Determine whether each ordered pair is a solution to the inequality [latex]y>x+4[/latex]:

a) [latex](0,0)[/latex]

[latex]\begin{array}{lrcl} (0,0) & y & > & x+4\\ \text{Substitute } {0} \text{ for x and } {0} \text{ for y.} & {0} & \stackrel{?}{>} & {0}+4 \\ \text{Simplify.} & 0 & \ngtr & 4\\ \text{So, } (0,0) \text{ is not a solution to } y>x+4. \\ \end{array}[/latex]

Now graph the inequality [latex]y> 2x-1[/latex].

The solution and steps are given below.

The line [latex]y=2x-1[/latex] is the boundary line. On one side of the line are the points with [latex]y > 2x-1[/latex], and on the other side of the line are the points with [latex]y \lt 2x-1[/latex].

Let’s test the point [latex](0,0)[/latex] and see which inequality describes its position relative to the boundary line.

At [latex](0,0)[/latex], which inequality is true: [latex]y > 2x-1[/latex] or [latex]y \lt 2x-1[/latex]?

[latex]\begin{array}{rclcccrcl} y & > & 2x-1 & & & & y & \lt & 2x-1\\ 0 & \stackrel{?}{>} & 20-1 && & & 0 & \stackrel{?}{\lt} & 20-1\\ 0 & > & -1 & & & & 0 & \lt & -1\\ \end{array}[/latex]

Since [latex]y > 2x-1[/latex] is true, the side of the line with [latex](0,0)[/latex] is the solution. The shaded region shows the solution of the inequality [latex]y > 2x-1[/latex].

Since the boundary line is graphed with a solid line, the inequality includes the equal sign.

The graph shows the inequality [latex]y > 2x-1[/latex].

We could use any point as a test point, provided it is not on the line. Why did we choose [latex](0,0)[/latex]? Because it’s the easiest to evaluate. You may want to pick a point on the other side of the boundary line and check that [latex]y \lt 2x-1[/latex].

Graph the linear inequality: [latex]x-2y \lt 5[/latex].

First, we graph the boundary line [latex]x-2y = 5[/latex]. The inequality is [latex]\lt[/latex], so we draw a dashed line.

Then, we test a point. We’ll use [latex](0,0)[/latex] again because it is easy to evaluate and it is not on the boundary line.

Is [latex](0,0)[/latex] a solution of [latex]x-2y \lt 5[/latex]?

[latex]\begin{array}{rcl} 0-2(0) & \stackrel{?}{ \lt } & 5\\ 0-0 & \stackrel{?}{ \lt } & 5 \\ 0 & \lt & 5\\ \end{array}[/latex]

The point [latex](0,0)[/latex] is a solution of [latex]x-2y \lt 5[/latex], so we shade in that side of the boundary line.

All points in the shaded region, but not those on the boundary line, represent the solutions to [latex]x-2y \lt 5[/latex].

Graph the linear inequality: [latex]y \leq -4x[/latex].

First, we graph the boundary line [latex]y=-4x[/latex]. It is in slope–intercept form, with [latex]m=-4[/latex] and [latex]b=0[/latex]. The inequality is [latex]\leq[/latex], so we draw a solid line.

Now we need a test point. We can see that the point [latex](1,0)[/latex] is not on the boundary line.

Is [latex](1,0)[/latex] a solution of [latex]y \leq -4x[/latex]?

[latex]\begin{array}{rcl} 0 & \stackrel{?}{ \leq } & -4(1)\\ 0 & \stackrel{?}{ \nleq } & -4 \\ \end{array}[/latex]

The point [latex](1,0)[/latex] is not a solution to [latex]y \leq -4x[/latex], so we shade in the opposite side of the boundary line.

All points in the shaded region and on the boundary line represent the solutions to [latex]y \leq -4x[/latex].

Graph the linear inequality: [latex]y > 3[/latex].

First, we graph the boundary line [latex]y=3[/latex]. It is a horizontal line. The inequality is [latex]>[/latex], so we draw a dashed line.

We test the point [latex](0,0)[/latex].

So [latex](0,0)[/latex] is not a solution to [latex]y>3[/latex].

So we shade the side that does not include [latex](0,0)[/latex] as shown in this graph.

All points in the shaded region, but not those on the boundary line, represent the solutions to [latex]y>3[/latex].

Hilaria works two part-time jobs in order to earn enough money to meet her obligations of at least $240 a week. Her job in food service pays $10 an hour, and her tutoring job on campus pays $15 an hour. How many hours does Hilaria need to work at each job to earn at least $240?

a) Let [latex]x[/latex] be the number of hours she works at the job in food service, and let [latex]y[/latex] be the number of hours she works tutoring. Write an inequality that would model this situation.

We let [latex]x[/latex] be the number of hours she works at the job in food service and let [latex]y[/latex] be the number of hours she works tutoring.

She earns $10 per hour at the job in food service and $15 an hour tutoring. At each job, the number of hours multiplied by the hourly wage will give the amount earned at that job.

[latex]\begin{array}{cccc} \text{Amount she earned at } & \text{ } & \text{the amount} & \text{is at }\\ \text{the food service job} & \text{plus} & \text{earned tutoring} & \text{least} & $240\\ 10x & + & 15y & \ge & 240\\ \end{array}[/latex]

b) Graph the inequality.

To graph the inequality, we put it in slope–intercept form.

[latex]\begin{array}{rcl} 10x+15y & \geq & 240\\ 15y & \geq & -10x+240\\ y & \geq & -\frac{2}{3}x+16\\ \end{array}[/latex]

c) Find three ordered pairs [latex](x,y)[/latex] that would be solutions to the inequality. Then, explain what that means for Hilaria.

From the graph, we see that the ordered pairs [latex](15,10)[/latex], [latex](0,16)[/latex], and [latex](24,0)[/latex] represent three of infinitely many solutions. Check the values in the inequality.

[latex]\begin{array}{rclcrclcrcl} (15,10)& & & & (0,16) & & & & (24,0)\\ 10x-15y & \geq & 240 & & 10x+15y & \geq & 240 & & 10x+15y & \geq & 240\\ 10({15})-15({10}) & \geq & 240 & & 10({0})+15({16}) & \geq & 240 & & 10({24})+15({0}) & \geq & 240\\ 300 & \geq & 240 \text{ True} & & 240 & \geq & 240 \text{ True} & & 240 & \geq & 240 \text{ True}\\ \end{array}[/latex]

For Hilaria, it means that to earn at least $240, she can work 15 hours tutoring and 10 hours at her food service job, earn all her money tutoring for 16 hours, or earn all her money while working 24 hours at the job in food service.