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
You may enter any of these types of equations and avail the math equation help. You will get step by step solution to all your questions.

Math Equation Examples 

Following are the simple examples of math equation solved by our experts.

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 (x2 - 5x + 6)2 - 5(x2 - 5x + 6) + 6 = 0

Solution: -

Given equation is (x2 - 5x + 6)2 - 5(x2 - 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 x2 - 5x + 6 as u. Then the equation becomes u2 - 5u + 6= 0
Now we split the middle term
u2 - 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
x2 - 5x + 6 =  3
subtracting 3 on both sides, we get
x2 - 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
x2 - 5x + 6 =  2
subtracting 2 on both sides, we get
x2 - 5x + 4 = 0
Now we split the middle term
x2 - 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 ax = b ⇒ loga b = x
So we can write the above equation as
10x+3 = 103x
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