SIMULTANEOUS EQUATIONS INVOLVING ONE LINEAR AND ONE QUADRATIC

Subject: 

MATHEMATICS

Term:

FIRST TERM

Week:

WEEK 8

Class:

SS 2

Topic:

SIMULTANEOUS EQUATIONS

Previous lesson: 

The pupils have previous knowledge of

SIMULTANEOUS EQUATIONS

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• solve simple word problem questions on simultaneous equation
• Solve Simultaneous Equations Involving One linear and One quadratic.
• Solve Simultaneous Equations Using Graphical Method

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

Content:

WEEK EIGHT

TOPIC: SIMULTANEOUS EQUATIONS

CONTENT

• Solving Simultaneous Equations Involving One linear and One quadratic.
• Solving Simultaneous Equations Using Graphical Method

SIMULTANEOUS EQUATIONS INVOLVING ONE LINEAR AND ONE QUADRATIC

One of the equations is in linear form while the other is in quadratic form.

Note: One linear, one quadratic is only possible analytically using substitution method.

Examples:

1. Solve simultaneously for x and y (i.e. the points of their intersection)

3x + y = 10 & 2x² +y² = 19

Solution

3x + y = 10 ———– eq 1

2x² + y² = 19 ——— eq 2

Make y the subject in eq 1 (linear equation)

y = 10 – 3x ———- eq 3

Substitute eq 3 into eq 2

2x² + (10-3x) ² = 19

2x²₊ (10 – 3x) (10 – 3x) = 19

2x² + 100 – 30x – 30x + 9x² = 19

2x² + 9x² – 30x – 30x + 100 – 19 = 0

11x² – 60x + 81 = 0

11x² – 33x – 27x + 81= 0

11x (x-3) – 27 (x – 3) = 0

(11x – 27) (x – 3) = 0

11x – 27 = 0 or x-3 = 0

11x = 27 or x = 3

∴ x = 27/11 or 3

Substitute the values of x into eq 3.

When x = 3

y = 10 – 3(x)

y = 10 – 3(3)

y = 10 – 9 = 1

When x =27/11

y = 10 – 3(27/11)

y = 10 – 51/11

y = 110 – 51

11

y = 59/11

∴w hen x = 3, y = 1

x = 27 , y = 59

11 11

2. Solve the equations simultaneously 3x + 4y = 11 &xy = 2

solution

3x + 4y = 11 ——– eq 1

xy = 2 ——– eq 2

Make y the subject in eq 1

4y = 11 – 3x

y = 11 – 3x ………… eq3

4

substituteeq 3 into eq 2

x y = 2

x ( 11- 3x ) = 2

4

2

x (11-3x) = 2×4

11x – 3x² = 8

-3x² + 11x – 8 = 0

-3x² + 3x + 8x – 8 = 0

-3x (x-1) +8 (x-1) = 0

(-3x + 8) (x-1) = 0

-3x + 8 = 0 or x – 1 = 0

3x = 8 or x = 1

x = 8/3 or 1

Substitute the values of x into eq 3

y = 11- 3x

4

when x = 1

y = 11 – 3(1) = 11-3 = 8

4 4 2

y = 4

when x = 8/3

y = 11 – 3(8/3)

4

y = 33 – 24 = 9 = 3

12 12 4

∴ x = 1, y = 2

x = 8/3, y = 3/4.

Evaluation

Solve for x and y

1. 3x ² – 4y = -1 2. 4x² + 9y² = 20

2x – y = 1 2x – 9y = -2

MORE EXAMPLES

Solve simultaneously for x and y.

3x – y = 3 ——– eq 1

9x² – y ² = 45 ——— eq 2

Solution

From eq 2

(3x)² – y ² = 45

(3x-y) (3x+y) = 45 ———- eq 3

Substitute eq 1 into eq 3

3 (3x + y) = 45

3x + y = 15 ……………..eq4

Solve eq 1 and eq 4 simultaneously.

3x – y = 3 ——— eq 1

3x + y = 15 ——– eq 4

eq 1 + eq 4

6x = 18

x = 18/ 6

x = 3

Substitute x = 3 into eq 4.

3x + y = 15

3 (3) + y = 15

9 + y = 15

y = 15 – 9

y = 6

∴ x = 3, y = 6

Evaluation

Solve for x and y in the following pairs of equations

1. (a) 4x² – y² = 15 (b) 3x² +5xy –y² =3

2x – y = 5 x – y = 4

WORD PROBLEMS LEADING TO LINEAR AND QUADRATIC EQUATIONS

Example

The product of two numbers is 12. The sum of the larger number and twice the smaller number is 11. Find the two numbers.

Solution

Let x = the larger number

y = the smaller number

Product, x y = 12 …………….eq1

From the last statement,

x + 2y = 11 ………….. eq2

From eq2, x = 11 – 2y ……………eq3

Sub. Into eq1

y(11 – 2y) = 12

11y – 2y2 = 12

2y2 -11y + 12 = 0

2y2 – 8y – 3y + 12 = 0

2y(y-4) – 3(y-4) = 0

(2y-3)(y-4) =0

2y-3 =0 or y-4 =0

2y = 3 or y = 4

y= 3/2 or 4

when y = 3/2 when y=4

x = 11 – 2y x = 11- 2y

x = 11 – 2(3/2) x = 11 – 2(4)

x = 11 – 3 x = 11 – 8

x = 8 x = 3

Therefore, (8 , 3/2)(3 , 4)

Evaluation

Solve the following simultaneous equation

1. (a) 2^2x-3y = 32, 3^x-2y = 81 (b) 2^x+2y=1, 3^2x+y = 27

2. Bisi’s and Fibie’s ages add up to 29. Seven years ago Bisi was twice as old as Fibie. Find their present ages.

SOLVING SIMULTANEOUS EQUATIONS USING GRAPHICAL METHOD

Examples

Using the scale 2cm to 1 units on x-axis and 2cm to 2 unit on y-axis, draw the graph of y = x² – x – 1 and y = 2x – 1 (on the same scale and axis for values of x: – 3≤x< 4

Solution

Table of values for y = x² – x – 1

Table of values for y = 2x – 1

Evaluation

a. Copy and complete the table below of values for the relation y = 2x² – 3x – 7

b.Using a scale of 2cm to 1 unit on x-axis and 2cm to 5 unit on y-axis, draw the graph of the relation

y = 2x²-3x-7 for -3 < x ≤ 5

c.Using the same scale and axis, draw the graph of y = 2x-1

d. Use your graph to find the values of x and y.

GENERAL EVALUATION AND REVISION QUESTIONS

1. Solve the simultaneous equation: 3x² – 4y = -1 & 2x – y = 1

2. Five years ago, a father was 3 times as old as his son, now their combined ages amount to 110 years. How old are they?

3. Solve: 4x² – y² = 15 & 2x – y = 5

4. Seven cups and eight plates cost # 1750. Eight cups and seven plates cost #1700. Calculate the cost of a cup and of a plate.

WEEKEND ASSIGNMENT

Solve each of the following pairs of equations simultaneously,

1. xy = -12 ; x – y = 7 a. (3 , -4)(4 ,-3) b. (-2 ,4)(-3, -4) c.(-4, 5)(-2 , 3) d.(3 ,-3)(4,-4)

2. x – 5y = 5 ; x² – 25y² = 55 a (-8, 0)(3/5 , 0) b. (0, 0)(-8 , 3/5) c. (8 , 3/5) d. (0, 8)(0, 3/5)

3. y = x² and y = x + 6 (a).(0,6) (3,9) (b)(-3,0) (2,4) (c) (-2,4) (3,9) (d).(-2, 3), (-3,2)

4. x – y = -3/2 ; 4x² + 2xy – y² = 11/4 : a. (-1, 1/2)(1, 5/2). b. (3, 2/5) (1, 1/2) c.(3/2 , -1) (4,2) d.(-1 , -1/2)(-1 , 5/2)

5. m² + n² = 25 ; 2m + n – 5 = 0 : a. (0,5)(4, -3) b.(5,0)(-3,4)c.(4,0)(-3,5) d(-5,3)(0,4)

THEORY

1a. Find the coordinate of the points where the line 2x – y = 5 meets the curve 3x² – xy -4 =10

b. Solve the simultaneous equation: 2^2x+4y = 4, 3^{3x + 5y} – 81= 0

2. A woman is q years old while her son is p years old. The sum of their ages is equal to twice the difference of their ages. The product of their ages is 675.

Write down the equations connecting their ages and solve the equations in order to find the ages of the woman and her son. (WAEC)

Conclusion

The class teacher wraps up or concludes the lesson by giving out short notes to summarize the topic that he or she has just taught.

The class teacher also goes round to make sure that the notes are well copied or well written by the pupils.

He or she makes the necessary corrections when and where the needs arise.