Foundations of Mathematics: Linear Algebra Summary Template, Module 3
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This assignment provides a summary of key concepts covered in Module 3 of a linear algebra course, focusing on foundational mathematical principles. It begins with a review of expanding brackets using the distributive law, followed by an exploration of linear equations with one and two variables, i...

Summary Template – Module 3 Linear Algebra
Use the following statements to help you create your own summary of the topics covered in Week 3.
1. Expanding Brackets (The Distributive Law):
a. Complete the equation below showing the general form of the Distributive Law:
b. Use the Distributive Law to remove the brackets from the following:
1) 5x – 5
2) –gh -4g
3) 3a2 + 6a
2. Linear Equations (1 Variable):
a. In general, a linear equation with one variable can be represented by the
Give an example of a specific linear equation with one variable and identify the simple
variable (the term that can take a range of values) and the constants (just numbers!)
2Y + 3 = 4
b. Remembering that to solve an equation, ‘what you do to one side, you must do to the
other to keep things balanced’, solve:
i. 4 k = 40
Dividing both sides by 4,
4k/4 = 40/4
Hence K = 10
ii. 5 e + 7 = 29 – 6e
Rearranging the terms,
11e = 22
Dividing both sides by 11,
e = 2.
1
a a
Use the following statements to help you create your own summary of the topics covered in Week 3.
1. Expanding Brackets (The Distributive Law):
a. Complete the equation below showing the general form of the Distributive Law:
b. Use the Distributive Law to remove the brackets from the following:
1) 5x – 5
2) –gh -4g
3) 3a2 + 6a
2. Linear Equations (1 Variable):
a. In general, a linear equation with one variable can be represented by the
Give an example of a specific linear equation with one variable and identify the simple
variable (the term that can take a range of values) and the constants (just numbers!)
2Y + 3 = 4
b. Remembering that to solve an equation, ‘what you do to one side, you must do to the
other to keep things balanced’, solve:
i. 4 k = 40
Dividing both sides by 4,
4k/4 = 40/4
Hence K = 10
ii. 5 e + 7 = 29 – 6e
Rearranging the terms,
11e = 22
Dividing both sides by 11,
e = 2.
1
a a
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3. Linear Equations (2 Variables):
a. Linear Equations with 2 variables are often written in ‘Slope-Intercept’ form, where and
represent simple variables and and represent constants. Using these pro-numerals,
complete the general form of a linear equation with 2 variables below:
Y = 4x + 2
b. P=4
2p = 3q – 7
8 = 3q-7
3q = 15
Divide both sides by 5,
q = 15/3 = 5.
4. Simultaneous Equations:
a. Simultaneous equations are solved at the same time as the equations have
the same unknowns. .
They have the same unknowns.
b. There are a number of methods that can be used to solve equations simultaneously. In
your own words, summarise the steps in the Substitution and Elimination Methods.
Substitution methods:
Step 1:
Solve one of the equations for either x = or y = .
Step 2:
Substitute the solution from step 1 into the other equation.
Step 3:
Solve this new equation.
Step 4:
Solve for the second variable.
Elimination methods:
Step 1: Multiply each equation by a suitable number so that the two equations have
the same leading coefficient. ...
Step 2: Subtract the second equation from the first.
Step 3: Solve this new equation for y.
Step 4: Substitute y into either Equation 1 or Equation 2 above and solve for
x.
2
a. Linear Equations with 2 variables are often written in ‘Slope-Intercept’ form, where and
represent simple variables and and represent constants. Using these pro-numerals,
complete the general form of a linear equation with 2 variables below:
Y = 4x + 2
b. P=4
2p = 3q – 7
8 = 3q-7
3q = 15
Divide both sides by 5,
q = 15/3 = 5.
4. Simultaneous Equations:
a. Simultaneous equations are solved at the same time as the equations have
the same unknowns. .
They have the same unknowns.
b. There are a number of methods that can be used to solve equations simultaneously. In
your own words, summarise the steps in the Substitution and Elimination Methods.
Substitution methods:
Step 1:
Solve one of the equations for either x = or y = .
Step 2:
Substitute the solution from step 1 into the other equation.
Step 3:
Solve this new equation.
Step 4:
Solve for the second variable.
Elimination methods:
Step 1: Multiply each equation by a suitable number so that the two equations have
the same leading coefficient. ...
Step 2: Subtract the second equation from the first.
Step 3: Solve this new equation for y.
Step 4: Substitute y into either Equation 1 or Equation 2 above and solve for
x.
2

c. Describe the type of problem where the Substitution Method would be a good method
to use to find the solution(s).
Substitution method can be used when one of the variable is already solved in one (or in
both) equations. Also the technique will be helpful when there is coefficient of one for
one of the variable in the equation.
3
to use to find the solution(s).
Substitution method can be used when one of the variable is already solved in one (or in
both) equations. Also the technique will be helpful when there is coefficient of one for
one of the variable in the equation.
3

d. Solve
i. 4 x +y = 8 and 5x + 3y = 59 using the substitution method. (Hint: rewrite
the first equation with as the subject)
From first equation y = 8 – 4x
Hence from second equation 5x + 3y = 59
5x + 3(8-4x) = 59
5x + 24 – 12x = 59
← - 7 x = 35
← x = -5
And -20 + y =8 Hence y = 28
ii. 4 a+ 3b = 24 and 7a + 5b = 72 using the elimination method. (Hint: you need to
multiply both equations by a different constant before you can eliminate!)
Mulitply both the equations by 7 and 4
28a + 21b = 168
28a + 20b = 288
Subtract the second from the first,
B = -120
And substituting in the first equation, a = 96
5. Writing Algebraic Equations:
To write your own algebraic equations to solve worded problems you need to:
1) Identify the variables (choose pro-numerals to represent the unknowns)
2) How? (look for keywords)
3) Help yourself! (organise the information and write your equations)
4) Calculate (solve the equations using an appropriate method)
5) Check your answer:
Explain how would you check your answer for a problem involving
simultaneous equations?
Substituting the final answers in the given equation will prove whether the
answers obtained or right or wrong.
References:
Whitehead, A. N. (2017). An introduction to mathematics. Courier Dover
Publications.
Bell, E. T. (2014). Men of mathematics. Simon and Schuster.
4
i. 4 x +y = 8 and 5x + 3y = 59 using the substitution method. (Hint: rewrite
the first equation with as the subject)
From first equation y = 8 – 4x
Hence from second equation 5x + 3y = 59
5x + 3(8-4x) = 59
5x + 24 – 12x = 59
← - 7 x = 35
← x = -5
And -20 + y =8 Hence y = 28
ii. 4 a+ 3b = 24 and 7a + 5b = 72 using the elimination method. (Hint: you need to
multiply both equations by a different constant before you can eliminate!)
Mulitply both the equations by 7 and 4
28a + 21b = 168
28a + 20b = 288
Subtract the second from the first,
B = -120
And substituting in the first equation, a = 96
5. Writing Algebraic Equations:
To write your own algebraic equations to solve worded problems you need to:
1) Identify the variables (choose pro-numerals to represent the unknowns)
2) How? (look for keywords)
3) Help yourself! (organise the information and write your equations)
4) Calculate (solve the equations using an appropriate method)
5) Check your answer:
Explain how would you check your answer for a problem involving
simultaneous equations?
Substituting the final answers in the given equation will prove whether the
answers obtained or right or wrong.
References:
Whitehead, A. N. (2017). An introduction to mathematics. Courier Dover
Publications.
Bell, E. T. (2014). Men of mathematics. Simon and Schuster.
4
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