Question

Solve the equation.

Answer 1

TO SOLVE

$y - \frac{2y + 6}{3} = \frac{1}{4} (y - \frac{11 + y}{3} ) - 2$

• STEP 01 ( Simplification both the sides separately)

$y - \frac{2y + 6}{3} = \frac{1}{4} (y - \frac{11 + y}{3} ) - 2 \\ \\ \\ \implies \frac{3y - (2y + 6)}{3} = \frac{1}{4} ( \frac{3y - (11 + y)}{3} ) - 2 \\ \\ \\ \implies \frac{3y - 2y - 6}{3} = \frac{1}{4} ( \frac{3y - 11 - y}{3} ) - 2 \\ \\ \\ \implies \frac{y - 6}{3} = \frac{1}{4} ( \frac{2y - 11}{3} ) - 2$

Here we take LCM in both LHS and RHS and simplify the terms carefully.

• STEP 02 (Dealing with complex Brackets)

$\implies \frac{y - 6}{3} = \frac{ - 4y + 22}{12} \\ \\ \\ \implies \frac{y - 6}{1} = \frac{22 - 4y}{4}$

We have to deal with multiple brackets carefully in the RHS and in the denominator we cancel out 3 easily from LHS and RHS.

• STEP- 03 (Forming linear equation and solving it by transposing method)

$\implies y - 6 = \frac{22 - 4y}{4} \\ \\ \\ \implies 4(y - 6) = 22 - 4y \\ \\ \\ \implies 4y - 24 = 22 - 4y \\ \\ \\ \implies 4y + 4y = 22 + 24 \\ \\ \\ \implies 8y = 46\\ \\ \\ \implies y = \frac{46}{8} \\ \\ \\ \implies \boxed{y = \frac{23}{4} }$