Answer :
Answer:
[tex](x,y) = \left(\dfrac{18}{5},\dfrac{67}{15}\right)[/tex]
Step-by-step explanation:
First rearrange all equations into this format:
[tex]ax+by=c[/tex]
that will result in:
[tex]3x-4y=-1[/tex]
[tex]9x-7y=15[/tex]
then we'll multiply an equation with a number such that we have same numbers on both equations (with only different signs). So we can add them
Here you can see that we've multiplied the first equation with (-3) so that we have -9x on one equation and 9x on the other. Now when we add the two equations the 9x and -9x terms will cancel out (or eliminate)
[tex]\[\begin{array}{r@{}l@{\quad}l@{\quad}r@{}l@{}c}3x-4y&{}=-1&\xrightarrow{\times (-3)}&-9x +12y&{}=+3\\[\jot]9x -7y&{}=15&\xrightarrow{\phantom{\times (-3)}}&9x-y&{}=15&~\smash{\raisebox{.8\normalbaselineskip}{$+$}}\\\cline{4-5}&&&+5y&{}=18\\[\jot]&&&y&{}=\dfrac{18}{5}\\\end{array}\]\\[/tex]
now that we've found the value of y
we'll use this value in any of the two equations to get the value of x.
let's put y in the first equation:
[tex]3x-4y=-1[/tex]
[tex]3x-4\left( \dfrac{18}{5}\right)=-1[/tex]
[tex]3x=-1+4\left( \dfrac{18}{5}\right)[/tex]
[tex]3x=-1+\dfrac{72}{5}[/tex]
[tex]3x=\dfrac{67}{5}[/tex]
[tex]x=\dfrac{67}{3(5)}[/tex]
[tex]x=\dfrac{67}{15}[/tex]