Question

Solve the following pair of linear (simultaneous) equations by the method of elimination:

$x+8y=19$, $2x+11y=28$

• A. $x=3$ and $y=2$
• B. $x=12$ and $y=5$
• C. $x=1$ and $y=6$
• D. $x=9$ and $y=-6$

Answer 1

The correct option is A $x=3$ and $y=2$

Multiply the equation $x+8y=19$ by $2$ to make the coefficients of $x$ equal. Then we get the equations:

$2x+16y=\mathrm{38.........}\left(1\right)$

$2x+11y=\mathrm{28.........}\left(2\right)$

Subtract Equation (2) from Equation (1) to eliminate $x$, because the coefficients of $x$ are the same. So, we get

$(2x-2x)+(16y-11y)=38-28$

i.e. $5y=10$

i.e. $y=2$

Substituting this value of $y$ in the equation $x+8y=19$, we get

$x+16=19$

i.e. $x=19-16=3$

Hence, the solution of the equations is $x=3,y=2$.