Question 150418
Okay,
{{{5x+3y=-13}}} -------------> eqn 1
{{{7x-2y=19}}} --------------> eqn 2
In eqn 1 we get,
{{{y=(-13-5x)/3}}} -----------> eqn 3
Substitute eqn 3 in eqn 2,
{{{7x-2((-13-5x)/3)=19}}}
{{{7x+(26+10x)/3=19}}}
{{{(21x+26+10x)/3=19}}} ---------> {{{31x+26=19*3}}}
{{{31x=57-26}}} -----------> {{{cross(31)x/cross(31)=cross(31)/cross(31)}}}
{{{x=1}}}
.
For "y" go back eqn 3, {{{y=(-13-5*1)/3}}}
{{{y=-6}}}
See graph below for your reference.
Thank you,
Jojo
{{{graph(400,300,-10,10,10,-10,"x",1,"y",-6)}}}