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

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Answer 1 · 2017-08-24T13:14:33

See a solution process below:

Step 1) Because the first equation is already solved for $x$ we can substitute $(2y + 7)$ for $x$ in the second equation and solve for $y$ :

$3x - 2y = 3$ becomes:

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

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

$6y + 21 - 2y = 3$

$6y - 2y + 21 = 3$

$(6 - 2)y + 21 = 3$

$4y + 21 = 3$

$4y + 21 - color(red)(21) = 3 - color(red)(21)$

$4y + 0 = -18$

$4y = -18$

$(4y)/color(red)(4) = -18/color(red)(4)$

$(color(red)(cancel(color(black)(4)))y)/cancel(color(red)(4)) = -18/4$

$y = -9/2$

Step 2) Substitute $-9/2$ for $y$ in the first equation and calculate $x$ :

$x = 2y + 7$ becomes:

$x = (2 xx -9/2) + 7$

$x = (color(red)(cancel(color(black)(2))) xx -9/color(red)(cancel(color(black)(2)))) + 7$

$x = -9 + 7$

$x = -2$

The Solution Is: $x = -2$ and $y = -9/2$ or $(-2, -9/2)$

How do you solve x=2y+7x=2y+7 and 3x−2y=3 3x−2y=3 using substitution?