These eight questions use new numbers and get harder as you go, ending with a word problem where one root is rejected. Write your own working first, then open the answer.
They follow the lessons in systems of equations. Check each answer in every original equation before you move on.
Questions
Question 1. Solve x + y + z = 9, 2x − y + z = 5 and x + 2y − z = 4.
Answer
Add the first and third: 2x + 3y = 13. Add the second and third: 3x + y = 9, so y = 9 − 3x.
Then 2x + 27 − 9x = 13, so x = 2 and y = 3. From the first equation, z = 4. Check the second: 4 − 3 + 4 = 5. Answer: x = 2, y = 3, z = 4.
Question 2. Solve x − y = 2 and x² + y² = 10.
Answer
x = y + 2, so (y + 2)² + y² = 10, giving 2y² + 4y − 6 = 0, or y² + 2y − 3 = 0. Then (y + 3)(y − 1) = 0.
So y = 1 with x = 3, or y = −3 with x = −1. Check (−1, −3): −1 + 3 = 2 and 1 + 9 = 10. Answer: (3, 1) and (−1, −3).
Question 3. Solve 5x + 3y = 21 and 2x − 3y = 0. Name the method you would pick.
Answer
The y-coefficients are +3 and −3, so elimination is the natural choice. Add: 7x = 21, so x = 3.
Then 6 − 3y = 0 gives y = 2. Check: 15 + 6 = 21. Answer: x = 3, y = 2.
Question 4. Find the points where the line y = x + 1 meets the curve y = x² − 2x + 1.
Answer
Set the two expressions for y equal: x + 1 = x² − 2x + 1, so x² − 3x = 0, which gives x(x − 3) = 0.
So x = 0 with y = 1, or x = 3 with y = 4. Answer: (0, 1) and (3, 4).
Question 5. At a school concert, a group buys 15 tickets in total.
The adult tickets are RM12 each, the child tickets are RM7 each, and the total paid is RM145. How many of each did the group buy?
Answer
Let a adults and c children: a + c = 15 and 12a + 7c = 145. Substitute c = 15 − a: 12a + 105 − 7a = 145, so 5a = 40 and a = 8, c = 7.
Check: 12(8) + 7(7) = 96 + 49 = 145. Answer: 8 adults and 7 children.
Question 6. A rectangle has perimeter 34 cm and diagonal 13 cm. Find its sides.
Answer
Let the sides be x and y: x + y = 17 and x² + y² = 169. Substitute y = 17 − x: x² + 289 − 34x + x² = 169, so 2x² − 34x + 120 = 0, or x² − 17x + 60 = 0.
Then (x − 5)(x − 12) = 0, so x = 5 or 12, with y = 12 or 5. Both describe the same rectangle: 12 cm by 5 cm.
Question 7. The length of a rectangular field is 6 m more than its width, and its area is 91 m². Find the dimensions.
Answer
Let the width be w m. Then w(w + 6) = 91, so w² + 6w − 91 = 0, which factorises as (w + 13)(w − 7) = 0.
So w = −13 or w = 7. A width cannot be negative, so w = −13 is rejected. Answer: width 7 m and length 13 m. Check: 7 × 13 = 91.
Question 8. Three friends have 56 stamps in total. Ali has twice as many as Bala, and Chong has 4 fewer than Ali. How many does each have?
Answer
Let Bala have b stamps. Then Ali has 2b and Chong has 2b − 4. So b + 2b + (2b − 4) = 56, which gives 5b = 60 and b = 12.
Ali has 24 and Chong has 20. Check: 12 + 24 + 20 = 56. Answer: Ali 24, Bala 12, Chong 20.
If you got these wrong
- Three-unknown slips in Question 1: revisit three linear equations in three unknowns.
- Squared-term slips in Questions 2, 4 and 6: read one linear and one nonlinear equation.
- Unsure which method to use in Question 3 or 5: read choosing substitution or elimination.
- Wrong or unexplained answer in Question 7: read rejecting invalid solutions.
Log each slip in the mistake log, and try a timed practice session when you are ready. If you want a teacher to go through your working, see online one-to-one Additional Mathematics tuition.