# 4.7 Graphs of linear inequalities  (Page 2/10)

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Similarly, for an inequality in two variables, the boundary line is shown with a solid or dashed line to indicate whether or not it the line is included in the solution. This is summarized in [link]

 $Ax+By $Ax+By\le C$ $Ax+By>C$ $Ax+By\ge C$ Boundary line is not included in solution. Boundary line is included in solution. Boundary line is dashed. Boundary line is solid.

Now, let’s take a look at what we found in [link] . We’ll start by graphing the line $y=x+4$ , and then we’ll plot the five points we tested. See [link] .

In [link] we found that some of the points were solutions to the inequality $y>x+4$ and some were not.

Which of the points we plotted are solutions to the inequality $y>x+4$ ? The points $\left(1,6\right)$ and $\left(-8,12\right)$ are solutions to the inequality $y>x+4$ . Notice that they are both on the same side of the boundary line $y=x+4$ .

The two points $\left(0,0\right)$ and $\left(-5,-15\right)$ are on the other side of the boundary line     $y=x+4$ , and they are not solutions to the inequality $y>x+4$ . For those two points, $y .

What about the point $\left(2,6\right)$ ? Because $6=2+4$ , the point is a solution to the equation $y=x+4$ . So the point $\left(2,6\right)$ is on the boundary line.

Let’s take another point on the left side of the boundary line and test whether or not it is a solution to the inequality $y>x+4$ . The point $\left(0,10\right)$ clearly looks to be to the left of the boundary line, doesn’t it? Is it a solution to the inequality?

$\begin{array}{cccc}y>x+4\hfill & & & \\ 10\stackrel{?}{>}0+4\hfill & & & \\ 10>4\hfill & & & \text{So},\phantom{\rule{0.2em}{0ex}}\left(0,10\right)\phantom{\rule{0.2em}{0ex}}\text{is a solution to}\phantom{\rule{0.2em}{0ex}}y>x+4.\hfill \end{array}$

Any point you choose on the left side of the boundary line is a solution to the inequality $y>x+4$ . All points on the left are solutions.

Similarly, all points on the right side of the boundary line, the side with $\left(0,0\right)$ and $\left(-5,-15\right)$ , are not solutions to $y>x+4$ . See [link] .

The graph of the inequality $y>x+4$ is shown in [link] below. The line $y=x+4$ divides the plane into two regions. The shaded side shows the solutions to the inequality $y>x+4$ .

The points on the boundary line, those where $y=x+4$ , are not solutions to the inequality $y>x+4$ , so the line itself is not part of the solution. We show that by making the line dashed, not solid.

The boundary line shown is $y=2x-1$ . Write the inequality shown by the graph.

## Solution

The line $y=2x-1$ is the boundary line. On one side of the line are the points with $y>2x-1$ and on the other side of the line are the points with $y<2x-1$ .

Let’s test the point $\left(0,0\right)$ and see which inequality describes its side of the boundary line.

At $\left(0,0\right)$ , which inequality is true:

$\begin{array}{ccccc}\hfill y>2x-1\hfill & & \hfill \text{or}\hfill & & \hfill y<2x-1?\hfill \\ \hfill y>2x-1\hfill & & & & \hfill y<2x-1\hfill \\ \hfill 0\stackrel{?}{>}2·0-1\hfill & & & & \hfill 0\stackrel{?}{<}2·0-1\hfill \\ \hfill 0>-1\phantom{\rule{0.2em}{0ex}}\text{True}\hfill & & & & \hfill 0<-1\phantom{\rule{0.2em}{0ex}}\text{False}\hfill \end{array}$

Since, $y>2x-1$ is true, the side of the line with $\left(0,0\right)$ , is the solution. The shaded region shows the solution of the inequality $y>2x-1$ .

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

The graph shows the inequality $y\ge 2x-1$ .

We could use any point as a test point, provided it is not on the line. Why did we choose $\left(0,0\right)$ ? 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 $y<2x-1$ .

Write the inequality shown by the graph with the boundary line $y=-2x+3$ .

$y\ge -2x+3$

Write the inequality shown by the graph with the boundary line $y=\frac{1}{2}x-4$ .

$y<\frac{1}{2}x-4$

4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
At 1:30 Marlon left his house to go to the beach, a distance of 5.625 miles. He rose his skateboard until 2:15, and then walked the rest of the way. He arrived at the beach at 3:00. Marlon's speed on his skateboard is 1.5 times his walking speed. Find his speed when skateboarding and when walking.
divide 3x⁴-4x³-3x-1 by x-3
how to multiply the monomial
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike? Got questions? Get instant answers now!
how do u solve that question
Seera
Two sisters like to compete on their bike rides. Tamara can go 4 mph faster than her sister, Samantha. If it takes Samantha 1 hours longer than Tamara to go 80 miles, how fast can Samantha ride her bike?
Seera
Speed=distance ÷ time
Tremayne
x-3y =1; 3x-2y+4=0 graph
Brandon has a cup of quarters and dimes with a total of 5.55\$. The number of quarters is five less than three times the number of dimes
app is wrong how can 350 be divisible by 3.
June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?
Susanna if the first cooler holds five times the gallons then the other cooler. The big cooler holda 40 gallons and the 2nd will hold 8 gallons is that correct?
Georgie
@Susanna that person is correct if you divide 40 by 8 you can see it's 5 it's simple
Ashley
@Geogie my bad that was meant for u
Ashley
Hi everyone, I'm glad to be connected with you all. from France.
I'm getting "math processing error" on math problems. Anyone know why?
Can you all help me I don't get any of this
4^×=9
Did anyone else have trouble getting in quiz link for linear inequalities?
operation of trinomial