How do you solve this system of equations: $x + 2 y = 3 and 3 x + 5 y = 15$ $x +2y = 3 and 3x +5y=15$ ?

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Answer 1 · 2017-09-30T15:40:40

$y = - 6$ $y=-6$
$x = 15$ $x=15$

We will start by multiplying the first equation by 3 on both sides.

$3 (x + 2 y) = 3 (3)$ $3(x+2y)=3(3)$
$3 x + 6 y = 9$ $3x+6y=9$

Now that we have $3 x$ $3x$ in both equations we can subtract the equations form each other to eliminate the $x$ $x$ 's.

$3 x + 6 y = 9$ $color(red)(3x+6y=9)$
$-$ $-$
$3 x + 5 y = 15$ $color(blue)(3x+5y=15)$

$3 x - 3 x + 6 y - 5 y = 9 - 15$ $color(red)(3x)color(blue)(-3x)color(red)(+6y)color(blue)(-5y)=color(red)9color(blue)(-15)$
$y = - 6$ $y=-6$

We can substitute this back into the one of the original equations to find $x$ $x$ .

$x + 2 (- 6) = 3$ $x+2(-6)=3$
$x - 12 = 3$ $x-12=3$
$x = 15$ $x=15$

How do you solve this system of equations: x+2y=3and3x+5y=15 x +2y = 3 and 3x +5y=15 ?