How do I use elimination to find the solution of the system of equations $4 x + 3 y = 7$ $4x+3y=7$ and $3 x + 5 y = 8$ $3x+5y=8$ ?

Question

4669 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-19T07:30:13

$4 x + 3 y = 7$ $4x+3y = 7$ ..........equation $(1)$ $(1)$
$3 x + 5 y = 8$ $3x +5y =8$ ..........equation $(2)$ $(2)$

multiplying equation 1 by 3 and equation 2 by 4 :

$(4 x + 3 y = 7) \times 3$ $(4x+3y = 7) xx 3$
$12 x + 9 y = 21$ $12x + 9y = 21$ ........equation $(3)$ $(3)$

and:

$(3 x + 5 y = 8) \times 4$ $(3x +5y =8) xx 4$
$12 x + 20 y = 32$ $12x +20y = 32$ .......equation $(4)$ $(4)$

on subtracting equation $4$ $4$ from $3$ $3$ ,
$12 x$ $12x$ gets cancelled

$cancel12x + 9y = 21$
$-cancel12x -20y = -32$

$-11y= -11$
$y =1$

substituting this value of $y$ in equation 1:

$4x+3(1) = 7$
$4x = 7-3$
$4x = 4$
$x = 1$

the solutions for the system of equations are :
$color(blue)x=1$
$color(blue)y=1$

How do I use elimination to find the solution of the system of equations 4x+3y=74x+3y=74x+3y=7 and 3x+5y=83x+5y=83x+5y=8?