# 9 < y – 5x < 15 graph this inequalitie

vixen999 | High School Teacher | (Level 2) Adjunct Educator

Posted on

We can split this compound inequality into two separate inequalities:

9 < y - 5x       and      y - 5x <15

Let's look at the one on the left first.  We want to move it into slope intercept form, but with an inequality and not an equation.

9 < y - 5x  which we can rewrite as y - 5x > 9

Now, add 5x to both sides and get y > 5x + 9.  Now we have it in the format that we want.  We want to graph the line that would be represented by the graph y = 5x + 9.  The y-intercept is at (0,9) and the slope would be 5/1.  We would start at (0.9) and go over 1 to the right and up 5 to get to (1,14).  (I wish I could attach a picture for you.) Because the graph does not have a "bar" under the greater than sign, we use a dotted line for our line.  (If we had a greater than or equal to bar, then we would use a solid line because points on the line would represent solutions.)  Points on this line do not satisfy the inequality.  Next, we test a point on one side of the line.  I always test (0,0) if I can (the only reason I would not be able to is if it were on the line).  We will plug 0 in for X and 0 in for Y.  If our answer is true, we will color the side with the point (0,0).  If it is not true, we would color the other side in. 0 > 5(0) + 9.  Thus, we get 0 > 9.  This is not true, so we color in the other side of the line.  In this case, the entire part of the graph on the left side of the line is colored in.

We have now finished, half of our graph.  We need to follow the same procedure with the other inequality.  We can rewrite y - 5x <15 as y <5x + 15.  We can then graph the equation y = 5x + 15. We start at our y-intercept (0, 15).  Then, we will use the slope 5/1 to go over 1 and up 5 to (1, 20).   Once again, since we have just a less than sign and not a less than or equal to sign, we use a dotted line.  We can use (0,0) to test again.  0 < 5(0) + 15 which gives us 0 < 15 which is true.  Thus, we will color in the part of the graph to the right of this line.  Our solution is the part of the graph that has been colored in twice.  Basically, since we have two lines with the same slope, they are parallel and in our case, the solution is the space between them.

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

9 < y – 5x < 15

9<y-5x

9+5x<y-5x+5x

9+5x<y

So we have one inequality that 9+5x<y.

y-5x<15

y-5x+5x<15+5x

y<15+5x

Now we have another inequality y<15+5x

Lets graph the function y = 9+5x and y= 15+5x

The black graph shows y=9+5x graph and red one shows y=15+5x.

What we want is the area greater than 9+5x and less than 15+5x.

This happens between the range of x=-2 and x=-3. But x is not equal to either -2 or -3.

So the answer is `x in (-3,-2)`. This is the range of x that satisfy;

9 < y – 5x < 15

neela | High School Teacher | (Level 3) Valedictorian

Posted on

9 < y-5x < 15...(1) is the given inequality.

From  the left inequality, 9 < y-5x , we get  y > 5x+9...(2)

From the right inequality,y-5x < 15, we get y < 5x+15....(3)

So for all (x,y) coordinates of the points between the || lines at (2) and (3) satisfy the given inequality.

Take (x,y) = (5,37). We substitute in (1) and we get:

9 < 37-5*5 < 15

=> 9 < 12 < 15. So (x,y) = (5,37) satisfies 9 < y-5x < 15.

Therefore all the (x,y) pairs of coordinates of any point between the || lines y = 5x+9 and the line y = 5x+15 satisfy the inequality.