SIMULTANEOUS EQUATIONS

Subject: 

MATHEMATICS

Term:

FIRST TERM

Week:

WEEK 7

Class:

SS 2

Topic:

SIMULTANEOUS EQUATIONS

 

Previous lesson: 

The pupils have previous knowledge of

QUADRATIC 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

 

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 SEVEN

TOPIC: SIMULTANEOUS EQUATIONS

CONTENT

• Solving Simultaneous Equations Using Elimination and Substitution Method
• Solving Equations Involving Fractions.
• Word problems.

SIMULTANEOUS LINEAR EQUATIONS

Methods of solving Simultaneous equations

i. Elimination method

ii. Substitution method

iii. Graphical method

ELIMINATION METHOD

One of the unknowns with the same coefficient in the two equations is eliminated by subtracting or adding the two equations. Then the answer of the first unknown is substituted into either of the equations to get the second unknown.

Example

Solve for x and y in the equations 2x + 5y = 1 and 3x – 2y = 30

Solution

To eliminate x multiply equation 1 by 3 and equation 2 by 2

2x + 5y = 1 ………. eqn 1 (x (3)

3x – 2y = 30 ………… eqn 2 (x (2)

Resulting into,

6x + 15 y = 3 ………. eqn 3

6x – 4y = 60 ……….. eqn 4

Subtract eqn 3 from eqn 4

6x – 6x + 15y – (- 4y) = 3 – 60

19y = -57 3

19 19

y = -3

Substitute y = – 3 into eqn 1

2x + 5 (-3) = 1

2x = 1 + 15

2x =16

2 2

x = 8

∴ y = -3 and x = 8

Evaluation

Using elimination method to solve the simultaneous equations.

1. 5x – 4y = 38 and x + 3y = 22

2. 2c-3d= -4 and 4c-3d= -14

SUBSTITUTION METHOD

One of the unknowns (preferably the one having 1 has its coefficient) is made the subject of the formula in one of the equations and substituted into the other equation to obtain the value of the first unknown which is then substituted into either of the equations to get the second unknown.

Example: Solve the simultaneous equation 2x + 5y = 1 and 3x – 2y = 30

Solution

2x + 5y = 1……………. eq 1

3x – 2y = 30 ………….. eq 2

Make x the subject in eqn 1

2x = 1 – 5y

2 2

x = 1 – 5y ………… eqn 3

2

Substitute eq3 into eqn 2

3 (1-5y) – 2y = 30

2

Multiple through by 2 or find the LCM and cross multiply.

3 – 15y – 4y = 30

2

3 – 15y – 4y = 60

3 – 19y = 60

-19y = 60 – 3

-19y = 57 3

-19 -19

y = – 3

Substitute y = -3 into eq 3

x =1 – 5y

2

x = 1 – 5 (-3)= 1 + 15 = 16

2 2 2

x = 8

∴ x = 8, y = -3

Evaluation

Solve for x and y in the equations

1. x + 2y = 10 and 4x + 3y = 20

2. 4x-y=8 and 5x+y=19

SIMULTANEOUS EQUATIONS INVOLVING FRACTIONS

Example

1. Solve the following equations simultaneously

2 – 1 = 3 and 4 + 3 = 16

x y x y

Solution

2 – 1 = 3

x y

4 + 3 = 16

x y

Instead of using x and y as the unknown, let the unknown be (¹/_x) an (¹/_y).

2(¹/_x) – (¹/_y) = 3 ……………. eqn 1

4 (¹/_x) – 3 (¹/_y) = 16 …………… eqn 2

Using elimination method, multiply equation 1 by 2 to eliminate x.

4(¹/_x) – 2(¹/_y) = 6 …………….. eqn 3

4 (¹/_x) + 3(¹/_y) = 16 ……………. eqn 4

-5 (¹/_y) = -10

-5 -5

1 = 2

y

∴ y = ½

Substitute (1/y) = 2 into eqn 1

2 (¹/_x) – (¹/_y) = 3

2 (¹/_x) – (2) = 3

2(¹/_x) = 3 + 2

2 (¹/_x) = 5

1 = 5

x 2

∴ x = ²/₅

∴ y = ½, x = ²/₅

Evaluation

I. Solve for x and y simultaneously, II. Solve the pair of equations for x and y

x +y = 1 respectively.

2 2 2x^-1 – 3y^-1 = 4

x – y = 1½ 4x^-1 + y^-1 = 1

2 6

FURTHER EXAMPLES

Solve for x and y simultaneously: 2x – 3y + 2 = x + 2y – 5 = 3x + y.

Solutions

2x – 3y + 2 = x + 2y – 5 = 3x + y

Form two equations out of the question

2x – 3y + 2 = 3x + y

x + 2y – 5 = 3x + y

OR

2x – 3y + 2 = x + 2y – 5 ————- eq 1

x + 2y – 5 = 3x + y ————– eq 2

Rearrange the equations to put the unknown on one side and the constant at the other side.

2x – 3y – x – 2y = – 5 – 2

2x – x – 3y – 2y = -7

x – 5y = -7 —————- eq 3

From eqn 2

x – 3x + 2y – y – 5

– 2x + y = 5 ————- eq 4

Using substitution method solve eq 3 & 4

x – 5y = -7 —————- eq 3

-2x + y = 5 ————— eq 4

Make y the subject in eq 4.

y = 5 + 2x ————— eq 5

Substitute eqn 5 into eqn 3.

x – 5 (5 + 2x) = -7

x – 25 – 10x = -7

-9x – 25 = -7

-9x = -7 + 25

-9x = 18

x = 18

-9

X = -2

Substitute x = – 2 into eqn 5

y = 5 + 2x

y = 5 + 2(-2)

y = 5 – 4

y = 1

∴ x = -2, y = 1

Example

Solve the equations

5^{x – y/2} = 1 81^x = 27 ^{3x -y}

9

Solution

5^{x – y/2} = 1 ———– eq 1

81^x= 27^{3x -y} ———- eq 2

9

From eq 1 (using the law of indices)

5^{x – y/2} = 5⁰

x – y/2 = 0

2x – y = 0 ———— eq 3

From eq 2.

81^x= 27^{3x -y}

9

3 ^4x = 3 ^3(3x-y)

3 ²

3 ^4x-2 = 3 ^3(3x-y)

By comparison

4x – 2 = 9x – 3y

4x – 9x + 3y = 2

– 5x + 3y =2 ——— eq 4

Solve equation 3 and 4 simultaneously

2x – y = 0 ——— eq 3

-5x + 3y =2 ———- eq 4

Using elimination method: multiply equation 3 by 3

6x – 3y = 0 ——– eq 3

-5x + 3y = 2 ———- eq 4

eq 3 + eq 4

x = 2

Substitute x = 2 into eq 3

2x – y = 0

2 (2) – y = 0

4 – y = 0

4 = 0+y

4 = y

∴ x = 2, y=4

WORD PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS

Examples

1.Seven cups and eight plates cost N1750, eight cups and seven plates cost N1700. Calculate the cost of a cup and a plate

solution

Let a cup be x and plate be y

7x + 8y = 1750 ————– eq 1

8x + 7y = 1700 ————– eq 2

Multiply eq 1 by 8 and eq 2 by 7 to eliminate x (cups).

56x + 64y = 14000 ———- eq 3

56x + 49y = 11900 ———- eq 4

Subtracting eq 4 from eq 3

15y = 2100

y = 2100

15

Y = 140

Substitute y = 140 into eq 2

8x + 7y = 1700

8x + 7 (140) = 1700

8x + 980 = 1700

8x = 1700 – 980

8x = 720

x = 720

8

x = 90

∴ Each cup cost N90 and each plate cost N140

2. Find a two digit number such that two times the tens digit is three less than thrice the unit digit and 4 times the number is 99 greater than the number obtained by reversing the digit.

Solution

Let the two digit number be ab, where a is the tens digit and b is the unit digit

From the first statement,

2a + 3 = 3b

2a – 3b = -3 ………….eq1

From the second statement,

4(10a + b) – 99 = 10b + a

40a + 4b – 99 = 10b + a

40a – a + 4b – 10b = 99

39a – 6b = 99

Dividing through by 3

13a – 2b = 33 ………….eq2

Solving both equations simultaneously,

a = 3 , b = 3

Hence, the two digit number is 33

EVALUATION

1.The sum of two numbers is 110 and their difference is 20. Find the two numbers.

2.A pen a ruler cost #30.If the pen costs #8 more than the ruler, how much does each item cost ?

GENERAL EVALUATION AND REVISION QUESTION

1. Solve the following simultaneous equation: 3(2x – y) = x + y + 5 & 5(3x – 2y) = 2 (x –y) + 1

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

3. A doctor and three nurses in a hospital together earn #255 000 per month, while three doctors and eight nurses together earn #720 000 per month. Calculate (a) how much a doctor earns per month. (b) How much a nurse earns per month.

4. Solve simultaneously, 2^{x + 2y} = 1; 3^2x+y = 27

5. Solve: 2x – 2y + 5 = 3x – 4y + 2 = -1

WEEKEND ASSIGNMENT

1. If (x-y) log₁₀6 = log₁₀ 216 and 2 ^x+y =32 , calculate the values of x and y

a. x=1 , y=4 b. x= 4 , y =1 c. x=-4 , y= 1 d. x=4, y= -1

2. The point of intersection of the lines 3x- 2y =-12 and x + 2y = 4 is …

a. (5, 0) b. (3, 4) c. (-2, 5) d. (-2, 3)

3. Find the value of (x – y), if 2x + 2y =16 and 8x – 2y = 44 a. 2 b. 4 c. 5 d. 6

4. If 5 ^{(p +2q)} =5 and 4 ^(p+3q) =16, the value of 3^(p+q) is ….. a.0 b. -1 c.2 d. 1

5. Given 4x – 3y = 11 evaluate y² – 3x

7x – 4y 23 3 a. -2 b. 3 c. -3 d. 2

THEORY

1. Given that 2 ^{1- x/y} = 1/32, find x in terms of y, and hence solve the simultaneous equations

2x + 3y – 30 = 0 and 2^{1- x/y} = 1/32 (WAEC)

2. A number is made up of two digits. The sum of the digits is 11. If the digits are interchanged, the original number is increased by 9. Find the original number .

 

 

 

 

 

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.