How do you solve $10x + 10y = 1$ and $x = y - 3$ using substitution?

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Answer 1 · 2017-02-11T12:52:46

See the entire solution process below:

Step 1) Because the second equation is already solved for $x$ we can substitute $y - 3$ for $x$ in the first equation and solve for $y$ :

$10x + 10y = 1$ becomes:

$10(y - 3) + 10y = 1$

$10y - 30 + 10y = 1$

$10y + 10y - 30 = 1$

$20y - 30 = 1$

$20y - 30 + color(red)(30) = 1 + color(red)(30)$

$20y - 0 = 31$

$20y = 31$

$(20y)/color(red)(20) = 31/color(red)(20)$

$(color(red)(cancel(color(black)(20)))y)/cancel(color(red)(20)) = 31/20$

$y = 31/20$

Step 2) Substitute $31/20$ for $y$ in the second equation and calculate $x$ :

$x = y - 3$ becomes:

$x = 31/20 - 3$

$x = 31/20 - (20/20 xx 3)$

$x = 31/20 - 60/20$

$x = -29/20$

The solution is: $x = -29/20$ and $y = 31/20$ or $(-29/20, 31/20)$

How do you solve 10x + 10y = 110x + 10y = 1 and x = y - 3 x = y - 3 using substitution?