Find Solutions to Algebraic Inequalities

I divide both sides by 5 to get x=6.

First I subtract 3 from both sides to get [ggfrac]x/2[/ggfrac]=2, and then I multiply both sides by 2 to get x=4. I use inverse operations to solve for the unknown variable.

I can use substitution. Once I find the value of x, I can substitute the value for x in the equation and see if both sides are equal. [ggfrac]4/2[/ggfrac]+3=5, so I know that 2 is the correct value for x.

Less than and greater than. 6 < 8 and 8 > 6.

Students draw a number line with an open point at 5 and the arrow traveling right infinitely. There is not only one possible value for x. Any value that is greater than 5 makes this inequality true. x can be 6, 7, 9.6, 100, or many other values.

Only 14 and 17. x must be greater than 13, so 13 and all smaller numbers are not part of the solution.

15, 16, and 17. Since x must be less than or equal to 17, 17 itself is part of the solution set.

To solve, I treat the > like an equal sign and use inverse operations to solve. Since x is multiplied by 2 and then 3 is added to it, I subtract 3 on both sides and then divide both sides by 2 to get x>2.

Since x is divided by 3 and then 4 is subtracted from it, I first add 4 to both sides, and then multiply both side by 3, to get x≤15.

I can check by substitution. Substituting 6 for x, I get [ggfrac]6/3[/ggfrac]-4≤1, which simplifies to -2≤1. Since -2 is less than or equal to 1, I know that 6 satisfies the inequality.