Question

In the xy-plane, the point of intersection of the lines $x-7y=25$ and $7x+49y=31$ is given by

• A. $(0,-3)$
• B. $(4,3)$
• C. $(-3,4)$
• D. $(\frac{103}{7},\frac{-72}{49})$

Answer 1

The correct option is D $(\frac{103}{7},\frac{-72}{49})$

$x-7y=25$ ...(i)
$7x+49y=31$ ...(ii)

multiply (i) by 7 and
On applying $\left(i\right)-\left(ii\right)$, we get
$-98y=144$

$\Rightarrow y=\frac{-144}{98}=\frac{-72}{49}$

from (i); $x=25+7y=25+7\left(\frac{-72}{49}\right)$

$\Rightarrow x=\frac{103}{7}$

Hence, intersection point is $(\frac{103}{7},\frac{-72}{49})$