Mathematics · Algebra — Teacher Copy
Full Solutions
Formula shown first, then short numbered steps with the working on its own line, so you can glance down and pick up the thread instantly while teaching.
1Quadratic Equations
Question 1
Solve x² + x − 12 = 0 by factorisation.
Rule: find two numbers that multiply to c and add to b
1
a = 1, b = 1, c = −12. Find two numbers that multiply to −12 and add to 1
4 and −3 (4 × −3 = −12, 4 + −3 = 1)
2
Write as two brackets
(x + 4)(x − 3) = 0
3
Set each bracket to zero and solve
x = −4 or x = 3
✓x = −4 or x = 3
Teaching tip: Since a = 1 here, the two numbers go straight into the brackets — no middle-term splitting needed. Point this out as the “easy case” before moving to Question 3, where a ≠ 1.
Question 2
Solve x² − 4x − 3 = 0 by completing the square, giving your answer in surd form.
Method: (x + b/2)² − (b/2)² + c = 0, then isolate and square root
1
Half of −4 is −2, so complete the square
(x − 2)² − 4 − 3 = 0
2
Simplify the constants
(x − 2)² = 7
3
Take the square root of both sides — remember ±
x − 2 = ±√7
4
Add 2 to both sides
x = 2 ± √7
✓x = 2 + √7 or x = 2 − √7
Teaching tip: Ask the student to check (−2)² = 4 came from squaring HALF of the −4 coefficient, not the −4 itself — this is the step most students get wrong on their first few attempts.
Question 3
Solve 2x² + 3x − 4 = 0 using the quadratic formula, giving your answer in surd form.
Formula: x = (−b ± √(b² − 4ac)) / 2a
1
Identify a, b, c
a = 2, b = 3, c = −4
2
Calculate the discriminant
b² − 4ac = 9 − 4(2)(−4) = 9 + 32 = 41
3
Substitute into the formula
x = (−3 ± √41) / 4
✓x = (−3 + √41)/4 or x = (−3 − √41)/4
Teaching tip: Since 41 isn’t a perfect square, the answer must stay in surd form — a good moment to ask “could this have factorised nicely?” (check: no whole-number pair multiplies to −8 and adds to 3), reinforcing WHY the formula is needed here.
2Simultaneous Equations — One Linear, One Non-Linear
Question 4
Solve simultaneously: y = x + 1 and y = x² − 5.
Method: substitute the linear equation into the non-linear one, then solve the resulting quadratic
1
Both equations equal y, so set them equal to each other
x + 1 = x² − 5
2
Rearrange into standard quadratic form
0 = x² − x − 6
3
Factorise (numbers that multiply to −6, add to −1)
(x − 3)(x + 2) = 0 → x = 3 or x = −2
4
Find the matching y-value for EACH x, using y = x + 1
x=3: y=4 x=−2: y=−1
✓(3, 4) and (−2, −1)
Teaching tip: Emphasise finding a y for EVERY x — the single most common mark loss here is stopping after finding both x-values and forgetting to pair each with its own y.
Question 5
Solve simultaneously: y = x − 1 and x² + y² = 25.
Method: substitute the linear equation’s y into the circle equation
1
Substitute y = x − 1 into x² + y² = 25
x² + (x − 1)² = 25
2
Expand the bracket
x² + x² − 2x + 1 = 25
3
Collect terms and rearrange to standard form
2x² − 2x − 24 = 0 → x² − x − 12 = 0
4
Factorise
(x − 4)(x + 3) = 0 → x = 4 or x = −3
5
Find each matching y, using y = x − 1
x=4: y=3 x=−3: y=−4
✓(4, 3) and (−3, −4)
Teaching tip: Point out that dividing the whole equation by 2 in step 3 (2x²−2x−24=0 → x²−x−12=0) made the numbers much friendlier before factorising — always check for a common factor first.
Question 6
Solve simultaneously: y = x² and y = 3x − 2.
Method: set the two expressions for y equal to each other
1
Both equal y, so set them equal
x² = 3x − 2
2
Rearrange to standard form
x² − 3x + 2 = 0
3
Factorise
(x − 1)(x − 2) = 0 → x = 1 or x = 2
4
Find each matching y, using y = x²
x=1: y=1 x=2: y=4
✓(1, 1) and (2, 4)
Teaching tip: Have the student verify BOTH solutions in the other equation too (y = 3x−2): at x=1, y=3(1)−2=1 ✓; at x=2, y=3(2)−2=4 ✓. A quick double-check like this catches substitution slips immediately.
Question 21
Solve simultaneously: y = x² and y = −3x + 10.
Method: set the two expressions for y equal to each other
1
Set equal and rearrange
x² = −3x + 10 → x² + 3x − 10 = 0
2
Factorise
(x + 5)(x − 2) = 0 → x = −5 or x = 2
3
Find each y using y = x²
x=−5: y=25 x=2: y=4
✓(2, 4) and (−5, 25)
Question 22
Solve simultaneously: y = x² and y = 5x − 4.
Method: set the two expressions for y equal to each other
1
Set equal and rearrange
x² = 5x − 4 → x² − 5x + 4 = 0
2
Factorise
(x − 1)(x − 4) = 0 → x = 1 or x = 4
3
Find each y using y = x²
x=1: y=1 x=4: y=16
✓(1, 1) and (4, 16)
Question 23
Solve simultaneously: y = x² + 2x and y = −3x − 6.
Method: set the two expressions for y equal to each other
1
Set equal and rearrange
x² + 2x = −3x − 6 → x² + 5x + 6 = 0
2
Factorise
(x + 2)(x + 3) = 0 → x = −2 or x = −3
3
Find each y using y = x² + 2x
x=−2: y=4−4=0 x=−3: y=9−6=3
✓(−2, 0) and (−3, 3)
Teaching tip: Both x-values are negative here — a good moment to remind students not to assume answers must be positive, since this question (unlike Section 3’s word problems) has no real-world restriction on x.
Question 24
Solve simultaneously: y = x + 2 and x² + y² = 20.
Method: substitute the line into the circle equation
1
Substitute y = x + 2
x² + (x + 2)² = 20
2
Expand and simplify
2x² + 4x + 4 = 20 → x² + 2x − 8 = 0
3
Factorise
(x + 4)(x − 2) = 0 → x = −4 or x = 2
4
Find each y using y = x + 2
x=−4: y=−2 x=2: y=4
✓(2, 4) and (−4, −2)
Question 25
Solve simultaneously: y = x − 1 and x² + y² = 13.
Method: substitute the line into the circle equation
1
Substitute y = x − 1
x² + (x − 1)² = 13
2
Expand and simplify
2x² − 2x + 1 = 13 → x² − x − 6 = 0
3
Factorise
(x − 3)(x + 2) = 0 → x = 3 or x = −2
4
Find each y using y = x − 1
x=3: y=2 x=−2: y=−3
✓(3, 2) and (−2, −3)
Question 26
Solve simultaneously: y = x² − 4x and y = x + 6.
Method: set the two expressions for y equal to each other
1
Set equal and rearrange
x² − 4x = x + 6 → x² − 5x − 6 = 0
2
Factorise
(x − 6)(x + 1) = 0 → x = 6 or x = −1
3
Find each y using y = x + 6
x=6: y=12 x=−1: y=5
✓(6, 12) and (−1, 5)
Question 27
Solve simultaneously: y = 4 and y = x² + x − 2.
Method: the “line” here is horizontal (y = 4) — substitute directly
1
Substitute y = 4 into the curve’s equation
4 = x² + x − 2 → x² + x − 6 = 0
2
Factorise
(x + 3)(x − 2) = 0 → x = −3 or x = 2
3
y is already known for both — it’s 4 either way
y = 4 in both cases
✓(−3, 4) and (2, 4)
Teaching tip: A horizontal line is still a linear equation (y = 0x + 4) — use this to show that “one linear, one non-linear” doesn’t always mean a sloped line.
Question 28
Solve simultaneously: y = 2x − 5 and x² + y² = 25.
Method: substitute the line into the circle equation
1
Substitute y = 2x − 5
x² + (2x − 5)² = 25
2
Expand and simplify
x² + 4x² − 20x + 25 = 25 → 5x² − 20x = 0
3
Factorise (take out the common factor 5x)
5x(x − 4) = 0 → x = 0 or x = 4
4
Find each y using y = 2x − 5
x=0: y=−5 x=4: y=3
✓(0, −5) and (4, 3)
Teaching tip: No constant term this time (5x² − 20x = 0), so it factorises by taking out 5x rather than needing two brackets — worth flagging so students don’t force it into (x−a)(x−b) form unnecessarily.
Question 29
Solve simultaneously: y = x² − 6x + 5 and y = −2x + 5.
Method: set the two expressions for y equal to each other
1
Set equal and rearrange
x² − 6x + 5 = −2x + 5 → x² − 4x = 0
2
Factorise (take out the common factor x)
x(x − 4) = 0 → x = 0 or x = 4
3
Find each y using y = −2x + 5
x=0: y=5 x=4: y=−3
✓(0, 5) and (4, −3)
Question 30
Solve simultaneously: y = 3x − 10 and x² + y² = 10.
Method: substitute the line into the circle equation — watch what happens to the discriminant
1
Substitute y = 3x − 10
x² + (3x − 10)² = 10
2
Expand and simplify
x² + 9x² − 60x + 100 = 10 → 10x² − 60x + 90 = 0
3
Divide every term by 10
x² − 6x + 9 = 0
4
Factorise — notice this is a perfect square
(x − 3)² = 0 → x = 3 (repeated root)
5
Find y using y = 3x − 10
y = 3(3) − 10 = −1
✓Only one solution: (3, −1) — the line is a tangent to the circle
Teaching tip: This is the exception to “two solutions” that’s worth building into every unit on linear/non-linear simultaneous equations — a repeated root means the line just touches the circle at one point rather than crossing it at two. Sketch it to make the geometry click.
3Solving Problems Using Quadratic Equations
Question 7
A rectangle has a length 4 cm more than its width. Its area is 96 cm². Let the width be x cm. Form a quadratic equation and solve it to find the dimensions.
Setup: width = x, length = x + 4, area = width × length
1
Write the area equation
x(x + 4) = 96
2
Expand and rearrange to standard form
x² + 4x − 96 = 0
3
Factorise (numbers that multiply to −96, add to 4)
(x + 12)(x − 8) = 0 → x = −12 or x = 8
4
Reject the negative solution — a width cannot be negative
x = 8
5
Find the length
length = x + 4 = 12
✓Width = 8 cm, Length = 12 cm
Teaching tip: This is the question type where students most often forget to reject the negative root — ask “can a width be −12 cm?” out loud every time, so rejecting it becomes automatic rather than something they need reminding about.
Question 8
The product of two consecutive positive integers is 156. Find the two integers.
Setup: let the integers be x and x + 1
1
Write the product equation
x(x + 1) = 156
2
Expand and rearrange to standard form
x² + x − 156 = 0
3
Factorise (numbers that multiply to −156, add to 1)
(x − 12)(x + 13) = 0 → x = 12 or x = −13
4
Reject the negative solution — the integers must be positive
x = 12, so the integers are 12 and 13
✓12 and 13
Teaching tip: A quick sanity check — 12 × 13 = 156 ✓ — takes two seconds and builds the habit of verifying word-problem answers against the original wording, not just the equation.
Question 9
A cyclist travels 60 km at a steady speed. If she had travelled 5 km/h faster, the journey would have taken 1 hour less. Let her original speed be x km/h.
Setup: time = distance / speed; original time − faster time = 1 hour
1
Write expressions for both journey times
original time = 60/x faster time = 60/(x+5)
2
The faster journey takes 1 hour less, so set up the equation
60/x − 60/(x + 5) = 1
3
Multiply every term by x(x + 5) to clear the fractions
60(x + 5) − 60x = x(x + 5)
4
Expand both sides
60x + 300 − 60x = x² + 5x
5
Simplify the left side, and rearrange to standard form
300 = x² + 5x → x² + 5x − 300 = 0
6
Solve — discriminant = 25 + 1200 = 1225 = 35², so it factorises
(x − 15)(x + 20) = 0 → x = 15 or x = −20
7
Reject the negative solution — speed cannot be negative
x = 15
✓Original speed = 15 km/h
Teaching tip: This is the hardest question in the set — walk through WHY multiplying by x(x+5) clears both denominators at once before touching any algebra, otherwise students often multiply by just one denominator and get stuck.
Question 10
A right-angled triangle has a shorter leg 7 cm shorter than the other leg, and a hypotenuse of 13 cm. Let the shorter leg be x cm.
Setup: Pythagoras’ theorem — (shorter leg)² + (longer leg)² = (hypotenuse)²
1
The longer leg is x + 7. Write Pythagoras’ theorem
x² + (x + 7)² = 13²
2
Expand the bracket and the right side
x² + x² + 14x + 49 = 169
3
Collect terms and rearrange to standard form
2x² + 14x − 120 = 0 → x² + 7x − 60 = 0
4
Factorise (numbers that multiply to −60, add to 7)
(x − 5)(x + 12) = 0 → x = 5 or x = −12
5
Reject the negative solution — a length cannot be negative
x = 5, so the longer leg = x + 7 = 12
✓Legs are 5 cm and 12 cm (a 5-12-13 triangle)
Teaching tip: Once the answer comes out, point out this is the well-known 5-12-13 Pythagorean triple — recognising these (3-4-5, 5-12-13, 8-15-17) helps students sanity-check answers on sight in future questions.
Question 11
A rectangular garden is 5 m longer than it is wide. Its area is 84 m². Find its dimensions.
Setup: width = x, length = x + 5, area = width × length
1
Write the area equation
x(x + 5) = 84
2
Rearrange to standard form
x² + 5x − 84 = 0
3
Factorise
(x + 12)(x − 7) = 0 → x = −12 or x = 7
4
Reject the negative solution and find the length
x = 7, length = 12
✓Width = 7 m, Length = 12 m
Question 12
A number added to its own square gives 42. Find the possible value(s) of the number.
Setup: x + x² = 42
1
Rearrange to standard form
x² + x − 42 = 0
2
Factorise
(x + 7)(x − 6) = 0 → x = −7 or x = 6
✓x = 6 or x = −7
Teaching tip: Unlike Questions 7–11, there’s no length or speed here — “a number” can be negative, so BOTH roots are kept. Contrast this with the previous questions to reinforce that rejecting a root is context-dependent, not automatic.
Question 13
Two positive numbers differ by 3. Their product is 88. Find the two numbers.
Setup: let the smaller number be x, so the larger is x + 3
1
Write the product equation
x(x + 3) = 88
2
Rearrange to standard form
x² + 3x − 88 = 0
3
Factorise
(x + 11)(x − 8) = 0 → x = −11 or x = 8
4
Reject the negative solution — numbers must be positive
x = 8, other number = 11
✓8 and 11
Question 14
The product of two consecutive even numbers is 168. Find the two numbers.
Setup: consecutive even numbers differ by 2 — let them be x and x + 2
1
Write the product equation
x(x + 2) = 168
2
Rearrange to standard form
x² + 2x − 168 = 0
3
Factorise
(x + 14)(x − 12) = 0 → x = −14 or x = 12
4
Take the positive solution (or note −14, −12 is also a valid consecutive-even pair)
x = 12, other number = 14
✓12 and 14 (or −14 and −12, if negative numbers are allowed)
Teaching tip: Worth a brief discussion: unless the question specifies “positive,” −14 and −12 are ALSO two consecutive even numbers with product 168 — a nice one for a stronger student to spot.
Question 15
A picture frame is 3 cm wider than it is tall, and has an area of 154 cm². Find its height and width.
Setup: height = h, width = h + 3, area = height × width
1
Write the area equation
h(h + 3) = 154
2
Rearrange to standard form
h² + 3h − 154 = 0
3
Factorise
(h + 14)(h − 11) = 0 → h = −14 or h = 11
4
Reject the negative solution and find the width
h = 11, width = 14
✓Height = 11 cm, Width = 14 cm
Question 16
Priya is currently x years old. In 3 years, the product of her age then and her current age will be 180. Find her current age.
Setup: current age = x, age in 3 years = x + 3
1
Write the product equation
x(x + 3) = 180
2
Rearrange to standard form
x² + 3x − 180 = 0
3
Factorise
(x + 15)(x − 12) = 0 → x = −15 or x = 12
4
Reject the negative solution — an age cannot be negative
x = 12
✓Priya is currently 12 years old
Question 17
A boat travels 24 km upstream and 24 km back downstream on a river flowing at 2 km/h. The whole trip takes 5 hours. Find the boat’s speed in still water.
Setup: upstream speed = x − 2, downstream speed = x + 2; total time = 5 h
1
Write the total time equation
24/(x−2) + 24/(x+2) = 5
2
Multiply every term by (x−2)(x+2) = x² − 4
24(x+2) + 24(x−2) = 5(x² − 4)
3
Expand — the +48 and −48 cancel on the left
48x = 5x² − 20
4
Rearrange to standard form
5x² − 48x − 20 = 0
5
Solve — discriminant = 48² + 4(5)(20) = 2304 + 400 = 2704 = 52²
x = (48 ± 52)/10 → x = 10 or x = −0.4
6
Reject the negative solution — speed cannot be negative
x = 10
✓The boat’s speed in still water is 10 km/h
Teaching tip: Point out that (x−2)(x+2) expands to the “difference of two squares,” x²−4 — recognising this pattern speeds up clearing the fractions considerably.
Question 18
Two pipes together fill a tank in 12 hours. Working alone, one pipe would take 10 hours longer than the other. Find how long each pipe takes alone.
Setup: faster pipe = x hours alone, slower pipe = x + 10 hours alone; combined rate = 1/12 tank per hour
1
Write the combined rate equation
1/x + 1/(x+10) = 1/12
2
Multiply every term by 12x(x + 10)
12(x+10) + 12x = x(x+10)
3
Expand both sides
24x + 120 = x² + 10x
4
Rearrange to standard form
x² − 14x − 120 = 0
5
Solve — discriminant = 196 + 480 = 676 = 26²
x = (14 ± 26)/2 → x = 20 or x = −6
6
Reject the negative solution and find the other pipe’s time
x = 20, other pipe = 30
✓The two pipes take 20 hours and 30 hours alone
Teaching tip: “Rate” problems (pipes, workers, taps) all follow the same template as speed problems — 1/time plays the same role that speed/distance played earlier. Draw that parallel explicitly.
Question 19
A shop buys x identical calculators for a total of $240. If each calculator had cost $2 less, the shop could have bought 4 more calculators for the same total amount. Find x.
Setup: price per calculator = 240/x; new price = 240/x − 2; new quantity = x + 4
1
Write the “new total is still $240” equation
(240/x − 2)(x + 4) = 240
2
Expand the brackets
240 + 960/x − 2x − 8 = 240
3
The 240s cancel — simplify what remains
960/x − 2x − 8 = 0
4
Multiply every term by x to clear the fraction
960 − 2x² − 8x = 0
5
Divide by −2 and rearrange to standard form
x² + 4x − 480 = 0
6
Solve — discriminant = 16 + 1920 = 1936 = 44²
x = (−4 ± 44)/2 → x = 20 or x = −24
7
Reject the negative solution — a quantity cannot be negative
x = 20
✓x = 20 calculators
Teaching tip: This is the most algebraically demanding question in the set — go slowly on step 2, since expanding brackets containing a fraction (240/x) is where most errors happen. Consider doing this one together as a class rather than leaving it fully independent.
Question 20
A 5 m ladder leans against a vertical wall. The foot of the ladder is 1 m further from the wall than the height it reaches up the wall. Find the height and the distance of the foot from the wall.
Setup: height = x, foot distance = x + 1, ladder (hypotenuse) = 5
1
Apply Pythagoras’ theorem
x² + (x + 1)² = 5²
2
Expand
x² + x² + 2x + 1 = 25
3
Collect terms and rearrange to standard form
2x² + 2x − 24 = 0 → x² + x − 12 = 0
4
Factorise
(x + 4)(x − 3) = 0 → x = −4 or x = 3
5
Reject the negative solution and find the foot distance
x = 3, foot distance = 4
✓Height = 3 m, Foot distance = 4 m (another 3-4-5 triangle)
Teaching tip: A second 3-4-5 triple appearing here (after the 5-12-13 in Question 10) is a good chance to ask the class if they notice the pattern before you point it out.
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