How do you solve $3x+3y=15$ and $2x+3y=-5$ ?

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Answer 1 · 2015-08-03T08:35:17

The solution for the system of equations is
$color(blue)(x=20,y=-15$

$3x+color(blue)(3y)=15$ .....equation $(1)$

$2x+color(blue)(3y)=−5$ ....equation $(2)$

Solving by elimination
Subtracting equation $2$ form $1$ results in elimination of $color(blue)(3y$ :

$3x+cancelcolor(blue)(3y)=15$

$-2x-cancelcolor(blue)(3y)=+5$

$x=color(blue)(20$

Substituting $x$ in equation $1$ to find $y$
$3y=15-3x$
$3y=15-3*(20)$

$3y=15-60$

$y=-45/3$

$color(blue)(y=-15$

How do you solve 3x+3y=153x+3y=15 and 2x+3y=-52x+3y=-5?