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

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

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Answer 1 · 2015-05-28T19:27:31

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

$\to 3 y = 3 \cdot (5 - y) + 1$ $->3y=3*(5-y)+1$
$\to 3 y = 15 - 3 y + 1$ $->3y=15-3y+1$
$\to 6 y = 16 \to y = \frac{16}{6} = \frac{8}{3}$ $->6y=16->y=16/6=8/3$

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

So $x = \frac{7}{3} and y = \frac{8}{3}$ $x=7/3 and y=8/3$

Check your answer with the second equation:

$3 \cdot \frac{8}{3} = 3 \cdot \frac{7}{3} + 1 ? \to 8 = 7 + 1$ $3* 8/3 = 3* 7/3 +1 ?-> 8=7+1$ Check!

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

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