Question

Solve the following pairs of linear (simultaneous) equation by the method of elimination:$6x=7y+7$, $7y-x=8$

• A. $x=3$, $y=\frac{11}{7}$
• B. $x=7$, $y=\frac{2}{3}$
• C. $x=\frac{1}{3}$ , $y=5$
• D. $x=\frac{2}{5}$ , $y=4$

Answer 1

The correct option is A $x=3$, $y=\frac{11}{7}$

The equation $6x=7y+7$ can be rewritten as $6x-7y=7$ and the equation $7y-x=8$ can be rewritten as $-x+7y=8$. Then we get the following equations:

$6x-7y=\mathrm{7.........}\left(1\right)$

$-x+7y=\mathrm{8.........}\left(2\right)$

Add equations 1 and 2 to eliminate $y$, because the coefficients of $y$ are the opposite. So, we get

$(6x-x)+(-7y+7y)=7+8$

i.e. $5x=15$

i.e. $x=3$

Substituting this value of $x$ in the equation $6x-7y=7$, we get

$18-7y=7$

i.e. $-7y=-11$

i.e. $y=\frac{11}{7}$

Hence, the solution of the equations is $x=3,y=\frac{11}{7}$.