Graph system of inequalities

Question

Graph the solution to this system of inequalities in the coordinate plane. 3y>2x+12 and 2x+y less than or equal to -5

Answer 1

To see how this graphs, I have made a Desmos graph with instructions: https://www.desmos.com/calculator/trrgzq32ow

Answer 2

Step 1: Rewrite the inequalities

Inequality 1:

3y > 2x + 12 \quad \text{→ divide all terms by 3}

y > \frac{2}{3}x + 4 ]

Inequality 2:

2x + y \leq -5 \quad \text{→ subtract 2x from both sides}

y \leq -2x - 5 ]

Step 2: Graph the boundary lines

For :

Use a dashed line because it's a strict inequality ( > )

Shade above the line

For :

Use a solid line because it's ≤

Shade below the line

Step 3: Find the solution

The solution region is where the shaded areas from both inequalities overlap. This is the area above the line and below or on the line .

Answer 3

Step 1 is to graph the equations. I like to put the equation in y=mx+b format

First Equation:

3y>2x+12 divide both sides by 3

y>2/3x+4

Start the graph at the y intercept (0,4)

For the next point, up 2, right 3 (3,6)

The line is dashed due to the inequality symbol, >

Now for the shading. use a test point (0,0) is the one I like. Plug it into the equation

y>2/3x+4

0>2/3(0)+4

0>4 false, shading does not include (0,0)

Second equation:

2x+y≤-5

Subtract 2x from both sides

y≤-2x-5

Start the graph at the y intercept (0,-5)

For the next point, down 2, right 1 (1,-7)

The line is solid due to the inequality symbol, ≤

Now for the shading. use a test point (0,0) is the one I like. Plug it into the equation

y≤-2x-5

0≤-2(0)-5

0≤-5 false, shading does not include (0,0)

The solution is the area shaded for both.