# How do you solve by substitution  x+-y=1 and -2x+y=4?

Sep 8, 2015

The problem has two solutions:
$x = - 1 , y = 2$.
$x = - 5 , y = - 6$.

#### Explanation:

To solve by substitution, we isolate one of the variables in one equation, and use that to solve the other equation.

$x \pm y = 1$ shows two cases: $x + y = 1$ or $x - y = 1$.

Case 1:
Our system is:

$x + y = 1$
$- 2 x + y = 4$

Since the first equation is easier to work with, we can isolate a variable there.
$x + y = 1$
$x = 1 - y$

Now, we substitute $1 - y$ into $x$ in the other equation, and simplify.

$- 2 x + y = 4$
$- 2 \left(1 - y\right) + y = 4$
$- 2 + 2 y + y = 4$
$3 y = 6$
$y = 2$

We now know the value of $y$, which we can use to solve for $x$ using either equation:

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

Therefore, the solution to Case 1 is $x = - 1 , y = 2$.

Case 2:
Our system is:

$x - y = 1$
$- 2 x + y = 4$

Like before, we isolate one of the variables in one equation and solve using the other equation.

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

$- 2 x + y = 4$
$- 2 \left(1 + y\right) + y = 4$
$- 2 - 2 y + y = 4$
$- y = 6$
$y = - 6$

At this point, we know the $y = - 6$, so we can solve for $x$ using either equation:

$x - y = 1$
$x + 6 = 1$
$x = - 5$

Therefore the solution to case 2 is $x = - 5 , y = - 6$.

The problem has two solutions:
$x = - 1 , y = 2$.
$x = - 5 , y = - 6$.