What is Algebra & Algebraic Expressions?

Algebra might seem like something difficult – given the name and the scientific sound words, but it’s really a way of using symbol to solve a problem where you have missing information.

Imagine you're planning a pizza party for your friends. You know that each pizza has 8 slices, and each person will likely eat 2 slices. But you don't know exactly how many friends are coming yet. This is where algebra can help!

That’s the everyday stuff, but the usefulness of algebra shouldn’t be underestimated. When you graduate, you might work in a field that requires algebra. For example, if you become a video game developer, algebra is used to calculate character movements, collision detection, projectile trajectories, and even the behaviour of artificial intelligence. For instance, making a character jump realistically involves using algebraic equations that factor in gravity, jump height, and speed.

Algebra is also used in the medical field - when you get an X-ray or MRI, the machines use algebraic algorithms to process the data and create images. Doctors use algebra to calculate the correct dosage of medication based on a patient's weight and other factors.

Here’s are the key words you need to know in algebra:

• Variables: Letters (like x, y, or a) that represent unknown numbers.
• Constants: Numbers that have a fixed value (like 5, 10, or -3).
• Coefficients: Numbers that are multiplied by variables (like the '3' in 3x).
• Terms: Parts of an expression separated by + or - signs.
• Expressions: Combinations of variables, constants, and coefficients connected by mathematical operations (+, -, ×, ÷).

Let’s have a look at algebra in action. This is an expression 2x + 5y – 7. x and y are variables. 7 is a constant. 2 and 5 are coefficients and 2x, 5y, and -7 are terms.

That’s the expression, but how could this be shown in an exam question?

Example 1: A store is having a promotion. You get two points for every item of type 'x' you buy, and five points for every item of type 'y' you buy. However, they deduct seven points from your total as a processing fee. What is the total number of points you will have in terms of x and y?

Example 2: The total cost is calculated by taking twice the number of apples, adding it to five times the number of oranges, and then reducing the sum by seven dollars.

Example 3: The result is obtained by multiplying a variable 'x' by two, adding the product of five and another variable 'y', and then subtracting a constant value of seven.

Let’s go through some basic questions now:

1. Identify the variables, constants, and coefficients in the expression 4a - 2b + 9.
2. How many terms are there in the expression 6x + y - 11z + 2?
3. Write an algebraic expression that represents "three times a number x plus seven."
4. What is the coefficient of y in the expression x - 8y + 1?
5. Write an expression with two variables, p and q, where the coefficient of p is 5, the coefficient of q is -2, and there is a constant term of 10.

Solutions:

1. Variables: a, b; Constant: 9; Coefficients: 4, -2
2. 4 terms
3. 3x + 7
4. -8
5. 5p - 2q + 10

Simplifying Expressions with the Four Operations

Sometimes, expressions can be long and convoluted like this: 3(2x - 4y + 5) - 2(x + 3y - 7) + 4(x - 2y + 1) - (5x - y + 9)

This type of expression makes it hard to see the relationship between terms. These types of expression could come about when you’re initially representing the problem you have at hand. To make it easier to work to expression, you’ll need to use distributive law along with simplify. Here’s how:

• Distribute and remove the parentheses. Multiplying a number by a sum (or difference) is the same as multiplying the number by each term inside the parentheses and then adding (or subtracting) the results. Use the distributive law to remove parentheses by multiplying the term outside the parentheses by each term inside.

1. 3(2x - 4y + 5) - 2(x + 3y - 7) + 4(x - 2y + 1) - (5x - y + 9)
2. (6x - 12y + 15) - (2x + 6y - 14) + (4x - 8y + 4) - (5x - y + 9)
3. 6x - 12y + 15 - 2x - 6y + 14 + 4x - 8y + 4 - 5x + y - 9

• Combine Like Terms - Terms that have the same variable(s) raised to the same power(s) are like terms. To combine, add or subtract the coefficients of like terms. Let’s do this for the above expression by (1) grouping the x terms, (2) grouping the y terms and then finally, grouping the constant terms:

1. (6x - 2x + 4x - 5x) = 3x
2. (-12y - 6y - 8y + y) = -25y
3. (15 + 14 + 4 - 9) = 24

We get: 3x - 25y + 24

Dividing

We’ve looked at multiplying and using the operators of addition and subtraction with algebra, how about division?

To divide, you first divide coefficients. Then divide variables – if dividing exponential terms with the same base, you subtract the exponents (Refer to checkpoint: Index notation).

Let’s do a quick example: Simplify 6ab / 2a = (6/2) x (a/ a) x b = 3b (Watch the video for workings!)

Solving Problems Involving Algebra: Simplify & Extend

Let’s now do some problems involving algebra! Part of the difficulty students face in the exam isn’t about not knowing formula and how to do calculations. It’s sometimes about comprehension of the question and then deciding how best to solve it, along with what knowledge would be needed to solve it.

Question 1 – The Garden

A farmer is planning a rectangular vegetable garden next to a barn. The garden’s width is w metres, and its length is three times the width. The farmer needs to put a fence around three sides of the garden (the barn acts as the fourth side along the length). After planning, the farmer decides to increase the width by 2 metres, which also changes the length. The total fencing needed for the three sides after the change is 15 metres, and all dimensions are whole numbers.

a. Write an expression for the total fencing needed after the change in terms of w, simplify it, and use it to find the original width of the garden in metres.
b. How much more or less fencing does the farmer need compared to the original plan?

Solution (Part A)

1. Original Dimensions: Width = w metres. Length = 3w metres (three times the width).
2. New Dimensions (after the change): New Width = w + 2 metres (original width increased by 2). New Length = 3 * (w + 2) = 3w + 6 metres (length is three times the new width)
3. Fencing Description: The barn acts as the fourth side along the length, so fencing is needed for three sides: two widths (the shorter sides) and one length (the side opposite the barn). Total fencing after the change = 2 * New Width + New Length.
4. Write the Expression: 2 * (w + 2) + (3w + 6)
5. Simplify the Expression: Expand: 2 * (w + 2) = 2w + 4. Add the length: 2w + 4 + 3w + 6. Combine like terms: 2w + 3w + 4 + 6 = 5w + 10
6. So, the simplified expression for the total fencing after the change is: 5w + 10
7. Set Up the Equation: The problem states the total fencing needed after the change is 15 metres: 5w + 10 = 15.
8. Solve for w: Subtract 10 from both sides: 5w + 10 - 10 = 15 - 10. Simplify: 5w = 5. Divide both sides by 5: w = 5 / 5 = 1

Solution (Part B)

1. Original Fencing: Original Width = 1 metre. Original Length = 3 metres. Original fencing = 2 * Original Width + Original Length = 2 * 1 + 3 = 2 + 3 = 5 metres
2. New Fencing: From part a, the fencing after the change is 15 metres.
3. Calculate the Change: Change in fencing = New Fencing - Original Fencing = 15 - 5 = 10 metres

Question 2: The Phone Plans

Two mobile phone companies offer different monthly plans: Company A: Charges a fixed fee of $20 plus $2 for each GB of data used. Company B: Charges a fixed fee of $10 plus $4 for each GB of data used.

a. Let 'd' represent the number of GB of data used in a month. Write an expression for the total monthly cost for each company in terms of 'd'.
b. If a customer uses 6 GB of data in a month, which company offers a cheaper plan?
c. For how many GB of data used will the total cost of both plans be the same?

Solution:

a. Company A: Total Cost = $20 + $2d. Company B: Total Cost = $10 + $4d
b. Company A (6 GB): Total Cost = $20 + ($2 * 6) = $20 + $12 = $32. Company B (6 GB): Total Cost = $10 + ($4 * 6) = $10 + $24 = $34
c. Set the two cost expressions equal to each other: $20 + $2d = $10 + $4d. Solve for 'd': Subtract $2d from both sides: $20 = $10 + $2d. Subtract $10 from both sides: $10 = $2d. Divide both sides by 2: d = 5

Practice time!

Now, it's your turn to practice.

The questions in this checkpoint are provided to give you an introduction to possible questions you may see in your exam. Don't worry too much as you'll continue to build your skills throughout the course.

Click on the button below and start your practice questions. We recommend doing untimed mode first, and then, when you're ready, do timed mode.

Every question has a suggested solutions videos after you complete the question. This video explains to you the steps to take to answer the question and provides tips and tricks.

Once you're done with the practice questions, move on to the next checkpoint.

Now, let’s get started on your practice questions.

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