## 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

**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 (x

^{2 }- 2x + 1)

^{2}- 8(x

^{2}- 2x + 1) + 16 = 0

**Solution: -**

Given equation is (x

^{2 }- 2x + 1)

^{2}- 8(x

^{2}- 2x + 1) + 16 = 0

We first try to convert the given equation into a quadratic equation. For this, we put x

^{2 }- 2x + 1 as u. Then the equation becomes u

^{2 }- 8u + 16= 0

Our next step is to split the middle term. We get

u

^{2}- 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

x

^{2 }- 2x + 1 = 4

subtracting 4 on both sides, we get

x

^{2}- 2x - 3 = 0

Here again we split the middle term, we get

x

^{2}- 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

2x

^{2}- 3xy - y

^{2}= 4 and 3x + y = 1

Solution: -

Given equation are 2x

^{2}- 3xy - y

^{2}= 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

2x

^{2}- 3x(1 - 3x) - (1 - 3x)

^{2}= 4

Expanding we get

2x

^{2}- 3x + 9x

^{2}- 1 + 6x - 9x

^{2}= 4

That is 2x

^{2}+ 3x - 1 = 4

Subtracting 4 on both sides, we get

2x

^{2}+ 3x - 5 = 0

This is a quadratic equation. We solve using grouping method.

Splitting the middle term, we get

2x

^{2}+ 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**