# Solve my Math Equation

## Math equation

An expression involving mathematical operations is called mathematical expression. When two mathematical expressions are equated with an "=" sign, we get a math equation. One or both sides of the equation may contain variables. By solving a math equation, we are trying to find a single value which satisfies the math equation. That is when we substitute the value of the solution for the variable, the equation balances.

## Solve this Math Equation

Some examples of solved problems are given below: -
Example 1: -

Solve the equation | 5x - 4| = 2

Solution: -

Given equation is | 5x - 4| = 2
This is an absolute value equation. So to solve this equation, first we split the original equation into two.
We get
5x - 4 = 2 and -(5x - 4) = 2
Now we solve the two equations obtained separately to get the required solution.

Lets consider 5x - 4 = 2
Adding 4 on both sides, we get 5x = 6
Now dividing both sides by 5, we get x = 6/5

Now consider the second equation
-(5x - 4) = 2
That is -5x + 4 = 2
Subtracting 4 on both sides, we get -5x = -2
Dividing by -5 on both sides, we get x = 2/5

Therefore solution is x = 6/5, 2/5.

Example 2: -

Solve the equation (x2 - 2x + 1)2 - 8(x2 - 2x + 1) + 16 = 0

Solution: -

Given equation is (x2 - 2x + 1)2 - 8(x2 - 2x + 1) + 16 = 0
We first try to convert the given equation into a quadratic equation.  For this, we put x2 - 2x + 1 as u. Then the equation becomes u2 - 8u + 16= 0
Our next step is to split the middle term.  We get
u2 - 4u - 4u + 16 = 0
Taking u common from first two terms and -4 common from last two terms, we get
u (u - 4) - 4 (u - 4) = 0
Now we take u - 4 common, we get
(u - 4) ( u - 4) = 0
Equating each factor to 0, we get
u - 4 = 0 ⇒ u = 4
and u - 4 = 0 ⇒ u = 4

When u = 4
x2 - 2x + 1 =  4
subtracting 4 on both sides, we get
x2 - 2x - 3 = 0
Here again we split the middle term, we get
x2 - 3x + x - 3 = 0
Taking x common from first two terms and 1 common from last two terms, we get
x (x - 3) + 1 (x - 3) = 0
Now we take x - 3 common, we get
(x - 3) ( x + 1) = 0
Equating each factor to 0, we get
x - 3 = 0 ⇒ x = 3
and x + 1 = 0 ⇒ x = -1

For the other u value also we get the same solution.

Therefore required solution is x = -1, -1, 3, 3

Example 3: -

Solve the simultaneous equations involving Quadratics
2x2 - 3xy - y2  =  4 and 3x + y = 1

Solution: -
Given equation are 2x2 - 3xy - y2  =  4 ...(1) and 3x + y = 1 ...(2)
To solve these equations we first find y from (2)
Using (2), we can write y = 1 - 3x
Substituting in (1), we get
2x2 - 3x(1 - 3x) - (1 - 3x)2 = 4
Expanding we get
2x2- 3x + 9x2 - 1 + 6x - 9x2  = 4
That is 2x2 + 3x - 1 = 4
Subtracting 4 on both sides, we get
2x2 + 3x - 5 = 0

This is a quadratic equation.  We solve using grouping method.
Splitting the middle term, we get
2x2 + 5x - 2x- 5 = 0
Taking x common from first two terms and -1 common from last two terms, we get
x (2x + 5) - 1 (2x + 5) = 0
Now we take 2x + 5 common, we get
(2x + 5) ( x - 1) = 0
Equating each factor to 0, we get
x - 1 = 0 ⇒ x = 1
and 2x + 5 = 0 ⇒ 2x = -5
⇒   x = -5/2

When x = 1, y = 1 -  3 x 1 = 1 - 3 = -2
When x = -5/2, y = 1 -  3 x (-5/2) = 1 + 15/2 = 17/2

Therefore solution x = 1, y = -2 and x = -5/2, y = 17/2