Question

Solve the following pairs of linear equations by elimination method:

$47x+31y=63$ and $31x+47y=15$

• A. $x=1,y=0$
• B. $x=2,y=-1$
• C. $x=3,y=4$
• D. $x=9,y=-7$

Answer 1

The correct option is B $x=2,y=-1$

Multiply the equation $47x+31y=63$ by $31$ and equation $31x+47y=15$ by $47$ to make the coefficients of $x$

equal. Then we get the equations:

$1457x+961y=\mathrm{1953.........}\left(1\right)$

$1457x+2209y=\mathrm{705.........}\left(2\right)$

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

$(1457x-1457x)+(2209y-961y)=32-27$

i.e. $1248y=-1248$

i.e. $y=-1$

Substituting this value of $y$ in the equation $47x+31y=63$, we get

$47x-31=63$

i.e. $47x=94$

i.e. $x=2$

Hence, the solution of the equations is $x=2,y=-1$.