# How do you solve algebraically x+y=2 and x-y=4?

Apr 7, 2015

If you have 2 equations you can add the left sides of the equations together and the right sides of the equations together to get a new valid equation

$\left(\begin{matrix}\null & x + y & = & 2 \\ + & x - y & = & 4 \\ \null & 2 x \pm 0 y & = & 6\end{matrix}\right)$

$2 x = 6$
$x = 3$

Substituting $x = 3$ back into either of our original allows us to solve for $y$
For example using $x + y = 2$
we get
$\left(3\right) + y = 2$

$y = - 1$

Apr 7, 2015
• A very simple way to solve this is by ADDING the two equations.

• The two equations are
$x + y = 2$ ----(1)
$x - y = 4$ ----(2)

Adding the Left Hand Sides will give us:

$\left(x + y\right) + \left(x - y\right)$
$= x + \cancel{y} + x - \cancel{y}$
$= 2 x$

Adding the Right Hand Sides will give us:
$2 + 4 = 6$

That gives us
$2 x = 6$
Dividing both sides by 2, we get
$\frac{\cancel{2} x}{\cancel{2}} = \frac{6}{2}$
$x = 3$

Substituting $x = 3$ in (1) we get
$3 + y = 2$
Subtracting 3 from both sides gives us
$\cancel{3} + y - \cancel{3} = 2 - 3$
$y = - 1$

The Solution for these two equations is $x = 3 \mathmr{and} y = - 1$

• Once we arrive at a solution, it is a good idea to VERIFY our answer

Substituting $x = 3 , y = - 1$ in (1) we get
Left Hand Side: $x + y = 3 + \left(- 1\right) = 2$(Right Hand Side)

Substituting $x = 3 , y = - 1$ in (2) we get
Left Hand Side: $x - y = 3 - \left(- 1\right) = 3 + 1 = 4$(Right Hand Side)

We have verified our answer, and we can be sure that it's correct.