Question

Solve the following simultaneous equations by the method of equating coefficients.

$\frac{2x-y}{3}+3x-2y=3;4x-3y+\frac{5x-4y}{4}=1$

• A. $x=4,y=5$
• B. $x=3,y=5$
• C. $x=7,y=5$
• D. $x=9,y=5$

Answer 1

The correct option is A $x=4,y=5$

$\frac{2x-y}{3}+3x-2y=3$ ....(i)

$4x-3y+\frac{5x-4y}{4}=1$ .....(ii)

Multiplying equation (i) by $3$ and equation (ii) by $4$. we get,

$2x-y+9x-6y=9\Rightarrow 11x-7y=9$ .......(iii)

$16x-12y+5x-4y=4\Rightarrow 21x-16y=4$ ....(iv)

Multiplying equation (iii) by $16$ and equation (iv) by $7$ . we get,

$176x-112y=144$

$\underset{–}{\underset{-}{1}47x\underset{+}{-}112y=\underset{-}{2}8}$
$29x=116\Rightarrow x=4$

Substitute $x=4$ in equation (iii). we get,

$11\left(4\right)-7y=9\Rightarrow 7y=35\Rightarrow y=5$

$\therefore x=4,y=5$