How do you solve 3r-5s=-35 and 2r-5s=-30?

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Answer 1 · 2015-09-05T07:19:51

The solution is:
$color(blue)(r=-5, s=4$

$3r-color(blue)(5s)=-35$ ....equation $1$
$2r-color(blue)(5s)=-30$ ......equation $2$

We can solve the system of equations by elimination as subtracting the two would eliminate $color(blue)(5s$

Subtracting equation $2$ from $1$
$3r-cancelcolor(blue)(5s)=-35$
$-2r+cancelcolor(blue)(5s)=+30$

$color(blue)(r=-5$

Finding $s$ by substituting the value of $r$ in equation $1$
$3r-5s=-35$

$3r+ 35=5s$

$3 * (color(blue)(-5))+ 35=5s$

$-15+ 35=5s$

$20 = 5s$

$20/5 = s$

$color(blue)(s=4$