Algebra and Pre-Algebra Lessons

Algebra 1 | Pre-Algebra | Practice Tests | Algebra Readiness Test

Algebra E-Course and Homework Information

Algebra E-course Info | Log In to Algebra E-course | Homework Calculator

Formulas and Cheat Sheets

Formulas | Algebra Cheat Sheets

Rewriting Inequalities

by Liz

Your site has really helped!
I just have a little question...

What happens when the inequality is like follows: x>11 + y? How do we rewrite the equation so that y is in front?

Karin from Algebra Class Says:

If you want to rewrite, x > 11 + y in slope intercept form, you must move the terms around using opposite operations. Take a look...

Step 1: Move 11 to the left hand side by subtracting 11 from both sides.

x - 11 > 11 - 11 + y


x - 11 > y

Now you have y by itself on the right hand side, but you might be most comfortable with the y on the left hand side. If this is the case, you can flip sides of the inequality, but you must also reverse the inequality symbol.

y < x - 11

Now the inequality is in slope intercept form.

I hope this helps,

Comments for Rewriting Inequalities

Average Rating starstarstarstarstar

Click here to add your own comments

Sep 11, 2011
Rewriting Inequalities
by: Karin

You have two inequality questions here:

The first refers to a bag of marbles where there are twice as many blue marbles as green marbles. So, let's start by defining our variables:
Let x = the number of green marbles
Let 2x = the number of blue marbles since there are twice as many.

Now the problems says that there are at least 33 in the bag. At least means greater than or equal to because the least you could have is 33, but you could have more. So, add the marbles in the bag and is must be greater than or equal to 33:

x + 2x >(or equal to) 33

Now solve:
3x > (or equal) 33

3x/3 >33/3
x >(or equal) 11

Therefore, there are atleast 11 green marbles in the bag.

Your second question states that one side of the rectangle is x and the other is 14 inches.
We know the perimeter of a rectangle is P= 2l +2w
Let the length = x in.
Let the width = 14 in.

Now substitute: 2x +2(14)= P

Since the perimeter is at most 80 inches, we will use the less than or equal to sign.

2x+28 <(or equal) 80

Now solve:
2x+28-28 < or = 80-28
2x < or = 52
2x/2 < or = 52/2
x < or = 26

The most x can be is 26, but it may also be less than 26.

Sorry about the greater than or less than symbol. In this application, it's not possible to type that symbol.

Hope this helps,

Sep 11, 2011
by: arlene345@hotmail.com

How could I resolve these 2 inequalities?

Use an inequality to solve the problem. Be sure to show the inequality and all of your work.
A bag of marbles has twice as many blue marbles as green marbles, and the bag has at least 33 marbles in it. At least how many green marbles does it have?

Use an inequality to solve the problem. Be sure to show the inequality and all of your work. One side of a rectangle is 14 inches and the other is x inches. What values of x will make the perimeter at most 80 inches?

Click here to add your own comments

Join in and write your own page! It's easy to do. How? Simply click here to return to Students.

Like This Page?

Top of the Page

Connect and Follow Algebra Class

Copyright © 2009-2015 Karin Hutchinson ALL RIGHTS RESERVED