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