Partial Variation deals with quantity or variable that depends on another independent variable where one of the independent variable remains the same always and the other changes according to the quantity of the independent variable used.

For example in the electricity bill, the total amount depends on two quantities rental charge and the number of wants consumed by the customer. The rental change is always a constant. The bill can be in the forms.

T = a + bw, where a and b are constants problems on partial variation always result in the solution of a pair of simultaneous equations. The equations are then solved to find the constant.

Having known this fact, let start solving problems.

**Example 1:**

The force E needed to make a machine pull a load is partly constant and partly varies as the load to be pulled itself. When the load is 20g, the force needed is 1.4N, whereas the force needed for 30g load is 2N. Find the

- Law connecting the load and the force
- Force for the load of 50g

**Solution**

This means that the force is constant while the load varies. The two constants are a and b

Force = E = a (Force is constant and ExL, E = bL (the force varies with the load)

Adding them together yields

E = a + bL

Where a and b are constant

When L = 20g, E = 1.4N

1.4 = a + 20b —– (1)

Where E = 2N, L = 30g

2 = a + 30b —– (2)

Solve equation 1 and two simultaneously by elimination method

Subtract equation 1 from 2

2 – 1.4 = 30b – 20b

0.6/10 = 10b/10

b = 0.06

Substitute b = 0.06 into equation 1

1.4 = a + 20b

1.4 = a + 20 x 0.06

1.4 = a + 1.2

a = 0.2

- The law connecting the load and the force is by substituting the value of a and b into the initial equation

The law is

E = 0.2 + 0.06L

- Force (E) when load (L) = 50g

E = 0.2 + 0.06 x 50

E = 0.2 + 3

E = 3.2N

**Example 2:**

S is partly constant and partly varies with T. where S = 530, T = 1600, when S = 730, T = 3600, Find:

- The formula connecting S and T
- S when T = 1300

Solution

S = a – b, S x T, 8 =

For S x T, remove the sign of proportionality X and put a constant b

S = bT — (2)

Add equation 1 and 2

S = a + bT —– (3)

When S = 530, T = 1600

530 = a + 1600b — (4)

When S = 730, T = 3600

730 = a + 3600b —- (5)

Subtract equation 4 from 5

730 – 530 = 3600b – 1600b

200/200 = 2000/2000

b = 1/10 = 0.1

Substitute b = 0.1 into equation 4

530 = a + 1600 x 0.1

530 = a + 160

a = 530 – 160

a = 370 and b = 1/10 = 0.1

- The formula connecting S and T

Is S = 370 + 1/10 T

- S When T = 1300

S = 370 + 0.1 x 1300

S = 370 + 130

S = 500

**Read Also:**

**How to Solve nth Term of G.P****See How to Construct Quadratic Equation****How to Solve Equation Involving Fractions****Sum of Terms in G.P****The best way to Solve Sequence****How to Solve Quadratic Equation by Factorization**

**Example 3:**

The charge C of a telephone company is partly constant and partly varies as the number of units of call U. the cost of 50 units is N2500 and the cost of 120 units is N3,000. Find.

- The formula connecting C and U
- U when C N4,000

Solution

C = charge, U = number of units

C = a and C x U

C = bu

The equation then becomes

C = a + bu—-(1)

When U = 50, C = N2500

2500 = a + 50b – (2)

When C = N3000, U = 120

3000 = a + 120b —- (3)

Subtract equation 2 from 3

3000 – 2500 – a – a + 120b – 50b

500/70 = 70b/70

b = 500/70 = 50/7

Substitute b = 50/7 into equation 2

2500 = a + 50 x 50/7

2500 = a + 2500/7

2500 = a + 2500/7

Multiply both sides of the equation by 7 to clear the fraction

7 x 2500 – 7xa + 7(2500)/7

17500 = 7a + 2500

Collect like terms

17500 – 2500 = 7a

1500/7 = 7a/7

a = 15000/7

- The formula connecting C and U is by substituting a = 15000/7 and b = 50/7into the initial formula (equation 1)

C= a + bU

C = 15000/7 + 50U/7

Multiply each term by 7 to clear the fraction

7xC = 7(15000)/7 + 500 x 7/7

7C = 15000 x 50U

Therefore, the formula connecting C and U is

7C = 15000 + 50U

- U when C = N4000

Substitute C = N4000 into the formula connecting C and U

7×4000 = 15000 + 50U

28,000 = 15000 + 50U

Collect like terms

28000-15000 = 50U

13000/50 = 50U/50

U = 260 units

**Example 4:**

The resistance r to the motion of a car is partly constant and partly proportional to the square of the speed V. when the sped is 30km/h, the resistance is 190 Newton and when the speed is 50km/h, the resistance is 350 Newton

Find for what speed the resistance is 302.5 Newton

Solution

Resistance = r

Speed = V

r = a and r x r^{2}

r = br^{2}

the equation becomes

r = a + br^{2} —– (1)

when V = 30km/h, r = 190newton

190 = a + (30)^{2}b

190 = a + 900b —- (2)

When r = 50km/h, r = 350newton

350 = a + (50)^{2}b

350 = a + 2500b —- (3)

Subtract equation 2 from equation 3

350 – 190 = a – a + 2500b – 900b

160 = 1600b

b = 160/16000 = 1/10 = 0.1

Substitute b = 0.1 into equation 2

190 = a + 900 x 0.1

190 = a + 90

Collect like terms

a = 190 – 90

When r = 302.5 newton

302.5 = 100 + 0.1r^{2}

302.5 – 100 = 0.1r^{2}

202.5/0.1 = 0.1r^{2} /0.1

r^{2} = 2025

Take square root of both sides

r^{2×1/2} =

r = 45km/h

In conclusion, some believe partial variation is a very difficult aspect of variation. But today after this lesson I believe you can prove them wrong.

The key thing you need to know about partial variation is that there are two constant and it always leads to a pair of the simultaneous equation which requires using either substitution or elimination method to solve.

Just follow the pattern we have established today, I believe you will not have any challenge on this topic.

Correction to example 4.

The law connecting r and V should be : r=a+bV² √ and not r=a+br² ×