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