Change the Subject of a Formula | How to Change the Subject of a Formula?

Learn How to Change the Subject of a Formula and also know about finding the value of the variables by reading the complete article. We included solved examples with clear explanations for better understanding. Students can immediately practice all the questions available in this article and learn the best ways to Change the Subject of a Formula.

Subject of the Formula

Expressing one variable in terms of other variables is the main concept of the formula. The variable that expresses in other variables is called the subject of the formula. The subject of the formula will be written on the left side and other constants and variables are written on the right side of the equality sign in a formula.

Example:
z = xy, where z is the subject of the formula where it is expressed in terms of the product of the x and y.
If we want to change the subject of the formula to x, then the above expression will change into x = z/y.

How to Change the Subject of the Formula?

Changing the subject of a formula can be possible by rearranging the formula to get the required subject. To change the subject of the formula, firstly change its side and change the operation.

When one variable moved to the other side of the equal to sign, the operation becomes inverse. For example, if a variable is added to the subject of the formula, then it will be subtracted after moving to the other side of the equal to sign.

Examples:

1. Make ‘v’ the subject of the formula in u = v + as

Solution:
Given that u = v + as
as is added to v.
To find the subject of the v subtract the as from both sides.
u – as = v + as – as
u – as = v

The final answer is v = u – as

2. Make ‘t’ the subject of the formula, s = x + bt

Solution:
Given that s = x + bt
x is added to the bt.
Firstly, subtract x on both sides.
s – x = x – x + bt
s – x = bt
b is multiplied to t.
Divide b on both sides.
(s – x)/b = bt/b
(s – x)/b = t

The final answer is t = (s – x)/b.

Change the Subject of a Formula Solved Examples

1. The volume of a box is the product of the length and breadth of the box?

Solution:
Given that the volume of a box is the product of the length and breadth of the box.
The volume of a box = v
The length of the box = l
The breadth of the box = b
v = l × b
If the subject of the box is length, the answer is l = v/b
If the subject of the box is the breadth, the answer is b = v/l.

The answer is v = l × b, l = v/b, and b = v/l.

2. In the relation x/5 = make (s – 16)/7 make s as the subject.

Solution:
Given that In the relation x/5 = make (s – 16)/7 make s as the subject.
x/5 = (s – 16)/7
7 is dividing (s – 16)
Multiply 7 on both sides.
7x/5 = (s – 16)7/7
7x/5 = (s – 16)
16 is subtracted from the s.
Add 16 on both sides of the equation.
7x/5 + 16 = s – 16 + 16
7x/5 + 16 = s

The final answer is s = 7x/5 + 16.

3. Make t the subject of the formula s = (t + r)/(t – r)

Solution:
Given that s = (t + r)/(t – r)
Multiply (t – r) on both sided.
s(t – r) = (t + r)
st – sr = t + r
To find the t as subject of the formula, move t variables on left side.
Subtract t on both sides.
st – t – sr = t – t + r
st – t – sr = r
Take t as common from st – t
t (s – 1) – sr = r
Add sr on both sides
t(s – 1) – sr + sr = r + sr
t(s – 1) = r(1 + s)
Divide (s – 1) on both sides
t(s – 1)/(s – 1) = r(1 + s)/(s – 1)
t = r(1 + s)/(s – 1)

The final answer is t = r(1 + s)/(s – 1).

4. Write the formula for finding the area of the rectangle and indicate the subject in this formula. Also, make b as the subject. If A = 24 cm² and l = 4 cm, then find b.

Solution:
The area of the rectangle is A = l × b.
To make the b as the subject of the formula, divide l on both sides.
A/l = lb/l
A/l = b
b = A/l
Now, substitute the values of l and A.
b = 24/4
b = 6.

Therefore, b = 6 is the answer.

5. For a right angled triangle abc, square of the hypotenuse (h) is equal to the sum of squares of its other two sides (s, t).
• Frame the formula for the above statement and find out h if s = 3 and t = 2.
• Also, make ‘s’ the subject of the formula and find s if h = 8 and t = 6.

Solution:
From the given data, Frame the formula for the above statement and find out h if s = 3 and t = 2.
The formula is h² = s² + t²
Substitute s = 3 and t = 2 in h² = s² + t²
h² = 3² + 2² = 9 + 4 = 13
h = √13

The answer is h = √13

From the given data, make ‘s’ the subject of the formula and find s if h = 8 and t = 6.
By changing the subject to s, the h² = s² + t² becomes s² = h² – t²
Substitute h = 8 and t = 6 in s² = h² – t²
s² = 8² – 6²
s² = 64 – 36
s² = 28
s = √28

The answer is s = √28.

6. In the formula, t = s + (b – 1)d make d as the subject. Find d when t = 10, s = 2, b = 5.

Solution:
Given that t = s + (b – 1)d.
Subtract s on both sides
t – s = s – s + (b – 1)d.
t – s = (b – 1)d.
Divide (b – 1) on both sides
(t – s)/(b – 1) = d
d = (t – s)/(b – 1)
Substitute t = 10, s = 2, b = 5.
d = (10 – 2)/(5 – 1) = 8/4 = 2
d = 2.

The final answer is d = 2.

Division of Algebraic Fractions | How to Divide Algebraic Fractions?

The division is one of the arithmetic operations. The division of algebraic fractions is similar to the division of numbers. Here, you need to factorize the numerators and denominators of the fractions. Cancel the like factors in a fraction and reverse the denominator fraction and multiply it with numerator fraction to get the division value. Find the solved example questions on the division of algebraic fractions.

How to Divide Algebraic Fractions?

Check the simple and easy steps to divide algebraic fractions in the following sections.

  • Take two algebraic fractions. One in the numerator and the second in the denominator.
  • Find the factors of fractions.
  • Get first fraction x 1 / second fraction.
  • Cancel the like terms in the numerator and denominator.
  • Multiply the numerators, denominators to get the result.

Solved Examples on Dividing Algebraic Fractions

Example 1.

Determine the quotient of the algebraic fractions: x / (x² – x) ÷ 10x / (x² + x – 2)?

Solution:

Given that,

x / (x² – x) ÷ 10x / (x² + x – 2)

Factorize the fractions and cancel the common terms.

x / x (x – 1) ÷ 10x / (x² + 2x – x – 2)

= x / x (x – 1) ÷ 10x / (x(x + 2) -1(x + 2))

= x / x (x – 1) ÷ 10x / (x + 2) (x – 1)

= 1/(x – 1) ÷ 10x / (x + 2) (x – 1)

= 1/(x – 1) x (x + 2) (x – 1) / 10x

= (x + 2) (x – 1) / 10x (x – 1)

= (x + 2) / 10x

Example 2.

Divide the algebraic fractions and express in the lowest form: 9x²+12x+4 / 4x²-27x-7 ÷ 12x²+5x-2 / 16x²-1?

Solution:

Given that,

9x²+12x+4 / 4x²-27x-7 ÷ 12x²+5x-2 / 16x²-1

Factorize the fractions and cancel the common terms.

= 9x²+6x+6x+4 / 4x²-28x+x-7 ÷12x²+8x-3x-2 / (4x)²-1²

= 3x(3x+2)+2(3x+2) / 4x(x-7)+1(x-7) ÷ 4x(3x+2)-1(3x+2) / (4x+1)(4x-1)

= (3x+2)(3x+2) / (x-7)(4x+1) ÷ (3x+2)(4x-1) / (4x+1)(4x-1)

= (3x+2)(3x+2) / (x-7)(4x+1) ÷ (3x+2) / (4x+1)

= (3x+2)(3x+2) / (x-7)(4x+1) * (4x+1) / (3x+2)

= (3x+2)(3x+2)(4x+1) / (x-7)(4x+1)(3x+2)

= 3x+2 / x-7

Example 3.

Find the quotient of the algebraic fractions:

x²+11x+24 / x²-15x+56 ÷ x²-x-12 / x²-11x+28

Solution:

Given that,

x²+11x+24 / x²-15x+56 ÷ x²-x-12 / x²-11x+28

Factorize the algebraic fractions and cancel the common terms.

= x²+8x+3x+24 / x²-8x-7x+56 ÷ x²-4x+3x-12 / x²-7x-4x+28

= x(x+8)+3(x+8) / x(x-8)-7(x-8) ÷ x(x-4)+3(x-4) / x(x-7)-4(x-7)

= (x+8)(x+3) / (x-8)(x-7) ÷ (x-4)(x+3) / (x-4)(x-7)

= (x+8)(x+3) / (x-8)(x-7) ÷ (x+3) / (x-4)

= (x+8)(x+3) / (x-8)(x-7) * (x-4) / (x+3)

= (x+8)(x+3)(x-4) / (x-8)(x-7)(x+3)

= (x+8)(x-4) / (x-8)(x-7)

= x²+4x-32 / x²-15x+56

Multiplication of Algebraic Fractions | How to Multiply Algebraic Fractions?

Multiplication of algebraic fractions is not so difficult. You need to multiply the numerators, denominators of all fractions together to get its product. Before multiplying the algebraic fractions, factorize them. Check out the simple steps of multiplication of algebraic fractions and solved examples in the following sections.

How to find the Product of Algebraic Fractions?

Here we are giving the simple steps to calculate the multiplication of two or more algebraic fractions. Follow these instructions to get the product quickly.

  • Find the factors of numerators, denominators of algebraic fractions.
  • If there are any like factors, then cancel them.
  • Multiply the numerators of remaining factors and denominators.

Solved Examples on Multiplication of Algebraic Fractions

Example 1.

Simplify 5 / (a + a²) x (a³ – a) / ab?

Solution:

Given that,

5 / (a + a²) x (a³ – a) / ab

Get the factors of both fractions

= 5 / a (a + 1) x a (a² – 1) / ab

= 5 / a (a + 1) x [a (a + 1) (a – 1)] / ab

Multiply the two algebraic fractions.

= 5a (a + 1) (a – 1) / a (a + 1) ab

Cancel the terms a (a +1) in both denominator and numerator.

= 5 (a – 1) / ab

Example 2.

Find the product of the algebraic fraction [5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)?

Solution:

Given that,

[5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)]

The least common multiple of denominators of the first part is a(2a – 1) and the L.C.M of denominators of the second part is a + 2.

Therefore, [5a. a / a(2a – 1) – (a – 2) . (2a – 1) / a(2a – 1)] x [2a / (a+2) – 1 / (a+2)]

= [(5a² – (a – 2) (2a – 1)) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(5a² – 2a² + a + 4a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a² + 5a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a² + 6a – a – 2) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(3a (a + 2) -1(a + 2)) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(a + 2) (3a – 1) / a(2a – 1)] x [(2a – 1) / (a+2)]

= [(a + 2) (3a – 1) (2a – 1)] / [a(2a – 1) (a+2)]

Here the common factors in the numerator and denominator are (a+2), (2a-1). Cancel these factors in both to find the lowest form

= (3a – 1) / a

Therefore, [5a / 2a-1 – (a-2) / a] x [2a / (a+2) – 1 / (a+2)] = (3a – 1) / a.

Example 3.

Find the product and express in the lowest form: 5x² / (x² – 2x) x (x² – 4) / (x² + 2x)?

Solution:

Given that,

5x² / (x² – 2x) x (x² – 4) / (x² + 2x)

Ge the factors of both fractions.

= 5x² / x(x – 2) x (x² – 2²) / (x(x + 2))

Cancel the common term x in the first part.

= 5x / (x – 2) x (x + 2) ( x – 2) / (x(x + 2))

Cancel the common factor (x+2) in the second part.

= 5x / (x – 2) x (x – 2) / x

Multiply both numerators and denominators

= 5x . (x – 2) / x . (x – 2)

Cancel the common factor (x – 2) in both numerator and denominator.

= 5x/ x

= 5.

∴ 5x² / (x² – 2x) x (x² – 4) / (x² + 2x) = 5.

Example 4.

Find the product of the algebraic fractions in the lowest form:

[(x + 2y) / (2x + y)] x [(2x + 5y) / (x + y)]

Solution:

Given that,

[(x + 2y) / (2x + y)] x [(2x + 5y) / (x + y)]

= [(x + 2y) (2x + 5y)] / [(2x + y) (x + y)]

= [2x² + 5xy + 4xy + 10y²] / [2x² + xy + 2xy + y²]

= [2x² + 9xy + 10y²] / [2x² + 3xy + y²]

Example 5.

Simplify (4x² – 1) / (9x – 6) x (15x – 10) / (x + 4)?

Solution:

Given that,

(4x² – 1) / (9x – 6) x (15x – 10) / (x + 4)

Calculate the factors.

= ((2x)² – 1²) / 3(3x – 2) x 5(3x – 2) / (x + 4)

= (2x + 1) (2x – 1) / 3(3x – 2) x 5(3x – 2) / (x + 4)

Multiply numerators, denominators together.

= [(2x + 1) (2x – 1) . 5(3x – 2)] / [3(3x – 2) (x + 4)]

Cancel the terms (3x – 2)

= [5(2x + 1) (2x – 1)] / 3(x + 4)

∴ (4x² – 1) / (9x – 6) x (15x – 10) / (x + 4) = [5(2x + 1) (2x – 1)] / 3(x + 4).

Solving Algebraic Fractions | Simplify & Reduce Algebraic Fractions

Find the solved example questions on algebraic fractions. This article includes addition, subtraction, multiplication, division, simplification, and reducing a fraction to its lowest term. By reading this page, you can solve any type of algebraic fraction questions easily & quickly. So, have a look at all the questions and solutions provided below and learn the concepts involved easily.

Questions on Solving Algebraic Fractions

Example 1.

Simplify the algebraic fraction [1 + 1 / (x + 1)] / [x – 4/x]?

Solution:

Given fraction is [1 + 1 / (x + 1)] / [x – 4/x]

Find the L.C.M of denominators.

[(1. (x + 1)+ 1) / (x + 1)] / [(x² – 4) / x]

= [(x + 2) / (x + 1)] / [(x² – 2²) / x]

= [(x + 2) / (x + 1)] * [x / (x + 2) (x – 2)]

= [x(x + 2)] / [(x + 1) ( x – 2) (x + 2)]

= x / [(x + 1) (x – 2)]

∴[1 + 1 / (x + 1)] / [x – 4/x] = x / [(x + 1) (x – 2)]

Example 2.

Simplify the algebraic fraction [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]?

Solution:

Given algebraic fraction is [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)]

Find the L.C.M of denominators of the first fraction and simplify.

= [(k² + 1 – k (k – 1)) / (k – 1)] / [(k² – 1 + 1 (k + 1)) / (k + 1)]

= [(k² + 1 – k² + k)) / (k – 1)] / [(k² – 1 + k + 1)) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k² + k) / (k + 1)]

= [(k + 1) / (k – 1)] / [(k(k + 1) / (k + 1)]

= [(k + 1) / (k – 1)] / k / 1

= (k + 1) / k(k – 1)

= (k + 1) / (k² – 1²)

= (k + 1) / (k + 1) ( k – 1)

= 1 / (k – 1)

Simplification of the second fraction is

[1 – 2/(1 + 1/k)]= [1- 2 / k(k +1)]

= [k(k +1) – 2] / [k(k +1)]

= (k² + k – 2) / [k(k +1)]

= (k² + 2k – k – 2) / (k(k +1))

= (k (k + 2) – 1(k + 2)) / (k(k +1))

= [(k – 1) ( k + 2)] / (k(k +1))

Product of first and second fraction is

= 1 / (k – 1) * [(k – 1) ( k + 2)] / (k(k +1))

= (k + 2) / k(k +1)

∴ [((k² + 1 / k – 1) – k) / ((k² – 1 / k + 1) + 1)] [1 – 2/(1 + 1/k)] = (k + 2) / k(k +1)

Example 3.

Reduce the algebraic fractions [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Solution:

Given algebraic fraction is [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1]

Simplification of first fraction is

3 / √(1 + x) + √(1 – x)

= (3 + √(1 + x) * √(1 – x)) / √(1 + x)

= (3 + √(1 + x)(1 – x) / √(1 + x)

Simplification of the second fraction is

3 / √(1 – x²) + 1

= 3 + √(1 – x²) / √(1 – x²)

The division of algebraic fraction is

= [(3 + √(1 + x)(1 – x) / √(1 + x)] : [3 + √(1 – x²) / √(1 – x²)]

= [(3 + √(1 + x)(1 – x) (√(1 – x²))] / [√(1 + x) (3 + √(1 – x²))

= √(1 – x²) / √(1 + x)

= √(1 + x)(1 – x) / √(1 + x)

= √(1 – x)

∴ [3 / √(1+x) + √(1-x)] : [3 / √(1 – x²) + 1] = √(1 – x)

Example 4.

Reduce to lowest terms — if possible 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab².

Solution:

Given fraction is 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab²

Find the L.C.M of all terms denominators.

L.C.M of 4a²b, 6ab⁵, 2ab² is 12a²b⁶.

= [3x . 3b⁵]/12a²b⁶ – [7 . 2a]/12a²b⁶ – [5x . 6ab⁴]/12a²b⁶

= [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

∴ 3x / 4a²b – 7 / 6ab⁵ – 5x / 2ab² = [9xb⁵ – 14a – 30xab⁴]/ 12a²b⁶

Compound Interest when Interest is Compounded Quarterly | Quarterly Compounded Interest Solved Examples

In this article, you will learn how to find Compound Interest when Interest is Compounded Quarterly. You may feel the process of calculating the Compound Interest using the growing principal a bit difficult if the time duration is long. Refer to Solved Examples on finding Quarterly Compounded Interest and learn how to solve related problems. We even provided the solutions for the sample problems on calculating the Compound Interest when Interest is Compounded Quarterly in the coming modules.

How to find Compound Interest when Interest is Compounded Quarterly?

If the Rate of Interest is Annual and Interest is Compounded Quarterly then the number of years is multiplied by 4 i.e. 4n and the annual interest rate is cut down by one-fourth. In such cases, Formula for Quarterly Compound Interest is given as under

Let us assume the Principal = P, Rate of Interest = r/4 %, and time = 4n, Amount = A, Compound Interest = CI then

A = P(1+(r/4)/100)4n

In the above formula rate of interest is divided by 4 whereas the time is multiplied by 4.

We know CI = A – P

= P(1+(r/4)/100)4n – P

= P{1+(r/4)/100)4n – 1}

If you are aware of any of the three then you can automatically find the other one.

Solved Problems on finding Compound Interest when Compounded Quarterly

1. Find the compound interest when $1,00, 000 is invested for 6 months at 5 % per annum, compounded quarterly?

Solution:

Principal Amount = $1,00, 000

Rate of Interest = 5% per annum

n = 6 months = 1/2 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)4n

Substitute the Inputs in the above formula to find the amount

A = 1,00,000(1+(5/4)/100)4*1/2

= 1,00,000(1+5/400)2

= $ 1,02,515

CI = A – P

= $ 1,02,515 – $ 1,00,000

=$2515

2. Find the amount and the compound interest on Rs. 12,000 compounded quarterly for 9 months at the rate of 10% per annum?

Solution:

Principal Amount = Rs.12, 000

Rate of Interest = 10% per annum

n = 9 months = 9/12 = 3/4 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)4n

Substitute the Input Values in the above formula to find the amount

A= 12,000(1+(10/4)/100)4*3/4

= 12,000(1+10/400)3

= 12,000(1+0.025)3

= 12,000(1.025)3

= Rs. 12922

CI = A – P

= 12922 – 12000

= Rs. 922

3. Calculate the compound interest (CI) on Rs. 4000 for 1 year at 10% per annum compounded quarterly?

Solution:

Principal Amount = Rs. 4,000

Rate of Interest = 10% per annum = 10/4 %

n = 1 year

Since Interest Rate is Compounded Quarterly divide the interest rate by 4 i.e. r/4 and multiply the time by 4 i.e. 4n

Amount A = P(1+(r/4)/100)4n

Substitute the Input Values in the above formula to find the amount

A = 4000(1+(10/4)/100)4*1

= 4000(1+10/400)4

= 4000(1.1038)

= Rs. 4415.25

CI = A – P

= 4415.25 – 4000

= Rs. 415.25