How do you solve the system $2 x + 5 y = 38$ $2x+5y=38$ and $x - 3 y = - 3$ $x-3y=-3$ using substitution?

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Answer 1 · 2017-08-23T17:29:27

$x = 9$ $x=9$

$y = 4$ $y=4$

First, let's take the second equation and rearrange it to make $x$ $x$ the subject:

$x - 3 y = - 3 \to x = 3 y - 3$ $x-3y=-3 -> x=3y-3$

Insert the rearranged equation in the first equation, in place of $x$ $x$ :

$2 (3 y - 3) + 5 y = 38$ $2(3y-3)+5y=38$

Now calculate $y$ $y$ :

$6 y - 6 + 5 y = 38$ $6y-6+5y=38$

$11 y = 44$ $11y=44$

$y = 4$ $y=4$

Now substitute the value of $y$ $y$ into the rearranged equation to find $x$ $x$ :

$x = 3 (4) - 3$ $x=3(4)-3$

$x = 12 - 3 \to x = 9$ $x=12-3 -> x=9$

How do you solve the system 2x+5y=382x+5y=38 and x−3y=−3x-3y=-3 using substitution?