Linear equations is important math topic and occur frequently in most of the other maths topics and their applications in accountancy, engineering and physics. First we have to understand, what is a linear equation? A linear equation in one variable is an algebraic equation with one variable. It always makes a straight line when on a coordinate system. It can be written in the form ax+b=0 where a $\neq$0 and a and b are real numbers. In this section we have listed a list of questions for practice along with the solution of some of the questions. As Practice is very important to get master on the math problems, without practice and lack of concept knowledge no matter how long you have been trying nothing could be learnt. We provide detailed explanations for all the questions on linear equations. Practice below questions and excel your knowledge.

Solution of Linear Equations in One Variable

Linear equations with one variable have one of the below listed solutions:

 Solution Type Explanation  Example 
Identity 
Solution set for these type of equations consists of all values which make the equation true 4x = x + 3x
or 4x = 4x
Above equation is true for all x values
 Conditional Such equations are true for only selected variable values  x + 2 = 2x - 4
=> x - 2x = -4 - 2
=> -x = -6
or x = 6
 Only solution for the problem
 Inconsistent Leads to a false statement  2x - 3 = 2(x - 5)
=> 2x - 3 = 2x - 10
=> -3 = -10
Which is not true. -3 $\neq$ -10

Questions

Example 1: Solve $\frac{y}{4}$ + 7 = 14
Solution:
$\frac{y}{4}$ + 7 = 14
Step 1: Subtract 7 from both the sides
$\frac{y}{4}$ + 7 - 7 = 14 - 7
$\frac{y}{4}$ = 7

Step 2:Simplify by multiplying both the sides by 4
$\frac{y}{4}$ $\times$ 4 = 7 $\times$ 4

y = 28. Answer!

Example 2: Simplify 15y - 2 = $\frac{y}{2}$ + 3
Solution: 15y - 2 = $\frac{y}{2}$ + 3
Add 2 to both sides
15y - 2 + 2 = $\frac{y}{2}$ + 3 + 2
15y = $\frac{y}{2}$ + 5
Multiply both sides by 2
15y $\times$ 2 = 2($\frac{y}{2}$ + 5)
30 y = y + 10
Subtract y from both the sides
30 y - y = y + 10 - y
29 y = 10
Divide each side by 29 to isolate y, we get
29y/29 = 10/29
or y = 10/29

Example 3: Solve 10x - 1 = 3(12 + x) - 1
Solution: 10x - 1 = 3(12 + x) - 1
Expand the brackets
10x - 1 = 36 + 3x - 1
Combine like terms each side
10x  -1 = 35 + 3x
Add 1 to both sides
10x  -1 + 1 = 35 + 3x + 1
10x = 36 + 3x
Subtract 3x from both the sides
10x - 3x = 18 + 3x - 3x
7x = 36
Divide by 7 to find the value of x
7x/7 = 36 / 7
x = 36/7

Example 4: Find the value of m. 12m + m - 2 = 2(m+2) + 11
Solution: 12m + m - 2 = 2(m+2) + 11
12m + m - 2 = 2m + 4 + 11  (Expand 2(m+2))
13m - 2 = 2m + 15
Add 2 to both the sides, we get
13m - 2 + 2 = 2m + 15 + 2
Simplify
13m = 2m + 17
Subtract 2m from both sides
13m - 2m = 2m + 17 - 2m
11m = 17
Divide by 11 both sides
11m/11 = 17/11
or m = 17/11

Practice Problems

Practice Problem 1: Solve for t, 5t - 10 = 20 + 2t
Practice Problem 2: Solve m/11 - 11 = 15 - t
Practice Problem 3: Solve x - 1 = 12(2x + 10) + 1/2
Practice Problem 4: Solve m - 10 = 3m -1
Practice Problem 5: Solve x - x/2 + 10 = 12 - x