Question

Solve the system of equation 3x - 4y + 1 = 0 and 9x - 7y = 15 using elimination.

Answer 1

Answer:

$(x,y) = \left(\dfrac{18}{5},\dfrac{67}{15}\right)$

Step-by-step explanation:

First rearrange all equations into this format:

$ax+by=c$

that will result in:

$3x-4y=-1$

$9x-7y=15$

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)

$\[\begin{array}{r@{}l@{\quad}l@{\quad}r@{}l@{}c}3x-4y&{}=-1&\xrightarrow{\times (-3)}&-9x +12y&{}=+3\\[\jot]9x -7y&{}=15&\xrightarrow{\phantom{\times (-3)}}&9x-y&{}=15&~\smash{\raisebox{.8\normalbaselineskip}{ + }}\\\cline{4-5}&&&+5y&{}=18\\[\jot]&&&y&{}=\dfrac{18}{5}\\\end{array}\]$

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:

$3x-4y=-1$

$3x-4\left( \dfrac{18}{5}\right)=-1$

$3x=-1+4\left( \dfrac{18}{5}\right)$

$3x=-1+\dfrac{72}{5}$

$3x=\dfrac{67}{5}$

$x=\dfrac{67}{3(5)}$

$x=\dfrac{67}{15}$