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Introduction to Algebra / LEVEL 1 · DIFFICULTY 1/5

Equations and Linear Relationships

Solve balances, linked equations and changing quantities.

3 stages · 24 practice problems · two 6-question assessment forms

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MINI QUEST ALG 1.1Linear EquationsRead the lesson
MINI QUEST ALG 1.2Two Equations, Two UnknownsRead the lesson
MINI QUEST ALG 1.3Slope and Rate of ChangeRead the lesson
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STAGE ALG 1.1 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Linear Equations

Useful preparation: Solving & Checking Linear Equations

Goal: Solve a linear equation and verify the result.

Before you begin: linear equations

START WITH SOMETHING YOU CAN MOVE

What is inside the mystery box?

Every star weighs one unit. Each mystery box has the same weight. Drag stars into the tray, or tap a star and then a “Move selected” button.

One small move: The two sides start with equal weight. Now move ONE star from the mystery side into the tray.

Mystery side

1 box + 3 stars

Star side

7 stars

Star tray

Both sides start level. Nothing has been moved yet.

Next: connect equations to graphs

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Pick an x value. The rule tells you its y value. A point (x, y) shows both numbers together. Drag across the graph or use the point slider. This is an adjustable learning example; use the coefficients in your own practice question when solving it.

Parent equation: $\displaystyle y = x$

Start with $\displaystyle y = x$ . Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

Trace a point along the graph

Blue: $\displaystyle y = 2\, x + 1$

Moving point: (-4, -7)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2

Vertical intercept b: 1

Choose the point’s x: 1

Point (1, 3)

\displaystyle 2 \times \left(1\right) + \left(1\right) = 3

What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting $\displaystyle x = 0$ gives $\displaystyle y = 1$ , where the line meets the vertical axis.

Three points made by the blue rule
x	Substitute into the rule	y
-1	$\displaystyle 2 \times \left(-1\right) + \left(1\right)$	-1
0	$\displaystyle 2 \times \left(0\right) + \left(1\right)$	1
1	$\displaystyle 2 \times \left(1\right) + \left(1\right)$	3

Understand the idea

An equation says that two expressions have the same value. Applying the same reversible operation to both sides preserves every solution. The coefficient tells you how many copies of the unknown you have.

\displaystyle ax + b = c \Rightarrow x = \frac{c - b}{a}

, for

\displaystyle a \ne 0

Choose and carry out a method

Collect variable terms on one side and constants on the other. Undo addition before multiplication. If variables occur on both sides, subtract the smaller variable term first to make the arithmetic easier.

Check the reasoning

Replace the variable in the original equation with your answer. Both sides must match. If the variable terms cancel, distinguish an identity from a contradiction instead of dividing by zero.

WORKED EXAMPLE 1

Solve $\displaystyle 3\, x - 6 = 6$ for x.

Undo addition or subtraction before dividing by the coefficient.
$\displaystyle 3\, x = 6 + 6 = 12$ .
Dividing by 3 gives $\displaystyle x = 4$ ; substitution checks the original balance.

WORKED EXAMPLE 2

Solve $\displaystyle 4\, x - 7 = 13$ for x.

Undo addition or subtraction before dividing by the coefficient.
$\displaystyle 4\, x = 13 + 7 = 20$ .
Dividing by 4 gives $\displaystyle x = 5$ ; substitution checks the original balance.

Common pitfalls

Possible mix-up: Moving a term always makes it negative.

Think of subtracting or adding the same term on both sides, rather than changing a sign by habit.

Possible mix-up: Only the variable side needs dividing.

Divide the entire opposite side by the coefficient as well.

Explain it to yourself

Why does subtracting the same quantity from both sides keep the solution unchanged?

Preview the eight practice prompts

Problem 1
Solve $\displaystyle 6\, x - 9 = 33$ for x.
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Problem 2
Solve $\displaystyle 7\, x - 5 = 51$ for x.
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Problem 3
Solve $\displaystyle 8\, x - 6 = 66$ for x.
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Problem 4
Solve $\displaystyle 9\, x - 7 = 83$ for x.
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Problem 5
Solve $\displaystyle 10\, x - 8 = 102$ for x.
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Problem 6
Solve $\displaystyle 11\, x - 9 = 123$ for x.
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Problem 7New context
A workshop sells x tickets at $12 each and pays a fixed $5 fee. Its net income is $151. How many tickets were sold?
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Problem 8New context
A workshop sells x tickets at $13 each and pays a fixed $6 fee. Its net income is $176. How many tickets were sold?
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Open stage ALG 1.1 in the student workspace →

STAGE ALG 1.2 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Two Equations, Two Unknowns

Useful preparation: Linear Equations

Goal: Recover two quantities from two independent conditions.

Before you begin: Linear Equations

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Parent equation: $\displaystyle y = x$

Start with $\displaystyle y = x$ . Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

The orange line also starts from $\displaystyle y = x$ : reflect to $\displaystyle y = -x$ , then move up 4.

Trace a point along the graph

Blue: $\displaystyle y = 2\, x + 1$
Orange: $\displaystyle y = -x + 4$

Moving point: (-4, -7)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2

Vertical intercept b: 1

Choose the point’s x: 1

Point (1, 3)

\displaystyle 2 \times \left(1\right) + \left(1\right) = 3

What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting $\displaystyle x = 0$ gives $\displaystyle y = 1$ , where the line meets the vertical axis.

Three points made by the blue rule
x	Substitute into the rule	y
-1	$\displaystyle 2 \times \left(-1\right) + \left(1\right)$	-1
0	$\displaystyle 2 \times \left(0\right) + \left(1\right)$	1
1	$\displaystyle 2 \times \left(1\right) + \left(1\right)$	3

Understand the idea

A solution to a system must satisfy every equation at the same time. One equation may leave many possibilities; a second condition narrows them down.

Subtracting

\displaystyle 2\, \left(x + y\right) = 2\, s

from

\displaystyle 3\, x + 2\, y = v

gives

\displaystyle x = v - 2\, s

Choose and carry out a method

Use substitution when a variable is already isolated. Use elimination when multiplying an equation makes one pair of coefficients match. Subtract entire equations, including their constants.

Check the reasoning

Check the resulting pair in both equations. Parallel lines give no solution, while identical equations provide only one independent condition.

WORKED EXAMPLE 1

Solve $\displaystyle x + y = 7$ and $\displaystyle 3\, x + 2\, y = 16$ . Enter x only.

Eliminate one variable by comparing equal totals.
Subtract $\displaystyle 2\, \left(x + y\right) = 14$ from $\displaystyle 3\, x + 2\, y = 16$ .
This leaves $\displaystyle x = 2$ . Then $\displaystyle y = 5$ , which verifies both conditions.

WORKED EXAMPLE 2

Solve $\displaystyle x + y = 9$ and $\displaystyle 3\, x + 2\, y = 21$ . Enter x only.

Eliminate one variable by comparing equal totals.
Subtract $\displaystyle 2\, \left(x + y\right) = 18$ from $\displaystyle 3\, x + 2\, y = 21$ .
This leaves $\displaystyle x = 3$ . Then $\displaystyle y = 6$ , which verifies both conditions.

Common pitfalls

Possible mix-up: Solving one equation finishes a system.

The same pair must work in the second equation.

Possible mix-up: Elimination changes just one coefficient.

Multiply every term, including the right-hand side.

Explain it to yourself

How would you recognize that two equations describe the same line?

Preview the eight practice prompts

Problem 1
Solve $\displaystyle x + y = 13$ and $\displaystyle 3\, x + 2\, y = 31$ . Enter x only.
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Problem 2
Solve $\displaystyle x + y = 15$ and $\displaystyle 3\, x + 2\, y = 36$ . Enter x only.
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Problem 3
Solve $\displaystyle x + y = 17$ and $\displaystyle 3\, x + 2\, y = 41$ . Enter x only.
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Problem 4
Solve $\displaystyle x + y = 19$ and $\displaystyle 3\, x + 2\, y = 46$ . Enter x only.
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Problem 5
Solve $\displaystyle x + y = 21$ and $\displaystyle 3\, x + 2\, y = 51$ . Enter x only.
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Problem 6
Solve $\displaystyle x + y = 23$ and $\displaystyle 3\, x + 2\, y = 56$ . Enter x only.
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Problem 7New context
A 25-piece kit contains large clips weighing 3 g each and small clips weighing 2 g each. Total mass is 61 g. How many large clips are there?
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Problem 8New context
A 27-piece kit contains large clips weighing 3 g each and small clips weighing 2 g each. Total mass is 66 g. How many large clips are there?
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Open stage ALG 1.2 in the student workspace →

STAGE ALG 1.3 · 8 PRACTICE PROBLEMS · 2 NEW-CONTEXT APPLICATIONS

Slope and Rate of Change

Useful preparation: Two Equations, Two Unknowns

Goal: Calculate and interpret the rate of a linear relationship.

Before you begin: Two Equations, Two Unknowns

AN EQUATION IS A RULE FOR MAKING POINTS

Change the rule. Watch the graph.

Parent equation: $\displaystyle y = x$

Start with $\displaystyle y = x$ . Multiply each y value by 2, then shift up 1. A negative multiplier reflects across the x-axis.

Trace a point along the graph

Blue: $\displaystyle y = 2\, x + 1$

Moving point: (-4, -7)

Dashed gray: parent equation. Blue: current equation. View: −10 to 10 on both axes. The graph continues beyond this window. Purple marks your selected point.

Slope a: 2

Vertical intercept b: 1

Choose the point’s x: 1

Point (1, 3)

\displaystyle 2 \times \left(1\right) + \left(1\right) = 3

What does moving one step right do?

Each step of 1 to the right changes y by 2. The line rises. Setting $\displaystyle x = 0$ gives $\displaystyle y = 1$ , where the line meets the vertical axis.

Three points made by the blue rule
x	Substitute into the rule	y
-1	$\displaystyle 2 \times \left(-1\right) + \left(1\right)$	-1
0	$\displaystyle 2 \times \left(0\right) + \left(1\right)$	1
1	$\displaystyle 2 \times \left(1\right) + \left(1\right)$	3

Understand the idea

Slope compares how much the output changes with how much the input changes. Equal horizontal steps on a line produce equal vertical steps. Units belong to the ratio: meters per second is different from seconds per meter.

\displaystyle m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}

\displaystyle x_{2} \ne x_{1}

Choose and carry out a method

Label two points consistently, subtract their y-coordinates, and divide by the corresponding difference of x-coordinates. A line through one point can then be written as $\displaystyle y - y_{1} = m\, \left(x - x_{1}\right)$ .

Check the reasoning

A rising line has positive slope and a falling line has negative slope. A horizontal line has slope zero. A vertical line has no defined finite slope because its change in x is zero.

WORKED EXAMPLE 1

Find the slope of the line through (1,4) and (4,7).

Slope is change in output divided by change in input.
$\displaystyle m = \frac{7 - 4}{4 - 1} = \frac{3}{3}$ .
The slope is 1. Keep the coordinate order consistent in both differences.

WORKED EXAMPLE 2

Find the slope of the line through (2,5) and (5,10).

Slope is change in output divided by change in input.
$\displaystyle m = \frac{10 - 5}{5 - 2} = \frac{5}{3}$ .
The slope is $\displaystyle \frac{5}{3}$ . Keep the coordinate order consistent in both differences.

$\displaystyle \frac{5}{3}$

Common pitfalls

Possible mix-up: Slope is the second y-value divided by the second x-value.

Use differences unless the other point is the origin.

Possible mix-up: A vertical line has slope zero.

Its denominator is zero, so its slope is undefined.

Explain it to yourself

Why does switching both point labels leave the slope unchanged?

Preview the eight practice prompts

Problem 1
Find the slope of the line through (4,7) and (7,16).
Practise this exact problem →
Problem 2
Find the slope of the line through (5,8) and (8,19).
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Problem 3
Find the slope of the line through (6,9) and (9,22).
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Problem 4
Find the slope of the line through (7,10) and (10,25).
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Problem 5
Find the slope of the line through (8,11) and (11,28).
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Problem 6
Find the slope of the line through (9,12) and (12,31).
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Problem 7New context
A sensor reading rises from 13 to 34 as elapsed time rises from 10 to 13 seconds. What is the rate of change per second?
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Problem 8New context
A sensor reading rises from 14 to 37 as elapsed time rises from 11 to 14 seconds. What is the rate of change per second?
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Open stage ALG 1.3 in the student workspace →

Browse another level →

Equations and Linear Relationships

Linear Equations

What is inside the mystery box?

Change the rule. Watch the graph.

Parent equation: y=x\displaystyle y = xy=x

Understand the idea

Choose and carry out a method

Check the reasoning

Solve 3 x−6=6\displaystyle 3\, x - 6 = 63x−6=6 for x.

Solve 4 x−7=13\displaystyle 4\, x - 7 = 134x−7=13 for x.

Common pitfalls

Explain it to yourself

Two Equations, Two Unknowns

Change the rule. Watch the graph.

Parent equation: y=x\displaystyle y = xy=x

Understand the idea

Choose and carry out a method

Check the reasoning

Solve x+y=7\displaystyle x + y = 7x+y=7 and 3 x+2 y=16\displaystyle 3\, x + 2\, y = 163x+2y=16. Enter x only.

Solve x+y=9\displaystyle x + y = 9x+y=9 and 3 x+2 y=21\displaystyle 3\, x + 2\, y = 213x+2y=21. Enter x only.

Common pitfalls

Explain it to yourself

Slope and Rate of Change

Change the rule. Watch the graph.

Parent equation: y=x\displaystyle y = xy=x

Understand the idea

Choose and carry out a method

Check the reasoning

Find the slope of the line through (1,4) and (4,7).

Find the slope of the line through (2,5) and (5,10).

Common pitfalls

Explain it to yourself

Parent equation: $\displaystyle y = x$

Solve $\displaystyle 3\, x - 6 = 6$ for x.

Solve $\displaystyle 4\, x - 7 = 13$ for x.

Parent equation: $\displaystyle y = x$

Solve $\displaystyle x + y = 7$ and $\displaystyle 3\, x + 2\, y = 16$ . Enter x only.

Solve $\displaystyle x + y = 9$ and $\displaystyle 3\, x + 2\, y = 21$ . Enter x only.

Parent equation: $\displaystyle y = x$