Linear Equation

1. Solve: (2x + 5)/(x + 4) = 1

Solution:

(2x + 5)/(x + 4) = 1

⇒ 2x + 5 = 1(x + 4)

⇒ 2x + 5 = x + 4

⇒ 2x – x = 4 – 5   (Transferring positive x to the left hand side changes to negative x and again, positive 5 changes to negative 5)

⇒ x = -1

Therefore, x = – 1 is the required solution of the equation (2x + 5)/(x + 4) = 1

2. Solve: 6x – 19 = 3x – 10

Solution:

6x – 19 = 3x – 10

⇒ 6x – 3x = – 10 + 19   (Transferring 3x to L.H.S changes to negative 3x and -19 to R.H.S. changes to positive 19)

⇒ 3x = 9

⇒ 3x/3 = 9/3   (Dividing both sides by 3)

⇒ x = 3

3. Solve: 5 – 2(x – 1) = 4(3 – x) – 2x.

Solution:

5 – 2(x – 1) = 4(3 – x) – 2x

⇒ 5 – 2x + 2 = 12 – 4x – 2x   (Removing the brackets and then simplify)

⇒ 7 – 2x = 12 – 6x   (Transferring -6x to L.H.S. changes to positive 6x and 7 to R.H.S. changes to negative 7)

⇒ -2x + 6x = 12 – 7

⇒ 4x = 5

⇒ 4x/4 = 5/4

⇒ x = 5/4

4. x/2 + x/3 = x – 7

Solution:

x/2 + x/3 = x – 7

Least common multiple of2 and 3 is 6

⇒ (x/2 × 3/3) + (x/3 × 2/2) = x – 7

⇒ 3x/6 + 2x/6 = x – 7

⇒ (3x + 2x)/6 = x – 7

⇒ 5x/6 = x – 7

⇒ 5x = 6(x – 7)

⇒ 5x = 6x – 42   (Transferring 6x to L.H.S. changes to negative 6x)

⇒ 5x – 6x = -42

⇒ -x = -42

⇒ x = 42

5. 2/(x + 3) = 3/(5 – x)

Solution:

2/(x + 3) = 3/(5 – x)

⇒ 3(x + 3) = 2(5 – x)   (cross multiply and then remove the brackets)

⇒ 3x + 9 = 10 – 2x   (Transferring -2x to L.H.S. changes to positive 2x and 9 to R.H.S. changes to -9)

⇒ 3x + 2x = 10 – 9

⇒ 5x = 1

⇒ 5x/5 = 1/5   (Dividing both sides by 5)

⇒ x = 1/5