Chapter One: Algebra – Additional Mathematics Form Two
Introduction
The relationship between different quantities can easily be understood when symbols or alphabetical letters are used. For instance, if the price of an item and the number of items to be purchased are known, then the amount of money to be spent on the items can be determined.
The use of symbols or letters helps in formulating suitable simultaneous equations which describe the relationship among the quantities involved. The resulting equations can be solved to determine the value of every variable. The representation of quantities and numbers in symbols or letters is known as algebra.
In this chapter, you will learn how to solve simultaneous equations. The competencies developed will help you in solving different problems in business and financial management, cooking, sports, economics, engineering, health and fitness, among many other applications.
Simultaneous Equations
Simultaneous equations is a set of two or more equations each contains two or more variables whose values simultaneously satisfy each of the equations in the set. They are called simultaneous equations because their solution is obtained at the same time.
Algebraic Solutions
The solution of simultaneous equations involving linear and quadratic equations can be obtained algebraically by substitution method. In this case, one variable from one of the given equations is expressed in terms of the other and substituted into the unused equation. This results into one equation with only one variable which can be easily solved. The value obtained is substituted in one of the original equations to get the value of the second variable.
Activity 1: Solving simultaneous equations by substitution method
Individually or in a group, perform the following tasks:
- Write two variable simultaneous equations of your choice involving a linear and a quadratic equation.
- From the linear equation in task 1, make one of the variables the subject.
- Substitute the value of the variable obtained in task 2 into the quadratic equation in task 1.
- Solve the quadratic equation in task 3 to obtain the value(s) of the other variable.
- Substitute the value(s) obtained in task 4 into the equation obtained in task 2 to obtain the value(s) of the variable in task 2.
- Substitute the value(s) obtained in task 4 and 5 into the system of equations you obtained in task 1 to verify the solution.
- What conclusion can you draw from task 6?
- Identify activities that you think involves the concept of simultaneous equations in your everyday life.
- Share the results with other students through discussion for more inputs.
Example 1
Solve the following simultaneous equations:
x + y = 5
x² – y² = 5
Solution
Let the two equations be defined as follows:
x + y = 5 …… (i)
x² – y² = 5 …… (ii)
From equation (i) make x the subject of the formula to obtain,
x = 5 – y …… (iii)
Substituting equation (iii) into equation (ii) gives,
(5 – y)² – y² = 5 …… (iv)
Expand the first term of equation (iv) to obtain,
25 – 10y + y² – y² = 5
⇒ 25 – 10y = 5
⇒ 10y = 20
⇒ y = 2
Substituting y = 2 into equation (ii) gives
x² – 2² = 5
Rearrange the equation to get,
x² = 9
x = ±3
⇒ x = 3 and y = 2 or x = -3 and y = 2
But, the set of values x = -3 and y = 2 does not satisfy both equations.
Therefore, x = 3 and y = 2.
Example 2
Solve the following simultaneous equations:
y = x² + 1
y – x = 1
Solution
Let the two equations be defined as follows:
y = x² + 1 …… (i)
y – x = 1 …… (ii)
Substituting equation (i) into equation (ii) gives,
x² + 1 – x = 1
⇒ x² – x + 1 – 1 = 0
This implies that,
x² – x = 0
Factorize the quadratic equation to obtain,
x(x – 1) = 0
⇒ either x = 0 or x – 1 = 0
Thus, x = 0 or x = 1
Substituting x = 0 into equation (i) gives,
y = 0² + 1
⇒ y = 1
Substituting x = 1 into equation (i) gives,
y = 1² + 1
⇒ y = 2
Thus, x = 0, y = 1 or x = 1, y = 2.
Therefore, the solutions are x = 0, y = 1 and x = 1, y = 2.
Exercise 1
Solve each of the following simultaneous equations:
- x² + 8y = 13 and 2y + x = 2
- 3x² – 14x – 5 = y and y = 4x – 32
- x² + xy + y² = 28 and y = 2 – x
- y = x² – 2x + 1 and y = ½x
- x² + y² + 2x = 17 and y – x = 1
- xy = 12 and y = x – 1
- x² + y² – xy = 7 and y + x = 1
- x² + y² = 17 and y = x + 5
- x² + xy = 18 and 3x – y = 6
- x² – 4xy + 4y² = 4 and 3x + 2y = 10
- x² + 3x – y = 0 and x – y + 3 = 0
- x – y + 5 = 0 and x² + y² – 25 = 0
Graphical Solutions
Simultaneous equations involving one linear and one quadratic equation have one or two pairs of solutions. The solutions can be obtained graphically by plotting the graphs of linear and quadratic equations on the same xy-plane. The points of intersection of the two graphs give the solutions of the two equations.
Activity 2: Solving simultaneous equations graphically
Individually or in a group, perform the following tasks:
- Write two variable simultaneous equations of your choice involving a linear and a quadratic equation.
- Prepare a table of values for each equation in task 1.
- Use the tables of values obtained in task 2 to draw the graphs of the equations in task 1 on the same xy-plane.
- Locate and write the points of intersection of the graphs in task 3.
- Verify if the points obtained in task 4 are the solutions of the equations in task 1.
- Share your results with other students through discussion for more inputs.
From Activity 1.2, it can be observed that the points of intersection of the two graphs are the solutions to the given linear and quadratic equations.
Example 4
Use graphical method to find the value of x and y in the following system of simultaneous equations:
y = x² – 3x + 2
y = x – 1
Solution
First prepare a table of values for the given equations for few values of x as shown in the following table:
| x | -2 | -1 | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|---|---|
| y = x² – 3x + 2 | 12 | 6 | 2 | 0 | 0 | 2 | 6 | 12 |
| y = x – 1 | -3 | -2 | -1 | 0 | 1 | 2 | 3 | 4 |
The graphs of y = x² – 3x + 2 and y = x – 1 are as shown in the following figure:
Upon substitution of the points (1, 0) and (3, 2) into the given equations, it can be observed that the points satisfy the given equations.
Therefore, the solutions of the given simultaneous equations are x = 1, y = 0 and x = 3, y = 2.
Exercise 2
- Use graphical method to solve each of the following pair of simultaneous equations:
- x² – x + 2 = y and y = 3x – 1
- y = x² and y = x + 6
- x² – 3x + 4 = y and y = x – 1
- Use graphical method to find the points of intersection of the equations y = x² – 4 and y = x – 1.
- Draw the graphs of y = x² – 3x – 2 and y = x + 1 on the same xy-plane. Determine the points of intersection of the two graphs and use them to find the values of each of the following:
- 4(x + y)
- (2x – y)²
- Use graphical method to find the points of intersection of the circle x² + y² = 25 and the straight line x – y = 1.
- If y = x² and y = 2x – 1 meet at the point P(x, y), find the values of x and y using graphical method.
Word Problems Involving Simultaneous Equations
Some of everyday real life problems lead to a pair of simultaneous equations. In such cases, the problems are transformed into mathematical symbols, letters, and operation signs. The resulting simultaneous equation can be solved algebraically or graphically.
The following steps can be used when solving word problems involving simultaneous equations:
- Assign variables to the unknown quantities.
- Formulate the equations by expressing the conditions in the given problem in terms of the variables.
- Solve the resulting equations simultaneously.
Example 6
If the sum of two numbers is seven and their product is twelve, then find the difference between the two numbers.
Solution
Let the two numbers be x and y.
The word problem can be summarized using the following equations:
x + y = 7 …… (i)
xy = 12 …… (ii)
Equation (i) can be rearranged to give,
y = 7 – x …… (iii)
Substituting equation (iii) into equation (ii) gives,
x(7 – x) = 12
Expand the equation to get,
7x – x² = 12
The quadratic equation can be rearranged to obtain,
x² – 7x + 12 = 0
Solve for x by splitting the middle term of the quadratic equation, that is,
x² – 4x – 3x + 12 = 0
⇒ x(x – 4) – 3(x – 4) = 0
⇒ (x – 3)(x – 4) = 0
Thus, either x – 3 = 0 or x – 4 = 0.
This implies that, x = 3 or x = 4.
Substituting x = 3 into equation (iii) gives,
y = 7 – 3
⇒ y = 4
Also, substituting x = 4 into equation (iii) gives,
y = 7 – 4
⇒ y = 3
Thus, x = 3, y = 4 or x = 4, y = 3.
Hence, the two numbers are 3 and 4.
The difference between the two numbers is given by, 4 – 3 = 1 or 3 – 4 = -1.
Therefore, the difference between two numbers is 1 or -1.
Exercise 3
- The product of two numbers is twelve. The sum of the larger number and twice the smaller number is eleven. Find the two numbers.
- Five years ago, a father was three times as old as his son. If the product of their present ages is 525, find their present ages.
- A mother is p years old while her son is q years old. The sum of their ages is equal to twice the difference of their ages and the product of their ages is 675. Find the age of the son.
- If the product of two numbers is 168 and their sum is 26. Find the numbers.
- The area of rectangular farm is 264 m² and its perimeter is 68 m. Find its length and width.
- If the side of the first square is twice the side of a second square and the sum of their areas is 125 cm². Find the side of each square.
- The difference of two natural numbers is 3 and the difference of their squares is 39. Find the numbers.
- The demand and supply of maize in our village are given by the equations, pq = 100 and 20 + 3p = q, respectively, where p and q are the price and quantity, respectively. Find the equilibrium price and quantity.
- The perimeter of a rectangular piece of paper is 60 cm and its area is 200 cm². Find the length and width of the rectangular piece of paper.
- The base of a triangle is 5 cm less than the height and the area is 33 cm². Find the length of the base.
Chapter Summary
- Simultaneous equations is a set of two or more equations, each containing two or more variables whose values simultaneously satisfy each of the equations in the given set.
- The simultaneous equations involving one linear and one quadratic equation has either one or two pairs of solutions.
- The graphical solution of simultaneous equations involving a linear and a quadratic equation is given by the points of intersection(s) of the line and curve representing the equations, respectively.
Revision Exercise 1
- Solve each of the following simultaneous equations:
- y = 2x + 1 and x² + y² = 10
- y = x – 3 and x² + y² = 5
- y = 2x – 1 and x² + xy = 24
- y = 2x and y² – xy = 8
- 2x + y = 11 and xy = 15
- x² + y² = 17 and y = x + 5
- Use graphical method to determine the point of intersection of the curve x² – 5x – 12 = y and the straight line y = x – 5.
- Solve the following simultaneous equations using graphical method; x² – 2x + 1 = y and y = x + 1.
- Use graphical method to find the solution in each of the following:
- y = x² and y – x = 2
- x² = y + 1 and y + x = 1
- The difference between two positive numbers is 8 and their product is 105. Find the numbers.
- A rectangle has sides as shown in the following figure:
- Write down a pair of simultaneous equations using the information in the rectangle.
- Find the values of x and y of the simultaneous equations in (a).
- If the length of a classroom floor is 4 m more than the width and the area is 221 m², find the dimensions of the floor.
- Find the two consecutives even numbers whose product is 80.
- A piece of wire 56 cm long is bent to form a rectangle with an area of 171 cm². Find the dimensions of the rectangle.
- The perimeter of the rectangular room is 10 m. If the square of its length is reduced by 7 m² a square with a side y m is formed. By letting x and y be the length and width of the room, respectively.
- Form two equations relating to x and y.
- Find the length and width of the rectangle.
- In a School, there are 660 students in total. If there is 40% more boys than girls. Find the number of boys and girls in the school.
- The area of a rectangular pool is 64 m². If the length the pool is 60 m more than 4 times the width. Find the perimeter of the pool.
- The product of present ages of Asha and John is 175 more than the product of their ages 5 years ago. If Asha is 20 years older than John. What are their present ages?
- The length of the room is 8 m greater than the width. If both the length and the width are increased by 2 m, the area is increased by 60 m². Find the dimensions of the room.
- An adult and five kids paid a total of\ to attend the football game. Tsh 10 000 000 is the result of multiplying the adult and child ticket prices. What were the individual ticket prices for an adult and a child?