How do you solve the following system of equations? $3 x + 4 y = 15, 3 x + 2 y = 12$ $3x +4y = 15, 3x + 2y = 12$

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Answer 1 · 2018-03-14T02:35:05

$(3, \frac{3}{2})$ $(3,3/2)$

Because we have a $3 x$ $3x$ in both equations, we can solve this system using the elimination method.

$3 x + 4 y = 15$ $3x+4y=15$
$3 x + 2 y = 12$ $3x+2y=12$

You can subtract the bottom equation from the top one to get:

$2 y = 3$ $2y=3$
$y = \frac{3}{2}$ $y=3/2$

To solve for $x$ $x$ , you can plug $y$ $y$ into one of the equations. In this case, the bottom equation would be easier to use.

$3 x + 2 (\frac{3}{2}) = 12$ $3x+2(3/2)=12$
$3 x + 3 = 12$ $3x+3=12$
$3 x = 9$ $3x=9$
$x = 3$ $x=3$

$x = 3$ $x=3$ and $y = \frac{3}{2}$ $y=3/2$ , so the solution to this system is...
$(3, \frac{3}{2})$ $(3,3/2)$

How do you solve the following system of equations? 3x +4y = 15, 3x + 2y = 123x+4y=15,3x+2y=123x +4y = 15, 3x + 2y = 12