Question

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

Answer 1

`(x,y) = \left((18)/(5),(67)/(15)\right)`

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)

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

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( (18)/(5)\right)=-1`

`3x=-1+4\left( (18)/(5)\right)`

`3x=-1+(72)/(5)`

`3x=(67)/(5)`

`x=(67)/(3(5))`

`x=(67)/(15)`