## Math equation

When we equate two algebraic expressions with an "=" sign, we get a math equation. There may be variables on one or either sides of the equation. Solving math equation refers to find a single value which satisfies the math equation.

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There are different types of math equations.Following are few of them.

- Simple Linear equation
- Equation containing radical expressions
- Equations containing absolute values
- Quadratic equations
- Equations containing fractions
- Exponential equations
- Logarithmic equations

## Math Equation Examples

**Example 1: -**

Solve the equation 2(4x + 8) + 9( 8x + 5) = 3x+ 12(5x + 4)

**Solution: -**

Given equation is 2(4x + 8) + 9( 8x + 5) = 3x+ 12(5x + 4)

The equation is a linear equation with one variable. We first simplify the equation and then solve it. For this first we expand the brackets

We get

8x + 16 + 72 x + 45 = 3x + 60x + 48

That is

80x + 61 = 63x + 48

Now we just have to isolate the variable For this we take all variables to one side and all constants to other.

Lets first take the constants to the right side of the equation. For this, we subtract 61 on both sides, we get

80x = 63x - 13

Now we take the variables on the left side of the equation. For this, we subtract 63x on both sides, we get

17x = -13

Now we have variable on the left side and constant on the right side of the equation

Dividing by 17 on both sides, we get the required solution

x = -13/17

**Answer: x = -13/17**

**Example 2: -**

Solve the equation (x

^{2 }- 5x + 6)

^{2}- 5(x

^{2}- 5x + 6) + 6 = 0

**Solution: -**

Given equation is (x

^{2 }- 5x + 6)

^{2}- 5(x

^{2}- 5x + 6) + 6 = 0

We can see that only one variable is involved in this problem.. So to solve, we just have to isolate the variable.

Take x

^{2 }- 5x + 6 as u. Then the equation becomes u

^{2 }- 5u + 6= 0

Now we split the middle term

u

^{2}- 3u - 2u + 6 = 0

Taking u common from first two terms and -2 common from last two terms, we get

u (u - 3) - 2 (u - 3) = 0

Now we take u - 3 common, we get

(u - 3) ( u - 2) = 0

Equating each factor to 0, we get

u - 3 = 0 ⇒ u = 3

and u - 2 = 0 ⇒ u = 2

When u = 3

x

^{2 }- 5x + 6 = 3

subtracting 3 on both sides, we get

x

^{2}- 5x + 3 = 0

We use the quadratic formula to solve this equation.

x = [-(-5) ±√((-5)

^{2}- 4 x 1 x 3)] / 2

= (5 ±√13)/2

When u = 2

x

^{2 }- 5x + 6 = 2

subtracting 2 on both sides, we get

x

^{2}- 5x + 4 = 0

Now we split the middle term

x

^{2}- 4x - x + 4 = 0

Taking x common from first two terms and -1 common from last two terms, we get

x (x - 4) - 1 (x - 4) = 0

Now we take x - 4 common, we get

(x - 4) ( x - 1) = 0

Equating each factor to 0, we get

x - 4 = 0 ⇒ x = 4

and x - 1 = 0 ⇒ x = 1

**Answer : (5 ±√13)/2, 4, 1**

**Example 3: -**

Solve the equation log(x + 3) = log x + log 3

**Solution: -**

Given equation is log(x + 3) = log x + log 3

We can see that only one variable is involved in this problem.

We we try to get rid of the logarithm.

We know that log a + log b = log ab

So the original equation becomes

log (x + 3) = log 3x

Also we know that a

^{x}= b ⇒ log

_{a}b = x

So we can write the above equation as

10

^{x+3}= 10

^{3x}Equating the powers, we get

x + 3 = 3x

Now we have a simple linear equation. To solve this, we take the constants to one side of the equation and variables to the other side.

For this we subtract x on both sides. We get

3 = 2x

Now we divide both sides by 2 to isolate x

Dividing by 2, we get

x = 3/2

**Answer: x = 3/2**