Simultaneous Linear Equation – SS1 Mathematics Lesson Note (First Term)

Simultaneous linear equations are two equations with two unknowns, usually x and y, that are solved together to find the values that satisfy both equations.

Behavioural objectives

1. Solve simultaneous linear equations by the elimination method.
2. Solve simultaneous linear equations by the substitution method.
3. Solve simultaneous linear equations graphically.
4. Solve equations involving fractions by letting the unknowns be 1/x and 1/y.
5. Form and solve simultaneous equations from word problems.

What are simultaneous linear equations?

They are a pair of linear equations in two unknowns that have one common solution.

• Linear equation: an equation in which each unknown has power 1, such as 2x + 5y = 1.
• Unknown: a letter standing for a number we do not yet know.
• Coefficient: the number multiplying an unknown. In 3x, the coefficient is 3.
• Solution: the pair of values of x and y that makes both equations true.
• Why two equations: one equation with two unknowns has many answers. Two equations give one pair of answers.

What are the methods of solving simultaneous equations?

There are three methods: elimination, substitution and graphical.

What is the elimination method?

The elimination method removes one unknown by adding or subtracting the two equations after making its coefficients equal.

Steps of the elimination method

1. Write both equations in the form ax + by = c.
2. Multiply one or both equations so one unknown has the same coefficient in both.
3. If the equal terms have the same sign, subtract the equations.
4. If the equal terms have opposite signs, add the equations.
5. Solve for the unknown that remains.
6. Substitute its value into either equation to find the other unknown.
7. Check the answers in both equations.

Example: elimination method

Solve 2x + 5y = 1 and 3x – 2y = 30.

1. Label the equations: 2x + 5y = 1 (eqn 1) and 3x – 2y = 30 (eqn 2).
2. To eliminate x, multiply eqn 1 by 3 and eqn 2 by 2.
3. Eqn 3: 6x + 15y = 3.
4. Eqn 4: 6x – 4y = 60.
5. Subtract eqn 4 from eqn 3: 19y = -57.
6. Divide both sides by 19: y = -3.
7. Substitute y = -3 into eqn 1: 2x + 5(-3) = 1.
8. Simplify: 2x – 15 = 1, so 2x = 16 and x = 8.
9. Check in eqn 2: 3(8) – 2(-3) = 24 + 6 = 30. This is correct.

Answer: x = 8 and y = -3.

What is the substitution method?

The substitution method makes one unknown the subject of one equation and replaces it in the other equation.

Steps of the substitution method

1. Choose one equation and make one unknown the subject. Prefer the unknown with coefficient 1.
2. Substitute that expression into the other equation.
3. Solve the new equation, which now has one unknown.
4. Substitute the value found into the expression from step 1.
5. Check the answers in both equations.

Example: substitution method

Solve 2x + 5y = 1 and 3x – 2y = 30.

1. Label the equations: 2x + 5y = 1 (eqn 1) and 3x – 2y = 30 (eqn 2).
2. Make x the subject of eqn 1: 2x = 1 – 5y.
3. Divide both sides by 2: x = (1 – 5y)/2 (eqn 3).
4. Substitute eqn 3 into eqn 2: 3(1 – 5y)/2 – 2y = 30.
5. Multiply every term by 2: 3(1 – 5y) – 4y = 60.
6. Expand: 3 – 15y – 4y = 60, so 3 – 19y = 60.
7. Simplify: -19y = 57, so y = -3.
8. Substitute y = -3 into eqn 3: x = (1 – 5(-3))/2 = (1 + 15)/2 = 16/2.
9. So x = 8.

Answer: x = 8 and y = -3.

What is the graphical method?

The graphical method draws the line for each equation and reads the coordinates of the point where the lines meet.

Steps of the graphical method

1. Make a table of values for each equation.
2. Plot the points of each equation on the same axes.
3. Join the points of each equation with a straight line.
4. Read the x and y values at the point where the lines cross.

Example: graphical method

Solve x + y = 5 and x – y = 1 graphically.

• Crossing point: both lines pass through (3, 2).
• Answer: x = 3 and y = 2.

How do you solve simultaneous equations involving fractions?

Let the unknowns be 1/x and 1/y, and the equations become ordinary simultaneous linear equations.

Example: equations with fractions

Solve 2/x – 1/y = 3 (i) and 4/x + 3/y = 16 (ii).

1. Let the unknowns be (1/x) and (1/y).
2. Eqn 1: 2(1/x) – (1/y) = 3.
3. Eqn 2: 4(1/x) + 3(1/y) = 16.
4. Multiply eqn 1 by 2 to make the coefficients of 1/x equal.
5. Eqn 3: 4(1/x) – 2(1/y) = 6.
6. Subtract eqn 2 from eqn 3: -5(1/y) = -10.
7. Divide both sides by -5: 1/y = 2, so y = 1/2.
8. Substitute 1/y = 2 into eqn 1: 2(1/x) – 2 = 3.
9. Simplify: 2(1/x) = 5, so 1/x = 5/2.
10. Invert both sides: x = 2/5.
11. Check in eqn (ii): 4 ÷ (2/5) + 3 ÷ (1/2) = 10 + 6 = 16. This is correct.

Answer: x = 2/5 and y = 1/2.

• Common error: stopping at 1/x = 5/2. The question asks for x, so invert the fraction.

How do word problems lead to simultaneous equations?

A word problem leads to simultaneous equations when it gives two statements about two unknown quantities.

Steps for word problems

1. Choose a letter for each unknown quantity.
2. Write one equation from each statement in the problem.
3. Solve the equations by elimination or substitution.
4. Check the answers against the words of the problem.
5. State the answer in words, with units such as naira.

Example 1: cost of a cup and a plate

Seven cups and eight plates cost ₦1750. Eight cups and seven plates cost ₦1700. Find the cost of a cup and of a plate.

1. Let a cup cost x naira and a plate cost y naira.
2. Eq 1: 7x + 8y = 1750.
3. Eq 2: 8x + 7y = 1700.
4. Multiply eq 1 by 8 and eq 2 by 7 to eliminate x.
5. Eq 3: 56x + 64y = 14000.
6. Eq 4: 56x + 49y = 11900.
7. Subtract eq 4 from eq 3: 15y = 2100, so y = 2100/15 = 140.
8. Substitute y = 140 into eq 2: 8x + 7(140) = 1700.
9. Simplify: 8x + 980 = 1700, so 8x = 720 and x = 90.

Answer: a cup costs ₦90 and a plate costs ₦140.

Example 2: a two-digit number

Find a two-digit number such that two times the tens digit is three less than thrice the unit digit. Four times the number is 99 greater than the number got by reversing the digits.

1. Let the number be ab, where a is the tens digit and b is the unit digit.
2. The number is 10a + b. The reversed number is 10b + a.
3. First statement: 2a + 3 = 3b, so 2a – 3b = -3 (eq 1).
4. Second statement: 4(10a + b) – 99 = 10b + a.
5. Expand: 40a + 4b – 99 = 10b + a.
6. Collect terms: 39a – 6b = 99.
7. Divide through by 3: 13a – 2b = 33 (eq 2).
8. Multiply eq 1 by 2: 4a – 6b = -6 (eq 3).
9. Multiply eq 2 by 3: 39a – 6b = 99 (eq 4).
10. Subtract eq 3 from eq 4: 35a = 105, so a = 3.
11. Substitute a = 3 into eq 1: 6 – 3b = -3, so b = 3.
12. Check: 4 × 33 = 132, and 33 + 99 = 132. This is correct.

Answer: the number is 33.

Evaluation

1. Use the elimination method to solve 5x – 4y = 38 and x + 3y = 22.
2. Use the elimination method to solve 2c – 3d = -4 and 4c – 3d = -14.
3. Use the substitution method to solve x + 2y = 10 and 4x + 3y = 20.
4. Use the substitution method to solve 4x – y = 8 and 5x + y = 19.
5. The sum of two numbers is 20 and their difference is 6. Form two equations and find the numbers.

Assignment

Three mangoes and two oranges cost ₦170. Two mangoes and four oranges cost ₦180. Form two simultaneous equations, solve them, and find the cost of one mango and one orange.