# How do you solve x+y=3, y=2x-15 by graphing?

$\left(6 , - 3\right)$. See below for the graph.

#### Explanation:

We can graph them like this:

• $x + y = 3$ can be graphed by seeing that when $x = 0 , y = 3$ and when $x = 3 , y = 0$. Plot those two points and draw a line through them.

• y=2x-15 can be graphed by first plotting the y-intercept ($0 , - 15$) and then using the slope of 2 to move up two points and to the right 1 ($1 , - 13$), plotting those two points and drawing a line through them.

Together, they look like this:

graph{(y+x-3)(y-2x+15)=0 [-3.22, 14.56, -8.36, 0.53]}

And this gives us the point $\left(6 , - 3\right)$

Feb 11, 2017

$x = 6$ and $y = - 3$. See below for details.

#### Explanation:

To solve two linear equations in two variables, what you need is to draw their graph and as they represent two lines, look for their point of intersection, which gives the solution of the equations.

To draw the graph of lines, select three pair of solutions (as it is a linear equation two should satisfy, but we have mentioned three to confirm we are on right track for graph), which when represented on a Cartesian graph, will give the graph of the line. It would be preferable, if these points are wide apart.

For example, for first equation $x + y = 3$, we can have $\left(- 5 , 8\right)$, $\left(0 , 3\right)$ and $\left(5 , - 2\right)$. These when joined make the following graph.
graph{x+y=3 [-9.67, 10.33, -3.36, 6.64]}

For second equation, we can again choose $x = 0 , 5$ and $10$, which when put in equation $y = 2 x - 15$, gives $y = - 15 , - 5$ and $5$ and points as $\left(0 , - 15\right)$, $\left(5 , - 5\right)$ and $\left(10 , 5\right)$ and graph is as shown below.
graph{2x-15 [-36.55, 43.45, -20.32, 19.68]}

When the two graphs are drawn on the same graph, lines intersect at $\left(6 , - 3\right)$, which is the solution i.e. $x = 6$ and $y = - 3$.
graph{(x+y-3)(y-2x+15)=0 [-5.78, 14.22, -5.64, 4.36]}