# What are the points of intersection for y = 2x + 3 and y = x + 5?

Feb 23, 2016

Suppose we separated the variables into ${x}_{1}$, ${x}_{2}$, ${y}_{1}$, and ${y}_{2}$ labels, as a general case for if neither intersected the other.

$\setminus m a t h b f \left({y}_{1} = 2 {x}_{1} + 3\right)$

$\setminus m a t h b f \left({y}_{2} = {x}_{2} + 5\right)$

The point of intersection occurs when the two graphs have equal values of $x$ and $y$ at the same time. There is only one solution, because two straight lines can only intersect once. (On the other hand, two curved lines may intersect twice.)

The solution will be the coordinate (x,y) such that ${y}_{1} = {y}_{2}$ and ${x}_{1} = {x}_{2}$.

What we can do to proceed is assume that ${y}_{1} = {y}_{2}$ and ${x}_{1} = {x}_{2}$. Therefore, we get:

$2 {x}_{1} + 3 = {x}_{2} + 5$

$= {x}_{1} + 5$

Subtract ${x}_{1}$ from both sides to get:

${x}_{1} + 3 = 5$

Then I would subtract $3$ from both sides to get:

$\textcolor{b l u e}{{x}_{1} = {x}_{2} = 2}$

Now, since the solution coordinate requires that we have both $x$ and $y$, we need to solve for $y$.

$\textcolor{b l u e}{{y}_{1}} = 2 {x}_{1} + 3$

$= 2 \left(2\right) + 3 = \textcolor{b l u e}{7}$

And just to show that indeed ${y}_{1} = {y}_{2}$ if ${x}_{1} = {x}_{2}$:

$\textcolor{g r e e n}{{y}_{2}} = {x}_{2} + 5$

$= {x}_{1} + 5$

$= 2 + 5$

$= \textcolor{g r e e n}{7 = {y}_{1}}$

Finally, that means our solution coordinate is:

$\textcolor{b l u e}{\text{("2,7")}}$