How do you solve $\frac{1}{2} x + 2 y = 3, 6 y + 1 = x$ $1/2x+2y=3, 6y+1=x$ by substitution?

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How do you solve $\frac{1}{2} x + 2 y = 3, 6 y + 1 = x$ $1/2x+2y=3, 6y+1=x$ by substitution?

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Answer 1 · 2016-12-09T23:08:52

I found:
$x = 4$ $x=4$
$y = \frac{1}{2}$ $y=1/2$

Explanation:

Take the $x$ $x$ expression from the second equation and substitute into the first:
$\frac{1}{2} \cdot (6 y + 1) + 2 y = 3$ $1/2*(color(red)(6y+1))+2y=3$
rearrange:
$6 y + 1 + 4 y = 6$ $6y+1+4y=6$
$10 y = 5$ $10y=5$
$y = \frac{5}{10} = \frac{1}{2}$ $y=5/10=1/2$
go back to the second equation and substitute this value of $y$ $y$ :
$x = 6 \cdot \frac{1}{2} + 1 = 3 + 1 = 4$ $x=6*color(blue)(1/2)+1=3+1=4$

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How do you solve $\frac{1}{2} x + 2 y = 3, 6 y + 1 = x$ $1/2x+2y=3, 6y+1=x$ by substitution?

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How do you solve $\frac{1}{2} x + 2 y = 3, 6 y + 1 = x$ $1/2x+2y=3, 6y+1=x$ by substitution?