Solve Algebra Squared Fraction

Fractions:

In algebra, A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

In this article we are going to see how to solve algebra squared fraction and some example solved problems on algebra squared fraction.

Solve Algebra Squared Fraction :

Steps to solve algebra squared fraction:

Step 1: We know that ( a/b)^n = a^n / b^n . So we can take square for the numerator and denominator.

Step 2: If it is possible we can simplify it.

Let us see the example problems.
Example Problems on Solve Algebra Squared Fraction :

Problem 1:

Solve (4/5)^2

Solution:

Given , (4/5)^2

We need to squared the given fraction .

(4/5)^2

We know that ( a/b)^n  = a^n / b^n .

We can apply the above rule,

(4/5)^2 = 4^2 / 5^2

= 16 / 25

Answer: (4/5)^2 = 16 / 25

Problem 2:

Solve (3x / 4xy) ^ 2

Solution:

Given, (3x / 4xy) ^ 2

We need to find the squared the given fraction,

We know that (a/b)^n = a^n / b^n .

(3x / 4xy) ^ 2 = ((3x) ^ 2) / ( (4xy)^2)

= (9x^2) / ( 16x^2y^2)

now we can simplify it,

Divide by x^2 on both numerator and denominator,

(9x^2) / ( 16x^2y^2) = (9x^2)/x^2 / ( 16x^2y^2) / x^2

= 9 / 16y^2

Answer: (3x / 4xy) ^ 2   = 9 / 16y^2

Problem 3:

Solve (( x^2 - 4 ) / ( x + 2) )^2

Solution :

Given , (( x^2 - 4 ) / ( x + 2) )^2

We need to find the squared the given fraction,

We know that ( a/b)^n = a^n / b^n .

(( x^2 - 4 ) / ( x + 2) )^2 = ( x^2 - 4 )^ 2 / ( x + 2)^2

Before that we can simplify the given fraction,

(( x^2 - 4 ) / ( x + 2) )^2 = (((x+2)(x-2))/ ( x + 2))^ 2

= ((x - 2)^2)

= x^2 - 2x + 4

Answer: (( x^2 - 4 ) / ( x + 2) )^2 = x^2 - 2x + 4

Results and Discussion Section

Here we are going to see about the result and discussion, whether the solution to the given equation is correct or not. By simplifying the equation we find the value for the particular variable. when we substitute the value in the equation the equation balances on both sides.

For example: x – 4 = 0.

Here    x – 4 = 0

Add 4 on both sides. We get

x = 4

The solution is x = 4

According to the topic we have to check whether x = 4 is a solution to the given equation or not. By substituting the value the in equation we can see that both sides have same value.
Example Problem for Result and Discussion Section:

Given equation is discussion section  x + 7x + 5x + 3x – 5 + x = 29

Solution:

Given question is x + 7x + 5x + 3x – 5 + x = 29

Step 1:

Arrange the given question as per the x term and the numbers

x+ x + 3x + 5x + 7x – 5 = 29

Step 2:

Add the x term first

2x + 3x + 5x + 7x – 5 = 29

Step 3:

Add all the  ' x ' term in the equation

17x – 5 = 29

Step 4:

Add both sides by 5.Then we get

17x = 29 + 5

Step 5:

17 x = 34

Divide by 17 on both sides.

`(17x)/17` = `34/17`

x = 2

Discussion section about the result:

The result is 2. Here we are going to discuss about the result for given equation, whether the solution x =2 is correct or not.

Given question is x + 7x + 5x + 3x – 5 + x = 29

The result is x =2

Here we need to substitute the x value in the equation if that both side value is equal means the solution is correct. Otherwise the solution is not correct.

x + 7x + 5x + 3x – 5 + x = 29

Substitute the value x = 2 in the given equation

2 + 7( 2) + 5( 2) + 3(2) – 5 + 2 = 29

2 + 14 + 10 + 6 – 5 + 2 = 29

34 – 5 = 29

29 = 29

Both the sides are equal, so given solution is correct for the equation.
One more Example Problem for Result and Discussion Section:

Solve discussion section  16x - 12 = 20

Solution:

Add 12 on both sides. we get

16x - 12 + 12 = 20 + 12

16x  = 32

Divide by 16 on both sides.

`(16x )/ 16` = `32 / 16`

x = 2

whe we substitute x =2 in the equation 16x - 12 = 20. we get

16(2) - 12 = 20

32 - 12 = 20

20 = 20.

Thus the equation satisfies.