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# GRE Algebra | Solving linear Equations

In mathematics, a linear equation is an equation that may be put in the form,

`a1x1 +........... + anxn + b = 0 `

Where x1, xn are the variables, and a1, an, b are the coefficients, which are often real numbers.

To solve a linear equation means to find the values of all the variables that make the equation satisfy. Two equations said to be equivalent equation when both of them have same solution.

For example:
y + 4 = 16 and and 3y + 12 = 48 both are equivalent equation because both are true for (y = 12).

Rules for producing equivalent equations:

1. When same constant added to or subtracted from both sides of an equation then equality is preserved and new equation is equivalent to the original equation.
2. When the equation is multiplied or divided by the same non-zero constant on both sides, the equality is preserved and new equation is equivalent to the original equation.

Linear equations in one variable:
A linear equation in one variable, can be written in form,

`ax + b = c `

where a, b are real numbers.

Linear equation can be solved by combining all the like terms on each side of the equation. Then use all the rules of producing similar equivalent equations.

```Example-1: Solve  15x - 2x - 9 = 4x + 3(x - 5)
Solution:
13x - 9 = 4x + 3x - 15
13x - 9 = 7x - 15              (like terms combined)
13x - 9 -7x = 7x -15 - 7x      (-7x) added to both sides
6x - 9 = -15
6x - 9 + 9 = -15 + 9           (9) is added to both sides
6x = -6
6x/6 = -6/6                    both sides divided by 6
x  = -1 ```

Linear equations in two variables:
A linear equation in two variables x and y, can be written in the form,

`ax + by + c = 0    `

where a, b and c are real numbers but a and b not zero at the same time.

Two variable equation can be solved by two methods:

1. Substitution method
2. Elimination method

Example-2: Solve using substitution method 2x + 2y = 4 and 4x – 6y = 18
Solution: We can express x in terms of y in first equation as,

`x = 2 - y `

Then, substitute the (y-2) for x in the second equation to find the value of y.

```4(2 - y) - 6y = 18
8 - 4y - 6y = 18
-10y = 10
y = -1 ```

Substitute value of y in either equation to find the value of x.

```2x + 2(-1) = 4
2x = 6
x = 3 ```

Example-3: Solve using elimination method 2x + 2y = 4 and 4x – 6y = 18
Solution: Compare the coefficients of x and y and make equal for one variable.
Then subtract or add both the equations so that one variable can be eliminated.
Multiply first equation by 2 and subtract the second equation from first.

```4x + 4y = 8 ..........(1)
4x - 6y = 18 .........(2)
_______________
10y = -10
y = -1 ```

Hence, put value of y in first equation,

```2x + 2(-1) = 4
2x = 6
x = 3 ```