x + 20 = 100

2x = 100 - 20

2x = 80

x = 80/2

x = 40

So, the required number is 40.