Formulae
 A Formula is an expression or equation that expresses the relationship between certain quantities.
 For ExampleA = πr^{2 }is the formula to find the area of a circle of radius r units.
 From this formula, we can know the relationship between the radius and the area of a circle. The area of a circle varies directly as the square of its radius. Here π is the constant of variation.
Changing the Subject of a Formulae
Terminology
 In the formula
C =πd
Subject: C Rule: multiply πby diameter  The variable on the left, is known as the subject: What you are trying to find.
 The formula on the right, is the rule, that tells you how to calculate the subject.
 So, if you want to have a formula or rule that lets you calculate d, you need to make d, the subject of the formula.
 This is changing the subject of the formula from C to d.
 So clearly in the case above where
C = πd
We get C by multiplying π by the diameter
To calculate d, we need to divide the Circumference C by π
So d = ^{C}/_{π} and now we have d as the subject of the formula.
Method:
A formula is simply an equation, that you cannot solve, until you replace the letters with their values (numbers). It is known as a literal equation.
To change the subject, apply the same rules as we have applied to normal equations.
 Add the same variable to both sides.
 Subtract the same variable from both sides.
 Multiply both sides by the same variable.
 Divide both sides by the same variable.
 Square both sides
 Square root both sides.
Examples:
Make the letter in brackets the subject of the formula
 x + p = q [ x ]
Solution
(subtract p from both sides)
x = q – p  y  r = s [y ]
Solution
(add r to both sides)
y = s + r  P = RS[R]
Solution
(divide both sides by S)
R = ^{P}/_{S}  ^{A}/_{B}= L[A]
Solution
(multiply both sides by B)
A = LB  2w + 3 = y [w]
Solution
(subtract 3 from both sides)
2w = y – 3
(divide both sides by 2)
W = y – 3
2  P = ^{1}/_{3} Q[Q]
Solution
(multiply both sides by 3  get rid of fraction)
3P = Q  T = ^{2}/_{5} k[k]
Solution
(multiply both sides by 5 get rid of fraction)
5T = 2k
(divide both sides by 2)
^{5T}/_{2}= k Note that: ^{5T}/_{2 }is the same as ^{5}/_{2}T  A = πr^{2 }[r]
Solution
(divide both sides by pi)
^{A}/_{π}= r^{2}
(square root both sides)
√(^{A}/_{π}) = r  L = ½h −t[h]
Solution
(multiply both sides by 2)
2L = h −t
(add t to both sides)
2L + t = h
Example
Make d the subject of the formula G= √(^{d − x}/_{d − 1})
Solution
Squaring both sides
G^{2} = ^{d − x}/_{d − 1}
Multiply both sides by (d − 1)
G^{2}(d−1) = d − x
Expanding the L.H.S
dG^{2 }− G^{2 }= d − x
Collecting the terms containing d on the L.H.S
dG^{2 }− d = G^{2 }− x
Factorizing the L.H.S
d (G^{2 }− 1) = G^{2 }− x
Dividing both sides by
d = G^{2} − x
G^{2} − 1
Variation
In a formula some elements which do not change (fixed) under any condition are called constants while the ones that change are called variables. There are different types of variations.
 Direct Variation, where both variables either increase or decrease together
 Inverse or Indirect Variation, where when one of the variables increases, the other one decreases
 Joint Variation, where more than two variables are related directly
 Combined Variation, which involves a combination of direct or joint variation, and indirect variation
Examples
 Direct: The number of money I make varies directly (or you can say varies proportionally) with how much I work.
 Direct: The length of the side a square varies directly with the perimeter of the square.
 Inverse: The number of people I invite to my bowling party varies inversely with the number of games they might get to play (or you can say is proportional to the inverse of).
 Inverse: The temperature in my house varies indirectly (same as inversely) with the amount of time the air conditioning is running.
 Inverse: My school marks may vary inversely with the number of hours I watch TV.
Direct or Proportional Variation
When two variables are related directly, the ratio of their values is always the same. So as one goes up, so does the other, and if one goes down, so does the other.
Think of linear direct variation as a “y = mx” line, where the ratio of y to x is the slope (m). With direct variation, the yintercept is always 0 (zero); this is how it’s defined.
Direct variation problems are typically written:
→ y= kx where k is the ratio of y to x (which is the same as the slope or rate).
Some problems will ask for that k value (which is called the constant of variation or constant of proportionality ); others will just give you 3 out of the 4 values for x and y and you can simply set up a ratio to find the other value.
Remember the example of making ksh 1000 per week (y = 10x)? This is an example of direct variation, since the ratio of how much you make to how many hours you work is always constant.
Direct Variation Word Problem:
The amount of money raised at a school fundraiser is directly proportional to the number of people who attend. Last year, the amount of money raised for 100 attendees was $2500. How much money will be raised if 1 000 people attend this year?
Solution:
Let’s do this problem using both the Formula Method and the Proportion Method:
Formula Method  Explanation 
y = kx 2500= k100 k=25 
Since the amount of money is directly proportional (varies directly to the number who attend, we know that y = kx, where y = the amount of money raised and x = the number of attendees. Since the problem states that the amount of money is directly proportional to the number of attendees, we put the amount of money first, or as the y). 
y = 25x y=25(1000) y= 25000 
We need to fill in the numbers from the problem, and solve for k. We see that k = 25. So we have y = 25x. We plug the new x, which is 1000 We get the new y = 25000. So if 1000 people attend, $25,000 would be raised! 
Proportional Method  Explanation 
^{$$}/_{attendees }=^{$$}/_{attendees} ^{2500}/_{100} = ^{y}/_{100} 100y = 2500000 y = 25000 
We can set up a proportion with the y's on top (amount of money), and the x's on bottom number of attendees). We can then cross multiply to get the new amount of money (y). We get the new y = 25000. So if 1000 people attend, $25,000 will be raised! 
Direct Square Variation Word Problem
Again, a Direct Square Variation is when y is proportional to the square of x, or y= kx^{2}.
Example
If y varies directly with the square of x, and if y = 4 when x= 3, what is y when x= 2?
Solution:
Let’s do this with the formula method and the proportion method:
Formulae Method  notes 
y=kx^{2} y=^{4}/_{9}x^{2} 4 = k3^{2} y=^{4}/_{9}(2)^{2} k = ^{4}/_{9} y = ^{16}/_{9} 
Since y is directly proportional (varies directly) to the square of x, we know that y = kx^{2}. Plug in the first numbers we have for x and y to see that k =^{4}/_{9}
So we have y = ^{4}/_{9}x^{2}. We plug the new x, which is 2, and get the new y, which is ^{16}/_{9} 
Proportional method  notes 
y_{1} = y_{2} (x_{1})^{2} (x_{2})^{2} ^{4}/_{32} = ^{y}/_{22} y = 4 × 2^{2} = ^{16}/_{9} 3^{2} 
We can set up a proportion with the y's on top, and x^{2} on the bottom. We can plug in the numbers we have, and then cross multiply to get the new y. We then get the new y=^{16}/_{9} 
Example
The length (l) cm of a wire varies directly as the temperature T^{0}c.The length of the wire is 5 cm when the temperature is 65^{0}c. Calculate the length of the wire when the temperature is 69^{0}c.
Solution
l α T
Therefore l = KT
Substituting l =5 when T= 65^{0}c.
5 =k x 65
K = ^{5}/_{65} = ^{1}/_{13}
Therefore l = ^{1}/_{13}T
When T = 69
L = ^{1}/_{13} x69 = 5^{4}/_{13 }cm
Direct variation graph
Inverse or Indirect Variation
Inverse or Indirect Variation is refers to relationships of two variables that go in the opposite direction.
Let’s supposed you are comparing how fast you are driving (average speed) to how fast you get to your work.The faster you drive the earlier you get to your work. So as the speed increases time reduces and vice versa.
So the formula for inverse or indirect variation is:
→ y = ^{k}/_{x} or K = xy where k is always the same number or constant.
(Note that you could also have an Indirect Square Variation or Inverse Square Variation, like we saw above for a Direct Variation. This would be of the form → y = ^{k}/_{x2} or k= yx^{2} .)
Inverse Variation Word Problem:
So we might have a problem like this:
The value of y varies inversely with x, and y = 4 when x = 3. Find x when y = 6.
The problem can also be written as follows:
Let x_{1}= 3, y_{1}= 4, and y_{2}= 6. Let y vary inversely as x. Find x_{2}.
Solution:
We can solve this problem in one of two ways, as shown. We do these methods when we are given any three of the four values for x and y.
y_{1} = k

Since x and y vary inversely, we know that xy=k, or y = ^{k}/_{x} We first fill in the x and y values with x, and y, from the problem. Remember that the variables with the same subscript (x_{1}, y_{1}) stay together. We then solve for k, which is 12. The formula way may take a little more time, but you may be asked to do it this way, especially if you need to find k, and the equation of variation, which is y=^{12}/_{x}

Product Rule Method:
x_{1}y_{1}=x_{2}y_{2} (3)(4)= x_{2}(6) 12 = 6x_{2} 
We know that when you multiply the x's and y's (with the same subscript) we get a constant, which is k. You can see that k = 12 in this problem. So we can just substitute in all the numbers that we are given and solve for the number we want  in this case, x_{2}. This way is easier than the formula method, but, again, you will probably be asked to know both ways. 
Inverse Variation Word Problem:
Example 1
For the club, the number of tickets Moyo can buy is inversely proportional to the price of the tickets. She can afford 15 tickets that cost $5 each. How many tickets can she buy if each cost $3?
Solution:
Let’s use the product method:
x_{1}y_{1}=x_{2}y_{2 } (5(15)=x_{2}(3) 
We know that when you multiply the x's and y's we get a constant, which is k. So the number of tickets Allie can buy times the price of each ticket is k. We can let the x's be the price of the tickets. So we can just substitute in all the numbers that we are given and solve for the number we want. So we see that Allie can buy 25 tickets that cost $3. This makes sense, since we can see that she only can spend $75 (which is k!) 
Example 2
If 16 women working 7 hours day can paint a mural in 48 days, how many days will it take 14 women working 12 hours a day to paint the same mural?
Solution:
The three different values are inversely proportional; for example, the more women you have, the less days it takes to paint the mural, and the more hours in a day the women paint, the less days they need to complete the mural:
x_{1}y_{1}z_{1}=x_{2}y_{2}z_{2 } (16)(7)(48)=(14)(12)z_{2 } 
Since each woman is working at the same rate, we know that when we multiply the number of women (x) by the number of the hours a day (y) by the number of days they work (z), it should always be the same (a constant). (Try it yourself with some easy numbers). So we can just substitute in all the numbers that we are given and solve for the number we want (days). So we see that it would take 32 days for 14 women that work 12 hours a day to paint the mural. In this case, our kis 5376, which represents the number of hours it would take one woman alone to paint the mural. 
Joint Variation and Combined Variation
Joint variation is just like direct variation, but involves more than one other variable. All the variables are directly proportional, taken one at a time. Let’s do a joint variation problem:
Supposed x varies jointly with y and the square root of z. When x = –1 8 and y = 2, then z = 9. Find y when x = 10 and z = 4.
x=ky√z x=ky√z 
Again, we can set it up almost word for word from the word problem. For the words "varies jointly", just basically use the "=" sign, and everything else will fall in place. Solve for k first by plugging in variables we are given at first; we get k=3. Now we can plug in the new values of x and z to get the new y. So y will be ^{5}/3._{ }Really not that bad! 
Combined variation involves a combination of direct or joint variation, and indirect variation. Since these equations are a little more complicated, you probably want to plug in all the variables, solve for k, and then solve back to get what’s missing. Here is the type of problem you may get:
 y varies jointly as x and w and inversely as the square of z. Find the equation of variation when y = 100, x = 2, w = 4, and z = 20.
 Then solve for y when x = 1 , w = 5, and z = 4.
Solution:
y = kzw z^{2} 100 = k(2)(4) = 8k 8 y = 5000 x w (answer to a) z^{2} y= 5000(1)(5) 4^{2} y = 25000= 1562.5 (answer to b) 
Now this looks really complicated, and you may get "word problems" like this, but all we do is fill in all the variables we know, and then solve for k. We know that "the square of z" is a fancy way of saying z^{2}. Remember that what follows the "varies jointly as" is typically on the top of any fraction (this is like a direct variation), and what follows "inversely as" is typically on the bottom of the fraction. And always put k on the top. Now that we have the k, we have the answer to (a) above by plugging it in the original equation. Now we can get the new y when we have "new" x, w, and z values. So, for the second part of the problem, when x = 1, w = 5, and z= 4, y = 1562.5. (Just plug in). 
Example
The volume of wood in a tree (V) variesdirectly as the height (h) and inversely as the square of the girth (g). If the volume of a tree is 1 44 cubic meters when the height is 20 meters and the girth is 1.5 meters, what is the height of a tree with a volume of 1 000 and girth of 2 meters?
Solution:
V = k(height) 
We can set it up almost word for word from the word problem. For the words "varies directly", just basically use the "=" sign, and everything else will fall in place. Remember to put everything on top for direct variation (including k), unless the problem says "inversely as"; those go on bottom. Solve for k first; we get k = 16.2. Now we can plug in the new values to get the new height. So the new height is 246.1 meters.

Example
The average number of phone calls per day between two cities has found to be jointly proportional to the populations of the cities, and inversely proportional to the square of the distance between the two cities. The population of Charlotte is about 1,500,000 and the population of Nashville is about 1,200,000, and the distance between the two cities is about 400 miles. The average number of calls between the cities is about 200,000.
 Find the k and write the equation of variation.
 The average number of daily phone calls between Charlotte and Indianapolis (which has a population of about 1,700,000) is about 134,000. Find the distance between the two cities.
Solution:
It may be easier if you take it one step at a time.
C = kp_{1}p_{2}_{ } d^{2} 200000=k(1500000)(1200000) k = (200000)400^{2} = .01778 C=.01778p_{1}p_{2}_{ } ← answer to (a) 134000=.01778(1500000)(1700000) 134000d^{2 }=.01778(1500000)(1700000) d = 581.7 miles ← answer to (b) 
We can set it up almost word for word from the word problem. Solve for k first; we get k = .01778. Now we can plug in the new values to get the distance between the cities (d). We can actually cross multiply to get d, So the distance between Charlotte and Indianapolis is about 581.7 miles. In reality, the distance between these two cities is 585.6 miles, so we weren't too far off! 
Example
A varies directly as B and inversely as the square root of C. Find the percentage change in A when B is decreased by 10 % and C increased by 21 %.
Solution
A= KB ………………………(1)
√C
A change in B and C causes a change in A
A_{1 }= KB_{1 }………………………. (2)
√C_{1}
B_{1} =^{90}/_{100} B
= 0.9B
C_{1} = ^{121}/_{100}C
= 1 .21 C
Substituting B1 and C1 in equation (2)
A_{1} = K0.9B
√1.21C
=^{0.9}/_{1.1}(K^{B}/_{√C})
= ^{9}/_{11} A
Percentage change in A =A_{1−}A × 100%
A
=^{ 9}/_{11}A  A ×100%
A
=  18 ^{2}/_{11} %
Therefore A decreases 18 ^{2}/_{11} %
Partial Variation
The general linear equation y =mx +c, where m and c are constants, connects two variables x and y.in such case we say that y is partly constant and partly varies as x.
Example
A variable y is partly constant and partly varies as if x = 2 when y=7 and x =4 when y =1 1 , find the equation connecting y and x.
Solution
The required equation is y = kx + c where k and c are constants
Substituting x = 2 ,y =7 and x =4, y =1 1 in the equation gives ;
7 =2k +c …………………..(1 )
1 1 = 4k +c …………………(2)
Subtracting equation 1 from equation 2 ;
4 = 2 k
Therefore k =2
Substituting k =2 in the equation 1;
C =7 – 4
C =3
Therefore the equation required is y=2x +3
Past KCSE Questions on the Topic.
 The volume Vcm^{3 }of an object is given by
Express c in term of π r, s and V  Make V the subject of the formula
T = ^{1}/_{2}m(u^{2 }– v^{2})  Given that y = b – bx^{2 }make x the subject
cx^{2 }– a  Given that log y = log (10^{n}) make n the subject
 A quantity T is partly constant and partly varies as the square root of S.
 Using constants a and b, write down an equation connecting T and S.
 If S = 1 6, when T = 24 and S = 36 when T = 32, find the values of the constants a and b,
 A quantity P is partly constant and partly varies inversely as a quantity q, given that p = 10 when q = 1.5 and p = 20, when q = 1.25, find the value of p when q= 0.5
 Make y the subject of the formula p = ^{xy}/_{x−y}
 Make P the subject of the formula
P^{2 }= (P – q)(P – r)  The density of a solid spherical ball varies directly as its mass and inversely as the cube of its radius. When the mass of the ball is 500g and the radius is 5 cm, its density is 2 g per cm^{3}. Calculate the radius of a solid spherical ball of mass 540 density of 10g per cm^{3}
 Make s the subject of the formula
 The quantities t, x and y are such that t varies directly as x and inversely as the square root of y. Find the percentage in t if x decreases by 4% when y increases by 44%
 Given that y is inversely proportional to x^{n }and k as the constant of proportionality;

 Write down a formula connecting y, x, n and k
 If x = 2 when y = 12 and x = 4 when y = 3, write down two expressions for k in terms of n.
Hence, find the value of n and k.
 Using the value of n obtained in (a) (ii) above, find y when x = 5^{1}/_{3}

 The electrical resistance, R ohms of a wire of a given length is inversely proportional to the square of the diameter of the wire, d mm. If R = 2.0 ohms when d = 3mm. Find the vale R when d = 4 mm.
 The volume Vcm^{3 }of a solid depends partly on r and partly on r where rcm is one of the dimensions of the solid.
When r = 1 , the volume is 54.6 cm^{3 }and when r = 2, the volume is 226.8 cm^{3} Find an expression for V in terms of r
 Calculate the volume of the solid when r = 4
 Find the value of r for which the two parts of the volume are equal
 The mass of a certain metal rod varies jointly as its length and the square of its radius. A rod 40 cm long and radius 5 cm has a mass of 6 kg. Find the mass of a similar rod of length 25 cm and radius 8 cm.
 Make x the subject of the formula
P= xy
z + x  The charge c shillings per person for a certain service is partly fixed and partly inversely proportional to the total number N of people.
 Write an expression for c in terms on N
 When 100 people attended the charge is Kshs 8700 per person while for 35 people the charge is Kshs 1 0000 per person.
 If a person had paid the full amount charge is refunded. A group of people paid but ten percent of organizer remained with Kshs 574000.
Find the number of people.
 Write an expression for c in terms on N
 Two variables A and B are such that A varies partly as B and partly as the square root of B given that A=30, when B=9 and A=16 when B=14, find A when B=36.
 Make p the subject of the formula
A = −EP
(√P)^{2} + N
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