The problem asks to solve the following system of linear equations by graphing and to find the intersection point: $5x - 3y = 15$ $2x + y = -5$

1. Problem Description

The problem asks to solve the following system of linear equations by graphing and to find the intersection point:

$5x - 3y = 15$

$2x + y = -5$

2. Solution Steps

First, let's rewrite each equation in slope-intercept form ($y = mx + b$):

For the first equation, $5x - 3y = 15$, we want to solve for $y$.

Subtract $5x$ from both sides: $-3y = -5x + 15$

Divide both sides by $-3$: $y = \frac{-5x + 15}{-3} = \frac{5}{3}x - 5$

So the slope-intercept form of the first equation is $y = \frac{5}{3}x - 5$.

For the second equation, $2x + y = -5$, we want to solve for $y$.

Subtract $2x$ from both sides: $y = -2x - 5$

So the slope-intercept form of the second equation is $y = -2x - 5$.

Now we have two equations:

$y = \frac{5}{3}x - 5$

$y = -2x - 5$

To find the intersection point (the solution to the system), we set the two equations equal to each other:

$\frac{5}{3}x - 5 = -2x - 5$

Add 5 to both sides: $\frac{5}{3}x = -2x$

Add $2x$ to both sides: $\frac{5}{3}x + 2x = 0$

Convert 2x to have a denominator of 3: $2x = \frac{6}{3}x$

Then, $\frac{5}{3}x + \frac{6}{3}x = 0$

$\frac{11}{3}x = 0$

Multiply both sides by $\frac{3}{11}$: $x = 0$

Now, substitute $x = 0$ into either equation to find $y$. Let's use the second equation:

$y = -2(0) - 5$

$y = -5$

Therefore, the solution is $(0, -5)$.

3. Final Answer

(0, -5)