Just as we could manipulate equations to produce equivalent equations, so too can we manipulate inequalities to produce equivalent inequalities.
Consider the inequality $9<15$9<15. If we add or subtract any number on both sides, say $3$3, we can see that the resulting inequality remains true. More specifically we can write $9+3<15+3$9+3<15+3 and $93<153$9−3<15−3.
Adding $3$3 to $9$9 and $15$15. 
Subtracting $3$3 from $9$9 and $15$15. 
Addition property of inequality: Adding the same number to each side of an inequality produces an equivalent inequality.
Example:
If 
$x2$x−2  $<$<  $7$7 
Then 
$x2+2$x−2+2  $<$<  $7+2$7+2 
Subtraction property of inequality: Subtracting the same number to each side of an inequality produces an equivalent inequality.
Example:
If 
$x+5$x+5  $>$>  $7$7 

Then 
$x+55$x+5−5  $>$>  $75$7−5 

Let's apply our knowledge of inverses and the addition and subtraction properties of inequality to solve some inequalities.
Solve for $x$x in the inequality $x+3\ge15$x+3≥15, showing all of your work algebraically.
Think: The number $3$3 is being added to the variable $x$x. In order to undo the operation of adding $3$3, we should subtract $3$3 from both sides of the inequality.
Do:
$x+3$x+3  $\ge$≥  $15$15 
Write the original inequality. 
$x+33$x+3−3  $\ge$≥  $153$15−3 
Subtract $3$3 from each side. 
$x$x  $\ge$≥  $12$12 
Simplify by doing the subtraction. 
Reflect: We can check our answer by substituting a few values back into the original inequality. Since the solution was $x\ge12$x≥12, we can substitute values that are $12$12 or larger.
$x+3$x+3  $\ge$≥  $15$15 
Write the original inequality. 
$13+3$13+3  ?  $15$15 
Substitute $13$13 for $x$x. 
$16$16  $\ge$≥  $15$15 
The solution checks. 
Solve for $a$a in the inequality $10>a6$10>a−6, showing all of your work algebraically.
Think: The number $6$6 is being subtracted from the variable $a$a. In order to undo the operation of subtracting $6$6, we should add $6$6 to both sides of the inequality.
Do:
$10$10  $>$>  $a6$a−6 
Write the original inequality. 
$10+6$10+6  $>$>  $a6+6$a−6+6 
Add $6$6 to each side. 
$16$16  $>$>  $a$a 
Simplify by doing the addition. 
$a$a  $<$<  $16$16 
Reverse the inequality statement. 
Reflect: We can check our answer by substituting a few values back into the original inequality. Since the solution was $a<16$a<16, we can substitute values that are smaller than $16$16. Let's substitute the value $15$15 since $15<16$15<16.
$a6$a−6  $<$<  $10$10 
Write the original inequality. 
$156$15−6  ?  $10$10 
Substitute $15$15 for $a$a. 
$9$9  $<$<  $10$10 
The solution checks. 
We are familiar with being able to write an equation in two orders. For example, $x=10$x=10 and $10=x$10=x mean the same thing.
We can also write inequality statements in two orders, but we now need to be careful and switch the inequality sign being used as well.
For example, $x>10$x>10 means the same thing as $10
Consider the following statement: $7<10$7<10
Add $6$6 to both sides of the inequality and simplify.
After adding $6$6 to both sides, does the inequality still hold true?
Yes
No
Yes
No
Solve the following inequality: $9>y+4$9>y+4
Solve the following inequality: $x8\le4$x−8≤−4
Solve onestep linear inequalities in one variable, involving addition or subtraction, and graph the solution on a number line