Solve: 6 x + 1 3 + 1 = x - 3 6

Given, 6 x + 1 3 + 1 = x - 3 6 Multiplying both sides of the equation by 6 , 6 6 x + 1 3 + 6 × 1 = 6 x - 3 6 2 6 x + 1 + 6 = x - 3 12 x + 2 + 6 = x - 3 (opening the brackets) 12 x + 8 = x - 3 12 x - x = - 3 - 8 (Transposing x to L.H.S and 8 to the R.H.S) or 11 x = - 11 or x = - 1 (required solution) Check: L.H.S = 6 - 1 + 1 3 + 1 = - 6 + 1 3 + 1 = - 5 3 + 3 3 = - 5 + 3 3 = - 2 3 R.H.S = - 1 - 3 6 = - 4 6 = - 2 3 L.H.S = R.H.S. (as required)

Solve: 5 x - 2 2 x - 7 = 2 3 x - 1 + 7 2

Given, 5 x - 2 2 x - 7 = 2 3 x - 1 + 7 2 Let us open the brackets, L.H.S = 5 x - 4 x + 14 = x + 14 - - - - - - ( 1 ) R.H.S = 6 x - 2 + 7 2 = 6 x - 4 2 + 7 2 = 6 x + 3 2 - - - - - - - ( 2 ) From equation 1 and 2 , we get x + 14 = 6 x + 3 2 14 = 6 x - x + 3 2 14 = 5 x + 3 2 14 - 3 2 = 5 x (transposing 3 2 ). 28 - 3 2 = 5 x 25 2 = 5 x x = 25 2 × 1 5 = 5 2 Therefore, the required solution is x = 5 2 . Check: L.H.S = 5 × 5 2 - 2 5 2 × 2 - 7 = 25 2 - 2 5 - 7 = 25 2 - 2 - 2 = 25 2 + 4 = 25 + 8 2 = 33 2 R.H.S = 2 5 2 × 3 - 1 + 7 2 = 2 15 2 - 2 2 + 7 2 = 2 × 13 2 + 7 2 = 26 + 7 2 = 33 2 = L.H.S. (as required)

Solve: x + 1 2 x + 3 = 3 8

Observe that the equation is not a linear equation, since the expression on its LHS is not linear. But we can put it into the form of a linear equation. We multiply both sides of the equation by 2 x + 3 . x + 1 2 x + 3 × 2 x + 3 = 3 8 × 2 x + 3 Notice that 2 x + 3 gets cancelled on the L.H.S We have then, x + 1 = 3 2 x + 3 8 We have now a linear equation which we know how to solve. Multiplying both sides by 8 8 x + 1 = 3 2 x + 3 or 8 x + 8 = 6 x + 9 or 8 x = 6 x + 9 - 8 or 8 x = 6 x + 1 or 8 x - 6 x = 1 or 2 x = 1 or x = 1 2 Therefore, the solution is x = 1 2 . Check: Numerator of L.H.S = 1 2 + 1 = 1 + 2 2 = 3 2 Denominator of L.H.S = 2 x + 3 = 2 × 1 2 + 3 = 1 + 3 = 4 L.H.S = numerator ÷ denominator = 3 2 ÷ 4 = 3 2 × 1 4 = 3 8 L.H.S = R.H.S.