How do you solve the system $x = 5 - y$ and $3y = 3x + 1$ by substitution?

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Answer 1 · 2015-05-28T19:27:31

We can substitute the $x$ of the first equation into the second:

$->3y=3*(5-y)+1$
$->3y=15-3y+1$
$->6y=16->y=16/6=8/3$

Now plug that into the first equation:
$x=5-8/3 =15/3-8/3=7/3$

So $x=7/3 and y=8/3$

Check your answer with the second equation:

$3* 8/3 = 3* 7/3 +1 ?-> 8=7+1$ Check!

How do you solve the system x = 5 - yx = 5 - y and 3y = 3x + 13y = 3x + 1 by substitution?