Question

I need to solve each system by graphing. so pls help! This is Algebra 1

Answer 1

Given the system of inequalities:

2x + 3y < -6

-2x + 3y < 6

Let's solve the system by graphing.

To graph, rewrite the inequalities in slope-intercept form:

y = mx + b

Inequality 1:'

Subtract 2x from both sides:

2x - 2x + 3y < -2x - 6

3y < -2x - 6

Divide all terms by 3:

`\begin{gathered} (3y)/(3)<-(2x)/(3)-(6)/(3) \\ \\ y<-(2)/(3)x-2 \end{gathered}`

Inequality 2:

Add 2x to both sides:

-2x + 2x + 3y < 2x + 6

3y < 2x + 6

Divde all terms by 3:

`\begin{gathered} (3y)/(3)<(2x)/(3)+(6)/(2) \\ \\ y<(2)/(3)x+2 \end{gathered}`

Now, let's plot 3 points from each inequlality and connect using a straight edge.

Inequality 1:

When x = -3

Substitute -3 for x and solve for y:

`\begin{gathered} y<-(2)/(3)(-3)-2 \\ \\ y<2-2 \\ \\ y<0 \end{gathered}`

When x = 0:

`\begin{gathered} y<-(2)/(3)(0)-2 \\ \\ y<-2 \end{gathered}`

When x = 3:

`\begin{gathered} y<-(2)/(3)(3)-2 \\ \\ y<-2-2 \\ \\ y<-4 \end{gathered}`

From inequality 1, we have the points:

(x, y) ==> (-3, 0), (0, -2), (3, -4)

For inequlity 2:

When x = -3:

`\begin{gathered} y<(2)/(3)(-3)+2 \\ \\ y<-2+2 \\ \\ y<0 \end{gathered}`

When x = 0: