How do you solve the system $2 x + 2 y = 7$ $2x+2y=7$ and $x - 2 y = - 1$ $x-2y=-1$ using substitution?

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Answer 1 · 2016-11-09T23:03:24

$y = \frac{3}{2}$ $y = 3/2$ and $x = 2$ $x = 2$

Step 1) Solve the second equation for $x$ $x$ while keeping the equation balanced:

$x - 2 y + 2 y = - 1 + 2 y$ $x - 2y + 2y = -1 + 2y$

$x = 2 y - 1$ $x = 2y - 1$

Step 2) Substitute $2 y - 1$ $2y - 1$ for $x$ $x$ in the first equation and solve for $y$ $y$ while keeping the equation balanced.

$2 (2 y - 1) + 2 y = 7$ $2(2y - 1) + 2y = 7$

$4 y - 2 + 2 y = 7$ $4y - 2 + 2y = 7$

$6 y - 2 = 7$ $6y - 2 = 7$

$6 y - 2 + 2 = 7 + 2$ $6y - 2 + 2 = 7 + 2$

$6 y = 9$ $6y = 9$

$\frac{6 y}{6} = \frac{9}{6}$ $(6y)/6 = 9/6$

$y = \frac{3}{2}$ $y = 3/2$

Step 3) Substitute $\frac{3}{2}$ $3/2$ for $y$ $y$ in the second equation and solve for $x$ $x$ while keeping the equation balanced:

$x = 2 (\frac{3}{2}) - 1$ $x = 2(3/2) - 1$

$x = 3 - 1$ $x = 3 - 1$

$x = 2$ $x = 2$

How do you solve the system 2x+2y=72x+2y=7 and x−2y=−1x-2y=-1 using substitution?