Linear equations in one variable may take the form ax+b=0 and are solved using basic algebraic operations. Solving linear equations using below rules make your job easier and any ways you can use to find the solution for any problem that will help to sharp your skills.

**Rules for solving linear equations:**

If a = b then a + m = b + m; for any number m

If a = b then a - m = b - m; for any number m

If a = b then a * m = b * m; for any number m.

In this section, we will study about only linear equations having one or unique solution.

In this section, we will study about only linear equations having one or unique solution.

What does it mean to have one solution in a linear equation? Linear equations with one variable and two variables having only one solution satisfy below conditions:

Type |
General Form | Explanation |
Example |

Linear equations in one variable | ax+b = 0; a $\neq$0 | Conditional: Equation is true for only one value. |
2x + 4 = 5x - 5 => 2x - 5x = -5 - 4 => -3x = -9 or x = 3 Only solution for the problem |

Linear equations in two variables |
a$_1$x + b$_1$y + c$_1$ = 0 and a$_2$x + b$_2$y + c$_2$ = 0 | Unique solution for the set of two linear equations. If we graph both the equation, it will result as intersection of two lines at one point. |
Pair of linear equations: 8x - 6y = -20 and 16x - 7y = -30 The solution is $(-1, 2)$. *You can get explained solution under example section. |

2x + 6 = 0

Subtract 6 from both the side

2x + 6 - 6 = 0 - 6

2x = -6

To isolate x, divide above equation by 2, we get

Subtract 6 from both the side

2x + 6 - 6 = 0 - 6

2x = -6

To isolate x, divide above equation by 2, we get

2x/2 = - 6/2

or x = - 3

or x = - 3

The only solution to this equation is x = - 3

This solution can be represented on a number line.

**Example 2**: Solve for x, 4x - 27 = 5
**Solution**:

**Example 3**: Solve 3y - 10 = 44
**Solution:**

3y - 10 = 44

**Example 4**: Solve the following simultaneous set of two linear equation by substitution.

**Solution:**

**The solution is (-1, 2)**.

Graphical Representation:

We have 4x - 27 = 5

4x - 27 + 27 = 5 + 27

4x = 32

x = 32/4 = 8

Step 2: Add 10 to both the sides

3y - 10 + 10 = 44 + 10

3y = 54

Step 2: Divide both the sides by 3, to isolate variable y

y = 18

8x - 6y = -20

16x - 7y = -30

Since we are using substitution we can select any one variable, let us go with $x$ from the first equation. The second equation contains $16x$ which is a multiple of $8x$ which we have, this is a sure way to eliminate the fraction if it arises.

8x - 6y = -20

Add 6y to both the sides

8x = -20 + 6y

Divide by 8 on both the sides we get

x = $\frac{(-20 + 6y)}{8}$

We can factor out 2 from the numerator terms

x = $\frac{2(-10 + 3y)}{8}$

x = $\frac{(-10 + 3y)}{4}$

Now substituting this in the second equation

16x - 7y = -30

16 $\times$ $\frac{(-10 + 3y)}{4 }$ - 7y = -30

4(-10 + 3y) - 7y = -30

-40 + 12y - 7y = -30

-40 + 5y = -30

5y = -30 + 40

5y = 10

y = 2

Substitute for y in x = $\frac{(-10 + 3y)}{4}$

x = $\frac{(-10 + 3(2))}{4}$

x = $\frac{(-10+6)}{4}$

x = $\frac{-4}{4}$

x= -1

Graphical Representation: