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.