Answer key · Unit 4

Linear equations.

Each lesson’s practice questions, their correct answers and the mistakes they spot. Answers read off the Scene use the same setting the learner sees.

Misconception catalogue

Answer key 4

Lesson 1 of 4 · About 20 minutes

Solving an equation

An equation as two lines that meet, and solving it step by step by doing the same thing to both sides until x is on its own.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
Two sides, two lines
Practice · Calculate
Question
Solve the equation $2x + 1 = 7$ ⁠. What is $x$ ⁠?

Correct answer
3 ± 0.01
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
2fixed
Left: number
1fixed
Right: number
7fixed
The two sides / Where they meet
The two sides
Accepted within
± 0.01
Computed from the Scene at this step’s setting
X where the curve meets the second line (the smaller, when two); NaN when they never meet or are the same line/nodes/graph/outputs/crossX
Wrong answers it recognises
- 4 ± 0.01
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
- 6 ± 0.01
  Misconception: Stops at 2x = 6 and answers 6, forgetting to divide by the coefficient of x.
  Feedback from the catalogue
  $2x = 6$ ⁠ does not yet say what $x$ ⁠ is: it says what twice $x$ ⁠ is. The last step is to divide both sides by the coefficient of $x$ ⁠: $x = 6 \div 2 = 3$ ⁠. Check by substituting: $2 \cdot 6 + 1 = 13$ ⁠, not $7$ ⁠.
Feedback for a right answer
$x = 3$ ⁠: subtract $1$ ⁠ from both sides, $2x = 6$ ⁠, and divide by $2$ ⁠. Check: $2 \cdot 3 + 1 = 7$ ⁠. Choose Where they meet under Look inside and watch the lines meet at $(3, 7)$ ⁠.

Feedback for a wrong answer
Not yet. Try substituting your answer into $2x + 1$ ⁠: do you get $7$ ⁠?

Hint
What was done to $x$ ⁠? First it was multiplied by $2$ ⁠, then $1$ ⁠ was added. Undo them in reverse order.

Worked solution
1. The aim: leave $x$ ⁠ on its own on one side.
2. Subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠.
3. Divide both sides by $2$ ⁠: $x = 3$ ⁠.
4. Check: $2 \cdot 3 + 1 = 6 + 1 = 7$ ⁠. ✓
If you got $4$ ⁠, you added $1$ ⁠ instead of subtracting it ( $2x = 8$ ⁠). If you got $6$ ⁠, you stopped at $2x = 6$ ⁠ and forgot to divide by $2$ ⁠.
Step 2 of 4
Open this step
The same to both sides
Practice · Choose
Question
What is the simplest first step in solving $3x - 4 = 5$ ⁠?

Correct answer
- Correct Add $4$ ⁠ to both sides
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
3fixed
Left: number
-4fixed
Right: number
5fixed
The two sides / Where they meet
The two sides
Choices
Shown in a random order.
- Correct Add $4$ ⁠ to both sides
- Wrong Subtract $4$ ⁠ from both sides
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
- Wrong Divide $3x$ ⁠ by $3$ ⁠
  Misconception: Does an operation to one side, or to one term, only.
  Feedback from the catalogue
  An equation is like a pair of scales: whatever you do, you do to both sides, and to the whole of each side. Add $4$ ⁠ to one side only and the equality breaks, so the $x$ ⁠ you find no longer solves the original equation.
- Wrong Add $4$ ⁠ to the left side
  Misconception: Does an operation to one side, or to one term, only.
  Feedback from the catalogue
  An equation is like a pair of scales: whatever you do, you do to both sides, and to the whole of each side. Add $4$ ⁠ to one side only and the equality breaks, so the $x$ ⁠ you find no longer solves the original equation.
Feedback for a right answer
Adding $4$ ⁠ undoes subtracting $4$ ⁠: $3x - 4 + 4 = 5 + 4$ ⁠, that is, $3x = 9$ ⁠. Then divide by $3$ ⁠ to get $x = 3$ ⁠.

Hint
What was done to $x$ ⁠ last? Undo that first.

Worked solution
1. In $3x - 4$ ⁠, $x$ ⁠ was first multiplied by $3$ ⁠ and then $4$ ⁠ was subtracted. Undo in reverse order, so start with the subtraction.
2. Subtracting $4$ ⁠ is undone by adding $4$ ⁠, to both sides: $3x - 4 + 4 = 5 + 4$ ⁠, that is, $3x = 9$ ⁠.
3. Divide both sides by $3$ ⁠: $x = 3$ ⁠.
4. Check: $3 \cdot 3 - 4 = 5$ ⁠. ✓
Subtracting $4$ ⁠ would give $3x - 8 = 1$ ⁠, further from $x$ ⁠ on its own. And dividing $3x$ ⁠ alone, without the $-4$ ⁠ or the right side, breaks the equality.
Step 3 of 4
Open this step
x on both sides
Practice · Calculate
Question
Solve $4x - 3 = x + 3$ ⁠. What is $x$ ⁠?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
4fixed
Left: number
-3fixed
Right: number
3fixed
The two sides / Where they meet
The two sides
Other settings
M2/nodes/graph/props/m2 = 1
Accepted within
± 0.01
Computed from the Scene at this step’s setting
X where the curve meets the second line (the smaller, when two); NaN when they never meet or are the same line/nodes/graph/outputs/crossX
Wrong answers it recognises
- 1.2 ± 0.01
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
- 0 ± 0.01
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
Feedback for a right answer
$x = 2$ ⁠: $4x - x = 3 + 3$ ⁠, that is, $3x = 6$ ⁠. Check: the left side is $4 \cdot 2 - 3 = 5$ ⁠ and the right side $2 + 3 = 5$ ⁠. The lines meet at $(2, 5)$ ⁠.

Feedback for a wrong answer
Not yet. Substitute your answer into both sides: do they give the same number?

Hint
Subtract $x$ ⁠ from both sides, then add $3$ ⁠ to both sides.

Worked solution
1. Subtract $x$ ⁠ from both sides: $4x - x - 3 = 3$ ⁠, that is, $3x - 3 = 3$ ⁠.
2. Add $3$ ⁠ to both sides: $3x = 6$ ⁠.
3. Divide by $3$ ⁠: $x = 2$ ⁠.
4. Check: $4 \cdot 2 - 3 = 5$ ⁠ and $2 + 3 = 5$ ⁠. ✓
If you got $1.2$ ⁠, you moved $x$ ⁠ to the left with a plus ( $5x = 6$ ⁠). If you got $0$ ⁠, you moved the $-3$ ⁠ to the right with a minus ( $3x = 0$ ⁠). At every step, do the opposite operation, to both sides.
Step 4 of 4
Open this step
Check by substituting
Practice · Reach the target
Question
The left side is $2x + 1$ ⁠. Set Right: number so that the solution of the equation is $x = -2$ ⁠, then press Check.

Goal
- -2 ± 0.01
  At the start of the step: 3
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
2fixed
Left: number
1fixed
Right: number
7
The two sides / Where they meet
Where they meet
Read from the Scene at
X where the curve meets the second line (the smaller, when two); NaN when they never meet or are the same line/nodes/graph/outputs/crossX
One setting that reaches it
Right: number
-3
Feedback for a right answer
$2 \cdot (-2) + 1 = -3$ ⁠, so the equation $2x + 1 = -3$ ⁠ is solved by $x = -2$ ⁠. Substituting is also how you check any solution.

Feedback for a wrong answer
Not yet. The meeting point has to land above $x = -2$ ⁠. Move the pink line until it crosses the blue one there.

Hint
Substitute $x = -2$ ⁠ into the left side: what do you get? That is the number the right side needs.

Worked solution
1. For $x = -2$ ⁠ to solve the equation, the two sides must be equal when $x = -2$ ⁠.
2. Substitute into the left side: $2 \cdot (-2) + 1 = -4 + 1 = -3$ ⁠.
3. So the right side must be $-3$ ⁠: set Right: number to $-3$ ⁠.
4. Check: $2x + 1 = -3$ ⁠, $2x = -4$ ⁠, $x = -2$ ⁠. ✓

Lesson 2 of 4 · About 20 minutes

Inequalities

When one side is smaller than the other, how the solution becomes a whole stretch of the x-axis, and why multiplying or dividing by a negative number turns the sign round.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
Where is it smaller?
Practice · Predict, then look
Question
Which values of $x$ ⁠ solve $2x + 1 < 7$ ⁠?

Correct answer
- Correct Every $x$ ⁠ less than $3$ ⁠
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
2fixed
Right: number
7fixed
Sign
=fixed
The two sides / Where they meet
Where they meet
Other settings
C/nodes/graph/props/c = 1
Choices
- Correct Every $x$ ⁠ less than $3$ ⁠
- Wrong Every $x$ ⁠ greater than $3$ ⁠
  Feedback
  Try one number: at $x = 5$ ⁠ you get $2 \cdot 5 + 1 = 11$ ⁠, and $11$ ⁠ is not less than $7$ ⁠.
- Wrong Only $x = 3$ ⁠
  Feedback
  At $x = 3$ ⁠ the two sides are equal: that solves the equation, not the inequality. It is the boundary between the solutions and the rest.
- Wrong Every $x$ ⁠ less than $7$ ⁠
  Feedback
  $7$ ⁠ is the value of the right side, not a boundary for $x$ ⁠. Try $x = 5$ ⁠: $11 < 7$ ⁠ is false.
Once the prediction is locked in, the Scene changes
Sign
<fixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
To the left of $x = 3$ ⁠ the blue line is below the pink one, so there $2x + 1 < 7$ ⁠. The solution is $x < 3$ ⁠: a whole stretch of the $x$ ⁠-axis, not one number.

Worked solution
1. Solve it like an equation: subtract $1$ ⁠ from both sides, $2x < 6$ ⁠.
2. Divide by $2$ ⁠, a positive number, and the sign stays: $x < 3$ ⁠.
3. Check one number on each side: $x = 0$ ⁠ gives $1 < 7$ ⁠ ✓, and $x = 5$ ⁠ gives $11 < 7$ ⁠ ✗.
On the board: the blue line climbs, so it is below the pink one to the left of the meeting point.
Step 2 of 4
Open this step
Multiplying by a negative number
Practice · Predict, then look
Question
And now $-2x + 1 < 7$ ⁠. Which values of $x$ ⁠ solve it?

Correct answer
- Correct $x > -3$ ⁠
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
-2fixed
Right: number
7fixed
Sign
=fixed
The two sides / Where they meet
Where they meet
Other settings
C/nodes/graph/props/c = 1
Choices
- Correct $x > -3$ ⁠
- Wrong $x < -3$ ⁠
  Misconception: Divides an inequality by a negative number without turning the sign round.
  Feedback from the catalogue
  Multiplying or dividing by a negative number turns the inequality sign round. $2 < 5$ ⁠, but $-2 > -5$ ⁠. In $-2x < 6$ ⁠ you divide by $-2$ ⁠ and get $x > -3$ ⁠. Check one number: $x = 0$ ⁠ gives $0 < 6$ ⁠, which is true, and $0$ ⁠ is indeed greater than $-3$ ⁠.
- Wrong $x > 3$ ⁠
  Feedback
  $-2x < 6$ ⁠, then divide by $-2$ ⁠: $6 \div (-2) = -3$ ⁠, not $3$ ⁠.
- Wrong $x < 3$ ⁠
  Feedback
  That is the solution of $2x + 1 < 7$ ⁠. Here the coefficient of $x$ ⁠ is negative, and the line slopes down.
Once the prediction is locked in, the Scene changes
Sign
<fixed
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
$-2x < 6$ ⁠, and dividing by $-2$ ⁠ turns the sign round: $x > -3$ ⁠. On the board the blue line slopes down, so it is below the pink one to the right of the meeting point.

Worked solution
1. Subtract $1$ ⁠ from both sides: $-2x < 6$ ⁠.
2. Divide both sides by $-2$ ⁠. It is a negative number, so the sign turns round: $x > -3$ ⁠.
3. Check: $x = 0$ ⁠ gives $-2 \cdot 0 + 1 = 1 < 7$ ⁠ ✓, and $0 > -3$ ⁠ ✓. And $x = -5$ ⁠ gives $11 < 7$ ⁠ ✗.
If you chose $x < -3$ ⁠, you divided by $-2$ ⁠ without turning the sign round. The check at $x = 0$ ⁠ shows at once that something is off.
Step 3 of 4
Open this step
The boundary itself
Practice · Calculate
Question
What is the largest integer that solves $2x + 1 < 7$ ⁠?

Correct answer
2 ± 0.01
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
2fixed
Right: number
7fixed
Sign
<
The two sides / Where they meet
Where they meet
Other settings
C/nodes/graph/props/c = 1
Accepted within
± 0.01
A fixed value in the question
Wrong answers it recognises
- 3 ± 0.01
  Misconception: Counts the boundary of a strict inequality (< or >) as a solution.
  Feedback from the catalogue
  With $<$ ⁠ and $>$ ⁠ the boundary itself is not a solution: there the two sides are equal, and equal is not smaller. Only with $\le$ ⁠ and $\ge$ ⁠ is the boundary included. On the board: an empty ring for a boundary left out, a filled dot for one included.
Feedback for a right answer
$2$ ⁠: it is less than $3$ ⁠, and $2 \cdot 2 + 1 = 5 < 7$ ⁠. $3$ ⁠ itself gives $7 < 7$ ⁠, which is false. Switch Sign to ≤ and watch the empty ring fill in.

Hint
The solution is $x < 3$ ⁠. Is $3$ ⁠ itself included?

Worked solution
1. The solution of $2x + 1 < 7$ ⁠ is $x < 3$ ⁠.
2. $3$ ⁠ itself is not included: $2 \cdot 3 + 1 = 7$ ⁠, and $7$ ⁠ is not less than $7$ ⁠.
3. The largest integer less than $3$ ⁠ is $2$ ⁠. Check: $2 \cdot 2 + 1 = 5 < 7$ ⁠. ✓
If you answered $3$ ⁠, you included the boundary. That is right only for $2x + 1 \le 7$ ⁠.
Step 4 of 4
Open this step
Your turn
Practice · Choose
Question
Solve $5 - x > 2$ ⁠.

Correct answer
- Correct $x < 3$ ⁠
Setup, choices and feedback
The Scene at this step
Left: coefficient of x
-1fixed
Right: number
2fixed
Sign
=
The two sides / Where they meet
Where they meet
Other settings
C/nodes/graph/props/c = 5
Choices
Shown in a random order.
- Correct $x < 3$ ⁠
- Wrong $x > 3$ ⁠
  Misconception: Divides an inequality by a negative number without turning the sign round.
  Feedback from the catalogue
  Multiplying or dividing by a negative number turns the inequality sign round. $2 < 5$ ⁠, but $-2 > -5$ ⁠. In $-2x < 6$ ⁠ you divide by $-2$ ⁠ and get $x > -3$ ⁠. Check one number: $x = 0$ ⁠ gives $0 < 6$ ⁠, which is true, and $0$ ⁠ is indeed greater than $-3$ ⁠.
- Wrong $x > -3$ ⁠
  Feedback
  From $5 - x > 2$ ⁠ subtract $5$ ⁠: $-x > -3$ ⁠. Now multiply by $-1$ ⁠, and two things happen together: the signs of the numbers flip, and so does the inequality sign.
- Wrong $x < -3$ ⁠
  Feedback
  From $-x > -3$ ⁠, multiplying by $-1$ ⁠ flips the number too: $3$ ⁠, not $-3$ ⁠.
Feedback for a right answer
$-x > -3$ ⁠, and multiplying by $-1$ ⁠ turns the sign round: $x < 3$ ⁠. Check: $x = 0$ ⁠ gives $5 > 2$ ⁠ ✓.

Hint
Isolate $-x$ ⁠, then multiply by $-1$ ⁠. What happens to the sign?

Worked solution
1. Subtract $5$ ⁠ from both sides: $-x > -3$ ⁠.
2. Multiply both sides by $-1$ ⁠. It is a negative number, so the sign turns round: $x < 3$ ⁠.
3. Check: $x = 0$ ⁠ gives $5 - 0 = 5 > 2$ ⁠ ✓, and $x = 4$ ⁠ gives $1 > 2$ ⁠ ✗.
You can also avoid negatives: add $x$ ⁠ to both sides, $5 > 2 + x$ ⁠, then subtract $2$ ⁠: $3 > x$ ⁠, that is, $x < 3$ ⁠.

Lesson 3 of 4 · About 20 minutes

Rearranging a formula

A formula is an equation with several letters. You isolate one of them with the same steps, and get a new formula that answers the opposite question.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
A formula as a line
Practice · Calculate
Question
The plumber charges $y = 2x + 3$ ⁠ shekels for $x$ ⁠ hours. How many hours does a budget of $11$ ⁠ shekels buy?

Correct answer
4 ± 0.01
Setup, choices and feedback
The Scene at this step
Price per hour
2fixed
Call-out fee
3fixed
Budget
11fixed
The cost line / Where the budget runs out
The cost line
Accepted within
± 0.01
Computed from the Scene at this step’s setting
X where the curve meets the second line (the smaller, when two); NaN when they never meet or are the same line/nodes/graph/outputs/crossX
Wrong answers it recognises
- 8 ± 0.01
  Misconception: Stops at 2x = 6 and answers 6, forgetting to divide by the coefficient of x.
  Feedback from the catalogue
  $2x = 6$ ⁠ does not yet say what $x$ ⁠ is: it says what twice $x$ ⁠ is. The last step is to divide both sides by the coefficient of $x$ ⁠: $x = 6 \div 2 = 3$ ⁠. Check by substituting: $2 \cdot 6 + 1 = 13$ ⁠, not $7$ ⁠.
- 2.5 ± 0.01
  Misconception: Divides only one term of a side: (y − 3)/2 becomes y/2 − 3.
  Feedback from the catalogue
  When you divide a side by $2$ ⁠, you divide all of it: $\dfrac{y - 3}{2}$ ⁠, not $\dfrac{y}{2} - 3$ ⁠. Try a number: at $y = 11$ ⁠ the first gives $4$ ⁠ and the second $2.5$ ⁠. Only $4$ ⁠ gives back $11$ ⁠ in $y = 2x + 3$ ⁠.
Feedback for a right answer
$2x + 3 = 11$ ⁠, so $2x = 8$ ⁠ and $x = 4$ ⁠ hours. Check: $2 \cdot 4 + 3 = 11$ ⁠. Choose Where the budget runs out under Look inside and see the meeting point at $(4, 11)$ ⁠.

Hint
Solve $2x + 3 = 11$ ⁠.

Worked solution
1. The budget is the cost: $2x + 3 = 11$ ⁠.
2. Subtract $3$ ⁠ from both sides (the call-out fee): $2x = 8$ ⁠.
3. Divide by $2$ ⁠ (the price per hour): $x = 4$ ⁠.
4. Check: $2 \cdot 4 + 3 = 11$ ⁠. ✓
If you got $8$ ⁠, you forgot to divide by the price per hour. If you got $2.5$ ⁠, you divided only the $11$ ⁠ by $2$ ⁠ and then subtracted $3$ ⁠: $\tfrac{11}{2} - 3$ ⁠.
Step 2 of 4
Open this step
Isolate x
Practice · Choose
Question
Isolate $x$ ⁠ in the formula $y = 2x + 3$ ⁠.

Correct answer
- Correct $x = \dfrac{y - 3}{2}$ ⁠
Setup, choices and feedback
The Scene at this step
Price per hour
2
Call-out fee
3
Budget
11
The cost line / Where the budget runs out
Where the budget runs out
Choices
Shown in a random order.
- Correct $x = \dfrac{y - 3}{2}$ ⁠
- Wrong $x = \dfrac{y}{2} - 3$ ⁠
  Misconception: Divides only one term of a side: (y − 3)/2 becomes y/2 − 3.
  Feedback from the catalogue
  When you divide a side by $2$ ⁠, you divide all of it: $\dfrac{y - 3}{2}$ ⁠, not $\dfrac{y}{2} - 3$ ⁠. Try a number: at $y = 11$ ⁠ the first gives $4$ ⁠ and the second $2.5$ ⁠. Only $4$ ⁠ gives back $11$ ⁠ in $y = 2x + 3$ ⁠.
- Wrong $x = \dfrac{y + 3}{2}$ ⁠
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
- Wrong $x = 2y - 3$ ⁠
  Feedback
  Multiplying by $2$ ⁠ is undone by dividing by $2$ ⁠, not multiplying. Check: at $y = 11$ ⁠ you get $19$ ⁠ hours, but $2 \cdot 19 + 3 = 41$ ⁠.
Feedback for a right answer
$y - 3 = 2x$ ⁠, so $x = \dfrac{y - 3}{2}$ ⁠. Check: at $y = 11$ ⁠ you get $\dfrac{8}{2} = 4$ ⁠, as in the last step. Change Budget and check another number.

Hint
The same steps as for $2x + 3 = 11$ ⁠, only with $y$ ⁠ in place of $11$ ⁠.

Worked solution
1. Subtract $3$ ⁠ from both sides: $y - 3 = 2x$ ⁠.
2. Divide both sides by $2$ ⁠. Divide all of the left side: $\dfrac{y - 3}{2} = x$ ⁠.
3. Check with $y = 11$ ⁠: $\dfrac{11 - 3}{2} = 4$ ⁠, and the plumber does charge $2 \cdot 4 + 3 = 11$ ⁠ for $4$ ⁠ hours. ✓
The slip $\dfrac{y}{2} - 3$ ⁠ divides only $y$ ⁠: at $y = 11$ ⁠ it gives $2.5$ ⁠, and $2 \cdot 2.5 + 3 = 8$ ⁠, not $11$ ⁠.
Step 3 of 4
Open this step
Other formulas
Practice · Choose
Question
The perimeter of a rectangle is $P = 2a + 2b$ ⁠. Isolate $b$ ⁠.

Correct answer
- Correct $b = \dfrac{P - 2a}{2}$ ⁠
Setup, choices and feedback
The Scene at this step
Price per hour
2
Call-out fee
3
Budget
11
The cost line / Where the budget runs out
Where the budget runs out
Choices
Shown in a random order.
- Correct $b = \dfrac{P - 2a}{2}$ ⁠
- Wrong $b = \dfrac{P}{2} - 2a$ ⁠
  Misconception: Divides only one term of a side: (y − 3)/2 becomes y/2 − 3.
  Feedback from the catalogue
  When you divide a side by $2$ ⁠, you divide all of it: $\dfrac{y - 3}{2}$ ⁠, not $\dfrac{y}{2} - 3$ ⁠. Try a number: at $y = 11$ ⁠ the first gives $4$ ⁠ and the second $2.5$ ⁠. Only $4$ ⁠ gives back $11$ ⁠ in $y = 2x + 3$ ⁠.
- Wrong $b = \dfrac{P + 2a}{2}$ ⁠
  Misconception: Moves a term to the other side without changing its sign: 2x + 1 = 7 becomes 2x = 7 + 1.
  Feedback from the catalogue
  “Moving a term across” is shorthand for something you do to both sides. In $2x + 1 = 7$ ⁠ you subtract $1$ ⁠ from both sides: $2x + 1 - 1 = 7 - 1$ ⁠, that is, $2x = 6$ ⁠. The $+1$ ⁠ did not move across as it was; it was cancelled, and a $-1$ ⁠ appeared on the other side.
- Wrong $b = P - 2a$ ⁠
  Misconception: Stops at 2x = 6 and answers 6, forgetting to divide by the coefficient of x.
  Feedback from the catalogue
  $2x = 6$ ⁠ does not yet say what $x$ ⁠ is: it says what twice $x$ ⁠ is. The last step is to divide both sides by the coefficient of $x$ ⁠: $x = 6 \div 2 = 3$ ⁠. Check by substituting: $2 \cdot 6 + 1 = 13$ ⁠, not $7$ ⁠.
Feedback for a right answer
$P - 2a = 2b$ ⁠, so $b = \dfrac{P - 2a}{2}$ ⁠, which is also $\dfrac{P}{2} - a$ ⁠. Check: a $3$ ⁠ by $4$ ⁠ rectangle has perimeter $14$ ⁠, and $b = \dfrac{14 - 6}{2} = 4$ ⁠.

Hint
Treat $P$ ⁠ and $a$ ⁠ as if they were numbers: what is added to $2b$ ⁠, and what multiplies $b$ ⁠?

Worked solution
1. Subtract $2a$ ⁠ from both sides: $P - 2a = 2b$ ⁠.
2. Divide both sides by $2$ ⁠, all of the left side: $b = \dfrac{P - 2a}{2}$ ⁠.
3. You can also split it: $\dfrac{P}{2} - \dfrac{2a}{2} = \dfrac{P}{2} - a$ ⁠.
4. Check with a $3$ ⁠ by $4$ ⁠ rectangle: $P = 2 \cdot 3 + 2 \cdot 4 = 14$ ⁠, and $\dfrac{14 - 2 \cdot 3}{2} = 4$ ⁠. ✓
$\dfrac{P}{2} - 2a$ ⁠ gives $7 - 6 = 1$ ⁠ here: only $P$ ⁠ was divided by $2$ ⁠, not the $2a$ ⁠.
Step 4 of 4
Open this step
From hours to budget
Practice · Reach the target
Question
Set Budget so that it buys exactly $3.5$ ⁠ hours of work, then press Check.

Goal
- 3.5 ± 0.01
  At the start of the step: 4
Setup, choices and feedback
The Scene at this step
Price per hour
2fixed
Call-out fee
3fixed
Budget
11
The cost line / Where the budget runs out
Where the budget runs out
Read from the Scene at
X where the curve meets the second line (the smaller, when two); NaN when they never meet or are the same line/nodes/graph/outputs/crossX
One setting that reaches it
Budget
10
Feedback for a right answer
$y = 2 \cdot 3.5 + 3 = 10$ ⁠ shekels. The original formula answers “how much does it cost”, and the rearranged one, $x = \dfrac{y - 3}{2}$ ⁠, answers “how many hours”: $\dfrac{10 - 3}{2} = 3.5$ ⁠.

Feedback for a wrong answer
Not yet. The meeting point has to land above $x = 3.5$ ⁠, halfway between $3$ ⁠ and $4$ ⁠.

Hint
No need to isolate anything here: substitute $x = 3.5$ ⁠ into $y = 2x + 3$ ⁠.

Worked solution
1. You know the hours, $x = 3.5$ ⁠, and want the cost: substitute into the original formula.
2. $y = 2 \cdot 3.5 + 3 = 7 + 3 = 10$ ⁠.
3. Set Budget to $10$ ⁠.
4. Check with the rearranged formula: $x = \dfrac{10 - 3}{2} = 3.5$ ⁠. ✓

Lesson 4 of 4 · About 25 minutes

Two equations in two unknowns

Two lines, one point that satisfies both, and two ways to find it without the board, substitution and adding the equations. And what happens when the lines are parallel.

Open the lesson

The 4 steps and their answers

Step 1 of 4
Open this step
A point on both lines
Practice · Choose
Question
What is the solution of the system $y = x + 1$ ⁠ and $y = -x + 5$ ⁠?

Correct answer
- Correct $(2, 3)$ ⁠
Setup, choices and feedback
The Scene at this step
First: coefficient of x
1fixed
Second: coefficient of x
-1fixed
Second: number
5fixed
The two lines / Where they meet
The two lines
Other settings
C/nodes/graph/props/c = 1
Choices
Shown in a random order.
- Correct $(2, 3)$ ⁠
- Wrong $(3, 2)$ ⁠
  Misconception: Writes a point as (y, x), or gives x when asked for y.
  Feedback from the catalogue
  A point is always written $(x, y)$ ⁠: first the distance across, then the height. Check by substituting: put the first number in for $x$ ⁠ and see that you get the second.
- Wrong $(0, 1)$ ⁠
  Feedback
  $(0, 1)$ ⁠ is on the first line, but on the second line $x = 0$ ⁠ gives $y = 5$ ⁠. A solution of a system has to satisfy both equations.
- Wrong $(4, 0)$ ⁠
  Feedback
  $(4, 0)$ ⁠ is on the second line only: on the first, $x = 4$ ⁠ gives $y = 5$ ⁠.
Feedback for a right answer
$x + 1 = -x + 5$ ⁠ gives $2x = 4$ ⁠, $x = 2$ ⁠, and then $y = 2 + 1 = 3$ ⁠. Choose Where they meet under Look inside and watch the lines meet at $(2, 3)$ ⁠.

Hint
Where they meet, the two $y$ ⁠ values are equal: $x + 1 = -x + 5$ ⁠.

Worked solution
1. Where the lines meet, their $y$ ⁠ values are equal, so $x + 1 = -x + 5$ ⁠.
2. Add $x$ ⁠ to both sides: $2x + 1 = 5$ ⁠. Subtract $1$ ⁠: $2x = 4$ ⁠, so $x = 2$ ⁠.
3. Substitute into one equation: $y = 2 + 1 = 3$ ⁠.
4. Check in the other: $-2 + 5 = 3$ ⁠. ✓ The solution is $(2, 3)$ ⁠.
$(3, 2)$ ⁠ has the same numbers the other way round: on the first line $x = 3$ ⁠ gives $y = 4$ ⁠, not $2$ ⁠.
Step 2 of 4
Open this step
Substitution
Practice · Calculate
Question
Solve the system $y = 2x - 1$ ⁠ and $x + y = 8$ ⁠. What is $y$ ⁠?

Correct answer
5 ± 0.01
Setup, choices and feedback
The Scene at this step
First: coefficient of x
2fixed
Second: coefficient of x
-1fixed
Second: number
8fixed
The two lines / Where they meet
The two lines
Other settings
C/nodes/graph/props/c = -1
Accepted within
± 0.01
Computed from the Scene at this step’s setting
Y there/nodes/graph/outputs/crossY
Wrong answers it recognises
- 3 ± 0.01
  Misconception: Writes a point as (y, x), or gives x when asked for y.
  Feedback from the catalogue
  A point is always written $(x, y)$ ⁠: first the distance across, then the height. Check by substituting: put the first number in for $x$ ⁠ and see that you get the second.
- 3.667 ± 0.01
  Misconception: Drops the brackets when substituting an expression with a minus: x + 2x + 1 instead of x + (2x − 1).
  Feedback from the catalogue
  When you substitute an expression for a letter, put it in brackets: $y = 2x - 1$ ⁠ goes into $x + y = 8$ ⁠ as $x + (2x - 1) = 8$ ⁠. The minus belongs to the $1$ ⁠ and stays with it: $3x - 1 = 8$ ⁠, so $x = 3$ ⁠ and $y = 5$ ⁠.
Feedback for a right answer
$x + (2x - 1) = 8$ ⁠, $3x = 9$ ⁠, $x = 3$ ⁠, so $y = 2 \cdot 3 - 1 = 5$ ⁠. Check: $3 + 5 = 8$ ⁠.

Feedback for a wrong answer
Not yet. Find $x$ ⁠ first by substituting, then put it into $y = 2x - 1$ ⁠.

Hint
Put $2x - 1$ ⁠ in place of $y$ ⁠, in brackets: $x + (2x - 1) = 8$ ⁠. Find $x$ ⁠, then $y$ ⁠.

Worked solution
1. The first equation says $y = 2x - 1$ ⁠. Put that, in brackets, in place of $y$ ⁠ in the second: $x + (2x - 1) = 8$ ⁠.
2. Open the brackets: $3x - 1 = 8$ ⁠. Add $1$ ⁠: $3x = 9$ ⁠, so $x = 3$ ⁠.
3. Put $x = 3$ ⁠ into $y = 2x - 1$ ⁠: $y = 6 - 1 = 5$ ⁠.
4. Check in the second equation: $3 + 5 = 8$ ⁠. ✓
If you answered $3$ ⁠, you gave $x$ ⁠ rather than $y$ ⁠. If you got about $3.67$ ⁠, you lost the minus when substituting: $x + 2x + 1 = 8$ ⁠.
Step 3 of 4
Open this step
Adding the equations
Practice · Choose
Question
Add the equations $x + y = 5$ ⁠ and $x - y = 1$ ⁠, side by side. Which equation do you get?

Correct answer
- Correct $2x = 6$ ⁠
Setup, choices and feedback
The Scene at this step
First: coefficient of x
-1fixed
Second: coefficient of x
1fixed
Second: number
-1fixed
The two lines / Where they meet
Where they meet
Other settings
C/nodes/graph/props/c = 5
Choices
Shown in a random order.
- Correct $2x = 6$ ⁠
- Wrong $2x + 2y = 6$ ⁠
  Feedback
  $y + (-y) = 0$ ⁠: the $y$ ⁠ from the first equation and the $-y$ ⁠ from the second cancel out. That is exactly why you add them.
- Wrong $2x = 4$ ⁠
  Misconception: Does an operation to one side, or to one term, only.
  Feedback from the catalogue
  An equation is like a pair of scales: whatever you do, you do to both sides, and to the whole of each side. Add $4$ ⁠ to one side only and the equality breaks, so the $x$ ⁠ you find no longer solves the original equation.
- Wrong $2y = 4$ ⁠
  Feedback
  $2y = 4$ ⁠ comes from subtracting the equations, not adding them. That works too, but the question is about adding.
Feedback for a right answer
$(x + y) + (x - y) = 5 + 1$ ⁠, that is, $2x = 6$ ⁠ and $x = 3$ ⁠. Substitute: $3 + y = 5$ ⁠, so $y = 2$ ⁠. On the board the lines meet at $(3, 2)$ ⁠.

Hint
Add each side separately: $(x + y) + (x - y)$ ⁠ on the left, $5 + 1$ ⁠ on the right.

Worked solution
1. Add the left sides: $(x + y) + (x - y) = 2x + y - y = 2x$ ⁠.
2. Add the right sides: $5 + 1 = 6$ ⁠.
3. The new equation: $2x = 6$ ⁠, so $x = 3$ ⁠.
4. Substitute into $x + y = 5$ ⁠: $y = 2$ ⁠. Check in $x - y = 1$ ⁠: $3 - 2 = 1$ ⁠. ✓
$2x = 4$ ⁠ comes from subtracting the right sides while adding the left ones. Whatever you do to one side, you do to the other.
Step 4 of 4
Open this step
Parallel lines
Practice · Predict, then look
Question
How many solutions does the system $y = 2x + 1$ ⁠ and $y = 2x - 3$ ⁠ have?

Correct answer
- Correct No solution
Setup, choices and feedback
The Scene at this step
First: coefficient of x
2fixed
Second: coefficient of x
2fixed
Second: number
-3fixed
The two lines / Where they meet
The two lines
Other settings
C/nodes/graph/props/c = 1
Choices
- Correct No solution
- Wrong One solution
  Misconception: Thinks any two lines meet once, so every system has exactly one solution.
  Feedback from the catalogue
  Two lines with the same coefficient of $x$ ⁠ rise at the same rate, so the gap between them never changes: they are parallel and never meet, and the system has no solution. If the constant terms are the same too, it is the same line, and every point on it is a solution.
- Wrong Infinitely many solutions
  Feedback
  There are infinitely many solutions when the two equations are the same line. Here the constant terms differ: $+1$ ⁠ and $-3$ ⁠.
Once the prediction is locked in, the Scene changes
The two lines / Where they meet
Where they meet
The Scene then shows: 0 · How many times the curve meets the second line: 0, 1 or 2, Infinity when they are the same line/nodes/graph/outputs/crossCount
A prediction is never marked as a failure: the learner sees the outcome and the explanation either way.
Feedback for a right answer
At any $x$ ⁠ the first line is always $4$ ⁠ higher than the second, so they never meet. If you try to solve, $2x + 1 = 2x - 3$ ⁠ gives $1 = -3$ ⁠, which is false: there is no solution.

Worked solution
1. Set the $y$ ⁠ values equal: $2x + 1 = 2x - 3$ ⁠.
2. Subtract $2x$ ⁠ from both sides: $1 = -3$ ⁠.
3. The $x$ ⁠ has vanished and what is left is false, so no $x$ ⁠ satisfies both equations: there is no solution.
On the board: both lines rise $2$ ⁠ for every step to the right, and the vertical gap between them is always $4$ ⁠. Parallel lines never meet.

Across the unit · 9 misconceptions

Misconception catalogue

Each common mistake this unit recognises, in a few words. Open one for what the learner reads when an answer matches it, and where it is asked.

Index

Solving an equation

Two sides, two lines

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

The same to both sides

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

x on both sides

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Check by substituting

Question

Goal

The Scene at this step

One setting that reaches it

Hint

Worked solution

Inequalities

Where is it smaller?

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

Multiplying by a negative number

Question

Correct answer

The Scene at this step

Choices

Once the prediction is locked in, the Scene changes

Worked solution

The boundary itself

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Your turn

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Rearranging a formula

A formula as a line

Question

Correct answer

The Scene at this step

Wrong answers it recognises

Hint

Worked solution

Isolate x

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution

Other formulas

Question

Correct answer

The Scene at this step

Choices

Hint

Worked solution