How do you solve $7y - 2x = 10$ and $-3y + x = -3$ using substitution?

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Answer 1 · 2017-10-17T12:16:55

See a solution process below:

Step 1: Solve the second equation for $x$ :

$-3y + x = -3$

$color(red)(3y) - 3y + x = color(red)(3y) - 3$

$0 + x = 3y - 3$

$x = 3y - 3$

Step 2: Substitute $(3y - 3)$ for $x$ in the first equation and solve for $y$ :

$7y - 2x = 10$ becomes:

$7y - 2(3y - 3) = 10$

$7y - (2 * 3y) - (2 * -3) = 10$

$7y - 6y - (-6) = 10$

$7y - 6y + 6 = 10$

$(7 - 6)y + 6 = 10$

$1y + 6 = 10$

$y + 6 - color(red)(6) = 10 - color(red)(6)$

$y + 0 = 4$

$y = 4$

Step 3: Substitute $4$ for $y$ in the solution to the second equation at the end of Step 1 and calculate $x$ :

$x = 3y - 3$ becomes:

$x = (3 * 4) - 3$

$x = 12 - 3$

$x = 9$

The Solution Is: $x = 9$ and $y = 4$ or $(9, 4)$

How do you solve 7y - 2x = 107y - 2x = 10 and -3y + x = -3-3y + x = -3 using substitution?