The following examples come from RegentsPrep.org
Practice Problems:
Linear Equations Word Problems Inequalities
Solving linear equations is just a matter of undoing operations that are being done to the variable. The task is always to isolate the variable — get the variable ALONE on one side of the equal sign.
Remember when solving equations to“keep the equation balanced” by making the same changes to BOTH sides of the equal sign. 
Example 1: In a simple equation, you may only have to undo one operation to solve the equation.
Solve this equation for x: x + 3 = 8  
The variable is x and we need to get it alone. In the problem, 3 is being added to the variable, so to get rid of the added 3, we do the opposite — subtract 3. We are actually employing the additive inverse property to create a 0 since +3 – 3 = 0. Then the additive identity is used to get x alone since x + 0 = x. (Remember to subtract 3 from both sides of the equation to “keep the equation balanced”.) 
x + 3 = 8 
Check your answer: You will always know if your answer is correct by doing a simple “check” — substitute your answer into the original equation and see if the result is true. 
Check: x + 3 = 8 5 + 3 = 8 8 = 8 true 
Example 2: In an equation which has more than one operation, we have to undo the operations in the correct order. First, undo addition or subtraction, then undo multiplication or division.
Solve this equation for x: 5x – 2 = 13  
The variable is x. The question is multiplying x by 5, and then subtracting 2. First, undo the subtraction by adding 2. Then, undo the multiplication by dividing by 5. This process is actually employing the multiplicative inverse to create the value of 1 and then employing the multiplicative identity to isolate the x. 
(Remember to perform your changes to both sides of the equation to “keep the equation balanced”.) 5x – 2 = 13
+2 +2
5x = 15
5x = 15
5 5
x = 3
Check your answer:
Check:
5x – 2 = 13
5(3) – 2 = 13
15 – 2 = 13
13 = 13 true
Example 3: Suppose there are variables on both sides of the equation. The trick now, is to get the variables on the same side by adding them or subtracting them.
Solve this equation for x: 4x + 5 = x – 4 

This question has two terms with the variable; 4x and x. We need to get the variables combined into one term.

Subtract x from both sides.
Now we proceed as before.
(Remember to perform your changes to both sides of the equation to “keep the equation balanced”.)
4x + 5 = x – 4
–x –x
3x + 5 = 4
3x + 5 = 4
5 5
3x = 9
3 3
x = 3
Check your answer:
Check:
4x + 5 = x – 4
4(3) + 5 = 3 4
12 + 5 = 7
7 = 7 trueHint: Some students think of “moving” one variable to the other side of the equal sign as “moving” the variable over the “equal sign bridge”. Moving any term across the “equal sign bridge” changes the term’s sign (like paying a toll).
4x + 5 = x – 4
4x – x + 5 = 4 as the x moves to the left over the “equal sign bridge”, it changes its sign to negative.
Example 4: Sometimes there are equations which have multiple terms on the same side. The trick here is to combine all the similar terms before solving.
Solve this equation for y: 7y + 5 – 3y + 1 = 2y + 2 
First combine the similar terms on the left side. (Don’t forget to take the sign in front of the term. If there isn’t a sign in front of the term, it is considered +.) Combining: 7y – 3y = 4y +5 and +1 = +6Now proceed as before. 
(Remember to perform your changes to both sides of the equation to “keep the equation balanced”.)7y + 5 – 3y + 1 = 2y + 2
4y + 6 = 2y + 2
2y 2y
2y + 6 = 2
6 6
2y = 4
2 2
y = 2
Check your answer:
Check:
7y + 5 – 3y + 1 = 2y + 2
7(2) + 5 – 3(2) + 1 = 2(2) + 2
14 + 5 + 6 + 1 = 4 + 2
2 = 2 true
Example 5: There are also equations with parentheses. The first step in these problems is to multiply and get rid of the parentheses.
Solve this equation for n: 3(n – 1.8) = 2n + 1 
First distribute the 3 — multiply through the parentheses by 3. Now proceed as normal. 
(Remember to perform your changes to both sides of the equation to “keep the equation balanced”.)
3(n – 1.8) = 2n + 1
3n – 5.4 = 2n + 1
3n – 5.4 = 2n + 1
2n 2n
n – 5.4 = 1
+ 5.4 + 5.4
n = 6.4
Check your answer:
Check:
3(n – 1.8) = 2n + 1
3(6.4 – 1.8) = 2(6.4) + 1
3(4.6) = 12.8 + 1
13.8 = 13.8 true
Example 6: The last type of equation contains fractions.


Check your answer: 
Check: 2(5/6) + (1/3) = 2 (10/6) + (1/3) = 2 (5/3) + (1/3) = 2 6/3 = 2 2 = 2 true 

Solve for x: (method 2) This problem could also be solved by multiplying each term by the common denominator, 3, thus creating: 6x + 1 = 6 6x = 5 x = 5/6 


Solving linear inequalities is the same as solving linear equations… 


with one very important exception…

Inequalities with one variable: 
Consider:
Look at this true statement: What is the relationship between these two numbers ? 
5 > 3 (1)(5) ? (3)(1) 5 ? 3 


So, we must change the direction of the inequality when we multiply (or divide) by a negative number in order to get the correct answer. 
Before we begin our example problems, refresh your memory on what each inequality symbol means. It is helpful to remember that the “open” part of the inequality symbol (the larger part) always faces the larger quantity. 
SYMBOL 
MEANING 

less than 


greater than 


less than or equal to 


greater than or equal to 
Example 1 
Solve and graph the solution set of: 2x – 6 < 2
Add 6 to both sides. Divide both sides by 2. Open circle at 4 (since x can not equal 4) and an arrow to the left (because we want valuesless than 4). 
2x – 6 < 2 
Example 2 
Solve and graph the solution set of: 5 – 3x 13 + x
Subtract 5 from both sides. Subtract x from both sides. 
Divide both sides by 4, and don’t forget tochange the direction of the inequality !
(We divided by a negative.)
5 – 3x 13 + x
3x 8 + x
4x 8
x 2
Closed circle at 2 (since x can equal 2) and an arrow to the right (because we want values largerthan 2).
Example 3 
Solve and graph the solution set of: 3(2x + 4) > 4x + 10
Multiply out the parentheses. Subtract 4x from both sides. Subtract 12 from both sides. 
Divide both sides by 2, but don’t change the direction of the inequality, since we didn’t divide by a negative.
3(2x + 4) > 4x + 10
6x + 12 > 4x + 10
2x + 12 > 10
2x > 2
x > 1 Open circle at 1 (since x can not equal 1) and an arrow to the right (because we want values largerthan 1).