Simple Equations, Variations and Applications for SS 1

Simple Equations and Variations

Performance Objectives

By the end of the lesson, students should be able to:

1. change subject of the formula of given equations involving brackets, roots and indices;
2. substitute given values into re- arranged equations;
3. distinguish among the types of variation;
4. apply variation into real life situations.

Introduction to Equations and Formulas

In mathematics, an equation is a statement that shows two mathematical expressions are equal. It often contains one or more unknown values, represented by letters called variables. For example, $2x + 5 = 11$ is an equation where $x$ is the unknown variable.

A formula is a special type of equation that expresses a relationship between several quantities. It is a rule or principle often expressed in mathematical symbols. For example, the formula for the area of a rectangle is $A = lw$, where $A$ is the area, $l$ is the length, and $w$ is the width.

In a formula, one variable is usually expressed in terms of others. This variable is called the subject of the formula. For instance, in $A = lw$, $A$ is the subject of the formula because it is isolated on one side of the equation.

Change of Subject of a Formula

Changing the subject of a formula means rearranging the formula to make a different variable the subject. This involves using inverse operations to isolate the desired variable on one side of the equation. The key principle is to perform the same operation on both sides of the equation to maintain balance.

Changing the Subject Involving Basic Operations

To change the subject of a formula, operations like addition, subtraction, multiplication, and division are reversed. For example, if a term is added, subtract it from both sides. If a term is multiplied, divide both sides by it.

Example 1: Make $x$ the subject of the formula $y = mx + c$.

Solution:

1. Start with the formula: $y = mx + c$
2. Subtract $c$ from both sides: $y – c = mx$
3. Divide both sides by $m$: $\frac{y – c}{m} = x$
4. So, $x = \frac{y – c}{m}$

Example 2: Make $h$ the subject of the formula for the area of a trapezium, $A = \frac{1}{2}(a + b)h$.

Solution:

1. Start with the formula: $A = \frac{1}{2}(a + b)h$
2. Multiply both sides by 2: $2A = (a + b)h$
3. Divide both sides by $(a + b)$: $\frac{2A}{(a + b)} = h$
4. So, $h = \frac{2A}{a + b}$

Changing the Subject Involving Brackets

When brackets are present, it is often necessary to expand them first or treat the bracketed term as a single unit, depending on the position of the variable to be made the subject.

Example 3: Make $p$ the subject of the formula $Q = 3(p – r)$.

Solution:

1. Start with the formula: $Q = 3(p – r)$
2. Divide both sides by 3: $\frac{Q}{3} = p – r$
3. Add $r$ to both sides: $\frac{Q}{3} + r = p$
4. So, $p = \frac{Q}{3} + r$

Example 4: Make $x$ the subject of the formula $A = P(1 + \frac{x}{100})$.

Solution:

1. Start with the formula: $A = P(1 + \frac{x}{100})$
2. Divide both sides by $P$: $\frac{A}{P} = 1 + \frac{x}{100}$
3. Subtract 1 from both sides: $\frac{A}{P} – 1 = \frac{x}{100}$
4. Multiply both sides by 100: $100(\frac{A}{P} – 1) = x$
5. So, $x = 100(\frac{A}{P} – 1)$

Changing the Subject Involving Roots

To remove a square root, square both sides of the equation. To remove a cube root, cube both sides, and so on.

Example 5: Make $r$ the subject of the formula $T = \sqrt{\frac{2r}{g}}$.

Solution:

1. Start with the formula: $T = \sqrt{\frac{2r}{g}}$
2. Square both sides: $T^2 = \frac{2r}{g}$
3. Multiply both sides by $g$: $gT^2 = 2r$
4. Divide both sides by 2: $\frac{gT^2}{2} = r$
5. So, $r = \frac{gT^2}{2}$

Example 6: Make $x$ the subject of the formula $y = 5 + \sqrt{x – 2}$.

Solution:

1. Start with the formula: $y = 5 + \sqrt{x – 2}$
2. Subtract 5 from both sides: $y – 5 = \sqrt{x – 2}$
3. Square both sides: $(y – 5)^2 = x – 2$
4. Add 2 to both sides: $(y – 5)^2 + 2 = x$
5. So, $x = (y – 5)^2 + 2$

Changing the Subject Involving Indices

To remove a square (index of 2), take the square root of both sides. To remove a cube (index of 3), take the cube root of both sides, and so on.

Example 7: Make $a$ the subject of the formula $V = \frac{1}{3}\pi a^2 h$.

Solution:

1. Start with the formula: $V = \frac{1}{3}\pi a^2 h$
2. Multiply both sides by 3: $3V = \pi a^2 h$
3. Divide both sides by $\pi h$: $\frac{3V}{\pi h} = a^2$
4. Take the square root of both sides: $\sqrt{\frac{3V}{\pi h}} = a$
5. So, $a = \sqrt{\frac{3V}{\pi h}}$

Example 8: Make $t$ the subject of the formula $E = mc^2 + \frac{1}{2}mv^2$.

Solution:

This example seems to have a typo in the original problem statement as $t$ is not present. Let’s assume the formula is $E = \frac{1}{2}kt^2$ and we need to make $t$ the subject.

1. Start with the formula: $E = \frac{1}{2}kt^2$
2. Multiply both sides by 2: $2E = kt^2$
3. Divide both sides by $k$: $\frac{2E}{k} = t^2$
4. Take the square root of both sides: $\sqrt{\frac{2E}{k}} = t$
5. So, $t = \sqrt{\frac{2E}{k}}$

Substitution into Rearranged Equations

After changing the subject of a formula, the new formula can be used to calculate the value of the new subject by substituting given numerical values for the other variables. This is a practical application of rearranging formulas.

Example 9: Using the formula $y = mx + c$, if $y = 17$, $m = 3$, and $c = 2$, find the value of $x$. First, make $x$ the subject.

Solution:

1. First, make $x$ the subject of $y = mx + c$:
   $y – c = mx$
   $x = \frac{y – c}{m}$
2. Now, substitute the given values: $y = 17$, $m = 3$, $c = 2$.
   $x = \frac{17 – 2}{3}$
   $x = \frac{15}{3}$
   $x = 5$

Example 10: The formula for the volume of a cone is $V = \frac{1}{3}\pi r^2 h$. If $V = 154 \text{ cm}^3$, $\pi = \frac{22}{7}$, and $h = 12 \text{ cm}$, find the radius $r$. First, make $r$ the subject.

Solution:

1. First, make $r$ the subject of $V = \frac{1}{3}\pi r^2 h$:
   $3V = \pi r^2 h$
   $\frac{3V}{\pi h} = r^2$
   $r = \sqrt{\frac{3V}{\pi h}}$
2. Now, substitute the given values: $V = 154$, $\pi = \frac{22}{7}$, $h = 12$.
   $r = \sqrt{\frac{3 \times 154}{\frac{22}{7} \times 12}}$
   $r = \sqrt{\frac{462}{\frac{264}{7}}}$
   $r = \sqrt{462 \times \frac{7}{264}}$
   $r = \sqrt{\frac{3234}{264}}$
   $r = \sqrt{12.25}$
   $r = 3.5$
3. So, the radius $r = 3.5 \text{ cm}$.

Introduction to Variation

Variation describes how one quantity changes in relation to one or more other quantities. It helps to model real-life relationships where quantities are dependent on each other. The relationship is often expressed using a constant of proportionality, usually denoted by $k$.

Types of Variation

1. Direct Variation

Direct variation occurs when one quantity increases or decreases at the same rate as another quantity. This means that the ratio of the two quantities is constant.

If $y$ varies directly as $x$, it is written as $y \propto x$. The mathematical relationship is $y = kx$, where $k$ is the constant of proportionality. This constant $k$ can be found by dividing $y$ by $x$ ($k = \frac{y}{x}$).

Real-life Applications:

1. The cost of buying a certain number of identical items varies directly with the number of items. (Cost $\propto$ Number of items)
2. The distance covered by a car moving at a constant speed varies directly with the time taken. (Distance $\propto$ Time)
3. The amount of work done varies directly with the number of workers (assuming constant work rate). (Work $\propto$ Number of workers)

Example 11: If $y$ varies directly as $x$, and $y = 12$ when $x = 4$, find:

1. The constant of proportionality, $k$.
2. The equation connecting $y$ and $x$.
3. The value of $y$ when $x = 7$.

Solution:

1. Since $y$ varies directly as $x$, $y = kx$.
   Substitute $y = 12$ and $x = 4$:
   $12 = k \times 4$
   $k = \frac{12}{4}$
   $k = 3$
2. The equation connecting $y$ and $x$ is $y = 3x$.
3. To find $y$ when $x = 7$:
   $y = 3 \times 7$
   $y = 21$

Example 12: The cost $C$ of petrol varies directly as the number of litres $L$ bought. If 5 litres of petrol cost N1200, how much will 8 litres cost?

Solution:

1. Since $C$ varies directly as $L$, $C = kL$.
   Given $C = 1200$ when $L = 5$:
   $1200 = k \times 5$
   $k = \frac{1200}{5}$
   $k = 240$
2. The equation is $C = 240L$.
3. To find the cost of 8 litres, substitute $L = 8$:
   $C = 240 \times 8$
   $C = 1920$
4. So, 8 litres of petrol will cost N1920.

2. Inverse Variation

Inverse variation occurs when one quantity increases as the other quantity decreases, and vice versa. Their product remains constant.

If $y$ varies inversely as $x$, it is written as $y \propto \frac{1}{x}$. The mathematical relationship is $y = \frac{k}{x}$ or $xy = k$, where $k$ is the constant of proportionality.

Real-life Applications:

1. The time taken to complete a job varies inversely with the number of workers (assuming constant work rate). (Time $\propto \frac{1}{\text{Number of workers}}$)
2. The speed of a vehicle varies inversely with the time taken to cover a fixed distance. (Speed $\propto \frac{1}{\text{Time}}$)
3. The pressure of a gas varies inversely with its volume (at constant temperature). (Pressure $\propto \frac{1}{\text{Volume}}$)

Example 13: If $y$ varies inversely as $x$, and $y = 6$ when $x = 2$, find:

1. The constant of proportionality, $k$.
2. The equation connecting $y$ and $x$.
3. The value of $y$ when $x = 3$.

Solution:

1. Since $y$ varies inversely as $x$, $y = \frac{k}{x}$.
   Substitute $y = 6$ and $x = 2$:
   $6 = \frac{k}{2}$
   $k = 6 \times 2$
   $k = 12$
2. The equation connecting $y$ and $x$ is $y = \frac{12}{x}$.
3. To find $y$ when $x = 3$:
   $y = \frac{12}{3}$
   $y = 4$

Example 14: The time $T$ taken to travel a certain distance varies inversely as the speed $S$. If it takes 4 hours to travel at a speed of 60 km/h, how long will it take to travel the same distance at a speed of 80 km/h?

Solution:

1. Since $T$ varies inversely as $S$, $T = \frac{k}{S}$.
   Given $T = 4$ when $S = 60$:
   $4 = \frac{k}{60}$
   $k = 4 \times 60$
   $k = 240$
2. The equation is $T = \frac{240}{S}$.
3. To find the time when $S = 80$:
   $T = \frac{240}{80}$
   $T = 3$
4. So, it will take 3 hours to travel the same distance at 80 km/h.

3. Joint Variation

Joint variation occurs when one quantity varies directly as the product of two or more other quantities.

If $y$ varies jointly as $x$ and $z$, it is written as $y \propto xz$. The mathematical relationship is $y = kxz$, where $k$ is the constant of proportionality.

Real-life Applications:

1. The volume of a cylinder varies jointly as its height and the square of its radius. ($V \propto hr^2 \implies V = khr^2$)
2. The simple interest earned on an investment varies jointly as the principal amount, the interest rate, and the time. ($I \propto PRT \implies I = kPRT$)
3. The force exerted by a spring varies jointly as its stiffness constant and its extension. ($F \propto kx$) – this is a direct variation, not joint. Let’s use a better example.
4. The area of a rectangle varies jointly as its length and width. ($A \propto lw \implies A = klw$, where $k=1$)

Example 15: If $P$ varies jointly as $Q$ and $R$, and $P = 24$ when $Q = 3$ and $R = 2$, find:

1. The constant of proportionality, $k$.
2. The equation connecting $P$, $Q$, and $R$.
3. The value of $P$ when $Q = 5$ and $R = 4$.

Solution:

1. Since $P$ varies jointly as $Q$ and $R$, $P = kQR$.
   Substitute $P = 24$, $Q = 3$, and $R = 2$:
   $24 = k \times 3 \times 2$
   $24 = 6k$
   $k = \frac{24}{6}$
   $k = 4$
2. The equation connecting $P$, $Q$, and $R$ is $P = 4QR$.
3. To find $P$ when $Q = 5$ and $R = 4$:
   $P = 4 \times 5 \times 4$
   $P = 80$

Example 16: The volume $V$ of a rectangular prism varies jointly as its length $l$, width $w$, and height $h$. If a prism with $l = 5 \text{ cm}$, $w = 3 \text{ cm}$, and $h = 2 \text{ cm}$ has a volume of $30 \text{ cm}^3$, find the volume of a prism with $l = 6 \text{ cm}$, $w = 4 \text{ cm}$, and $h = 3 \text{ cm}$.

Solution:

1. Since $V$ varies jointly as $l$, $w$, and $h$, $V = klwh$.
   Given $V = 30$ when $l = 5$, $w = 3$, $h = 2$:
   $30 = k \times 5 \times 3 \times 2$
   $30 = 30k$
   $k = \frac{30}{30}$
   $k = 1$
2. The equation is $V = lwh$.
3. To find the volume when $l = 6$, $w = 4$, and $h = 3$:
   $V = 6 \times 4 \times 3$
   $V = 72$
4. So, the volume of the second prism is $72 \text{ cm}^3$.

4. Partial Variation

Partial variation occurs when a quantity is partly constant and partly varies directly with another quantity.

If $y$ varies partially as $x$, it means $y$ is made up of a constant part and a part that varies directly as $x$. The mathematical relationship is $y = a + kx$, where $a$ is a constant and $k$ is the constant of proportionality.

Real-life Applications:

1. The total cost of a taxi ride often consists of a fixed charge (constant) plus a charge per kilometre (varies directly with distance). (Cost $= \text{Fixed charge} + k \times \text{Distance}$)
2. A worker’s weekly wage might be a fixed basic salary plus a commission that varies directly with sales. (Wage $= \text{Basic salary} + k \times \text{Sales}$)
3. The cost of electricity might include a fixed service charge plus a charge per unit consumed. (Cost $= \text{Fixed charge} + k \times \text{Units consumed}$)

Example 17: $C$ varies partially as $n$. When $n = 2$, $C = 10$, and when $n = 5$, $C = 19$. Find:

1. The values of the constants $a$ and $k$.
2. The equation connecting $C$ and $n$.
3. The value of $C$ when $n = 8$.

Solution:

1. Since $C$ varies partially as $n$, the relationship is $C = a + kn$.
   Using the given values, we form two equations:
   When $n = 2$, $C = 10 \implies 10 = a + 2k$ (Equation 1)
   When $n = 5$, $C = 19 \implies 19 = a + 5k$ (Equation 2)
   Subtract Equation 1 from Equation 2:
   $(19 – 10) = (a – a) + (5k – 2k)$
   $9 = 3k$
   $k = \frac{9}{3}$
   $k = 3$
   Substitute $k = 3$ into Equation 1:
   $10 = a + 2(3)$
   $10 = a + 6$
   $a = 10 – 6$
   $a = 4$
   So, the constants are $a = 4$ and $k = 3$.
2. The equation connecting $C$ and $n$ is $C = 4 + 3n$.
3. To find $C$ when $n = 8$:
   $C = 4 + 3(8)$
   $C = 4 + 24$
   $C = 28$

Example 18: The cost of hiring a car, $N$, is partly constant and partly varies with the number of days, $d$, the car is hired. If it costs N5000 for 2 days and N8000 for 5 days, find the cost of hiring the car for 7 days.

Solution:

1. Let the relationship be $N = a + kd$.
   When $d = 2$, $N = 5000 \implies 5000 = a + 2k$ (Equation 1)
   When $d = 5$, $N = 8000 \implies 8000 = a + 5k$ (Equation 2)
   Subtract Equation 1 from Equation 2:
   $(8000 – 5000) = (a – a) + (5k – 2k)$
   $3000 = 3k$
   $k = \frac{3000}{3}$
   $k = 1000$
   Substitute $k = 1000$ into Equation 1:
   $5000 = a + 2(1000)$
   $5000 = a + 2000$
   $a = 5000 – 2000$
   $a = 3000$
2. The equation is $N = 3000 + 1000d$.
3. To find the cost for 7 days, substitute $d = 7$:
   $N = 3000 + 1000(7)$
   $N = 3000 + 7000$
   $N = 10000$
4. So, the cost of hiring the car for 7 days is N10,000.

5. Joint Partial Variation

Joint partial variation occurs when a quantity is partly constant and partly varies jointly with two or more other quantities.

If $y$ varies partly constantly and partly jointly as $x$ and $z$, the mathematical relationship is $y = a + kxz$, where $a$ is a constant and $k$ is the constant of proportionality.

Real-life Applications:

1. The total cost of manufacturing an item might be a fixed overhead cost plus a cost that varies jointly with the number of units produced and the complexity factor of the design. (Cost $= \text{Fixed overhead} + k \times \text{Units} \times \text{Complexity}$)
2. The total force required to move an object might be a constant initial friction plus a force that varies jointly with the mass of the object and its acceleration. (Force $= \text{Initial friction} + k \times \text{Mass} \times \text{Acceleration}$)

Example 19: $M$ varies partly constantly and partly jointly as $P$ and $Q$. When $P = 1$, $Q = 2$, $M = 10$. When $P = 2$, $Q = 3$, $M = 22$. Find:

1. The values of the constants $a$ and $k$.
2. The equation connecting $M$, $P$, and $Q$.
3. The value of $M$ when $P = 3$ and $Q = 4$.

Solution:

1. Since $M$ varies partly constantly and partly jointly as $P$ and $Q$, the relationship is $M = a + kPQ$.
   Using the given values, we form two equations:
   When $P = 1$, $Q = 2$, $M = 10 \implies 10 = a + k(1)(2) \implies 10 = a + 2k$ (Equation 1)
   When $P = 2$, $Q = 3$, $M = 22 \implies 22 = a + k(2)(3) \implies 22 = a + 6k$ (Equation 2)
   Subtract Equation 1 from Equation 2:
   $(22 – 10) = (a – a) + (6k – 2k)$
   $12 = 4k$
   $k = \frac{12}{4}$
   $k = 3$
   Substitute $k = 3$ into Equation 1:
   $10 = a + 2(3)$
   $10 = a + 6$
   $a = 10 – 6$
   $a = 4$
   So, the constants are $a = 4$ and $k = 3$.
2. The equation connecting $M$, $P$, and $Q$ is $M = 4 + 3PQ$.
3. To find $M$ when $P = 3$ and $Q = 4$:
   $M = 4 + 3(3)(4)$
   $M = 4 + 36$
   $M = 40$

Example 20: The cost $C$ of producing a certain item is partly fixed and partly varies jointly with the number of units produced $U$ and the quality index $I$. The fixed cost is N500. If it costs N1100 to produce 2 units with a quality index of 1, find the cost of producing 3 units with a quality index of 2.

Solution:

1. The relationship is $C = a + kUI$.
   Given the fixed cost $a = 500$. So, $C = 500 + kUI$.
   When $U = 2$, $I = 1$, $C = 1100$:
   $1100 = 500 + k(2)(1)$
   $1100 = 500 + 2k$
   $1100 – 500 = 2k$
   $600 = 2k$
   $k = \frac{600}{2}$
   $k = 300$
   So, the constants are $a = 500$ and $k = 300$.
2. The equation is $C = 500 + 300UI$.
3. To find the cost of producing 3 units with a quality index of 2, substitute $U = 3$ and $I = 2$:
   $C = 500 + 300(3)(2)$
   $C = 500 + 1800$
   $C = 2300$
4. So, the cost of producing 3 units with a quality index of 2 is N2300.