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Inverse operations are opposite operations. That is, one reverses the effect of the other.


In solving problems in algebra, we use inverse operations to isolate the given variable.
Example 1 :
Solve for k :
7k = 35
Solution :
Here 7 and k are multiplied.
So, we have to use the inverse operation of multiplication to solve for k.
Inverse operation of multiplication is division.
7k = 35
Divide both sides by 7.
k = 5
Example 2 :
Solve for k ;
k + 7 = 9
Solution :
Here 7 is added to k.
So, we have to use the inverse operation of addition to solve for k.
Inverse operation of addition is subtraction.
k + 7 = 9
Subtract 7 from each side.
k = 2
Example 3 :
Solve for b :
b/8 = 7
Solution :
Here b is divided by 8.
So, we have to use the inverse operation of division to solve for b.
Inverse operation of division is multiplication.
b/8 = 7
Multiply each side by 8.
b = 56
Example 4 :
Solve for m :
m - 15 = 9
Solution :
Here 15 is subtracted from m.
So, we have to use the inverse operation of subtraction to solve for b.
Inverse operation of subtraction is addition.
m - 15 = 9
Add 15 to each side.
m = 24
Example 5 :
Solve for z :
z + 17 = 23
Solution :
Here 17 is added with z.
So, we have to use the inverse operation of addition to solve for k.
Inverse operation of addition is subtraction.
z + 17 = 23
Subtract 17 from each side.
z = 6
Example 6 :
Solve for x :
2x + 5 = 23
Solution :
Here 5 is added to 2x.
So, we have to use the inverse operation of addition to solve for 2x.
Inverse operation of addition is subtraction.
2x + 5 = 23
Subtract 5 from each side.
2x = 18
Here 2 and x are multiplied.
Then, we have to use the inverse operation of multiplication to solve for x.
Inverse operation of multiplication is division.
2x = 18
Divide each side by 2.
x = 9
Example 7 :
Solve for p :
2p - 7 = 3
Solution :
Here 7 is subtracted from p.
So, we have to use the inverse operation of subtraction to solve for 2x.
Inverse operation of subtraction is addition.
2p - 7 = 3
Add 7 to each side.
2p = 10
Here 2 and p are multiplied.
Then, we have to use the inverse operation of multiplication to solve for x.
Inverse operation of multiplication is division.
2p = 10
Divide each side by 2.
p = 5
Example 8 :
Solve for r :
(r - 6)/2 = 3
Solution :
Here (r - 6) is divided by 2.
So, we have to use the inverse operation of division to solve for (r - 6).
Inverse operation of division is multiplication.
(r - 6)/2 = 3
Multiply each side by 2.
r - 6 = 6
Here 6 is subtracted from r.
Then, we have to use the inverse operation of subtraction to solve for r.
Inverse operation of subtraction is addition.
r - 6 = 6
Add 6 to each side.
r = 12
Example 9 :
Solve for p :
(p/3) - 4 = 0
Solution :
Here 4 is subtracted from (p/3).
So, we have to use the inverse operation of subtraction to solve for (p/3).
Inverse operation of subtraction is addition.
(p/3) - 4 = 0
Add 4 to each side.
p/3 = 4
Here p is divided by 3.
Then, we have to use the inverse operation of division to solve for p.
Inverse operation of division is multiplication.
p/3 = 4
Multiply each side by 3.
p = 12
Example 10 :
Solve for m :
10 - 3m = -5
Solution :
10 - 3m = -5
Add 3m to each side.
10 = -5 + 3m
Add 5 to each side.
15 = 3m
Divide each side by 3.
5 = m
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