PLEASE SOLVE

-9 LESS THAN OR EQUAL TO Y + 5X IS LESS THAN OR EQUAL TO 14

What are you solving for...y or x?

-9 ≤ y+5x ≤ 14

is this the correct set up of the problem

IT SAYS GRAPH THE INEQUALITY

okay first you will need to put the equation in y=mx+b form

HOW

You can separate the equation into two different ones

-9 ≤ y+5x
AND
y+5x ≤ 14

Then solve for y in both equations

To solve the inequality -9 ≤ y + 5x ≤ 14, you need to break it down into two separate inequalities and solve each one individually.

Starting with the left side of the inequality:

-9 ≤ y + 5x

To isolate y, we need to move 5x to the other side of the inequality:

-9 - 5x ≤ y

Now let's work on the right side of the inequality:

y + 5x ≤ 14

To isolate y, we need to move 5x to the other side of the inequality:

y ≤ 14 - 5x

So, we have the two inequalities:

-9 - 5x ≤ y
y ≤ 14 - 5x

Now, we can graph these inequalities on a coordinate plane to find the solution. However, since it is not specified whether the variables x and y are integers or real numbers, I will provide a general visual representation.

For the first inequality, -9 - 5x ≤ y, we can graph a line with a slope of -5 passing through the point (0, -9). This line would be a boundary for the solutions. To determine which side of the line to shade, we can choose a test point, such as (0, 0), and substitute it into the inequality:

-9 - 5(0) ≤ 0
-9 ≤ 0

Since -9 is less than 0, the shaded region will be below the line.

For the second inequality, y ≤ 14 - 5x, we can graph a line with a slope of -5 passing through the point (0, 14). Again, we can choose a test point, such as (0, 0), and substitute it into the inequality:

0 ≤ 14 - 5(0)
0 ≤ 14

Since 0 is less than 14, the shaded region will be above the line.

Now, we can find the overlapping region between the two shaded regions, which represents the solution to the original inequality -9 ≤ y + 5x ≤ 14.

Keep in mind that the specific values of x and y in this overlapping region will depend on the scale and units used for the graph.