How do you solve 1/2(x-y) = 2 and 1/2(x+y)+1 = 0?

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How do you solve 1/2(x-y) = 2 and 1/2(x+y)+1 = 0?

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Answer 1 · 2018-04-14T21:08:19

$x = 1$ $x=1$ and $y = - 3$ $y=-3$

Explanation:

Solve like simultaneous equations.

Equation 1: $\frac{1}{2} (x - y) = 2$ $1/2(x-y) = 2$
Expand the brackets to get $\frac{1}{2} x - \frac{1}{2} y = 2$ $1/2x-1/2y=2$

Equation 2: $\frac{1}{2} (x + y) + 1 = 0$ $1/2(x+y)+1 = 0$
Expand the brackets to get $\frac{1}{2} x + \frac{1}{2} y + 1 = 0$ $1/2x+1/2y+1=0$

$\frac{1}{2} x - \frac{1}{2} y = 2$ $1/2x-1/2y=2$

$\frac{1}{2} x + \frac{1}{2} y + 1 = 0$ $1/2x+1/2y+1=0$

Add the two equations together to get
$\frac{1}{2} x + \frac{1}{2} x + \frac{1}{2} y - \frac{1}{2} y + 1 = 2$ $1/2x+1/2x+1/2y-1/2y+1=2$

$x + 1 = 2$ $x+1=2$
$x = 1$ $x=1$

Substitute this value of $x$ $x$ into either Equation 1 or 2 and solve for $y$ $y$

Equation 2: $\frac{1}{2} (1) + \frac{1}{2} y + 1 = 0$ $1/2(1)+1/2y+1=0$
$\frac{1}{2} + \frac{1}{2} y + 1 = 0$ $1/2+1/2y+1=0$
$\frac{1}{2} y + 1 = - \frac{1}{2}$ $1/2y+1=-1/2$
$\frac{1}{2} y = - 1$ $1/2y=-1$ $\frac{1}{2}$ $1/2$
$y = - 3$ $y = -3$

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How do you solve 1/2(x-y) = 2 and 1/2(x+y)+1 = 0?

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