This set has twelve original questions, ordered from easier to harder, covering all five lessons in simultaneous linear and nonlinear models. Questions 1 to 4 substitute a line into a circle, 5 is a nonlinear pair, 6 and 7 are about tangents and missing intersections, 8 and 9 are word problems, and 10 to 12 use an unknown constant.
Attempt each question on paper and finish by substituting your answers back into both original equations. Only then open the answer.
Mark the ones you got wrong and use the routing list at the end. The line and circle intersection explorer and the quadratic structure explorer are useful for checking after you have tried by hand.
Questions
1. Find the points where y = x + 2 meets x² + y² = 10.
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Substitute: x² + (x + 2)² = 10, so 2x² + 4x + 4 = 10 and x² + 2x − 3 = 0. Then (x + 3)(x − 1) = 0, so x = −3 or x = 1.
Partners from the line: y = −1 and y = 3.
(−3, −1) and (1, 3). Check: 9 + 1 = 10 and 1 + 9 = 10.
2. Two numbers have a sum of 5 and a product of 6. Find them.
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x + y = 5 and xy = 6. Put y = 5 − x: x(5 − x) = 6, so x² − 5x + 6 = 0 and (x − 2)(x − 3) = 0.
The numbers are 2 and 3. Check: 2 + 3 = 5 and 2 × 3 = 6.
3. Solve 2x + y = 4 and x² + y² = 13.
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y = 4 − 2x, so x² + (4 − 2x)² = 13. Expand: x² + 16 − 16x + 4x² = 13, so 5x² − 16x + 3 = 0. Then (5x − 1)(x − 3) = 0, so x = 1/5 or x = 3.
Partners: when x = 3, y = −2. When x = 1/5, y = 4 − 2/5 = 18/5.
(3, −2) and (1/5, 18/5). Check the second: 1/25 + 324/25 = 325/25 = 13.
4. Find where y = x − 2 meets the circle (x − 1)² + (y + 2)² = 25.
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Since y + 2 = x, the circle becomes (x − 1)² + x² = 25. Expand: 2x² − 2x + 1 = 25, so x² − x − 12 = 0 and (x − 4)(x + 3) = 0. Then x = 4 or x = −3.
Partners: y = 2 and y = −5.
(4, 2) and (−3, −5). Check (−3, −5): (−4)² + (−3)² = 16 + 9 = 25.
5. Solve xy = 4 and x² + y² = 17.
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y = 4/x, so x² + 16/x² = 17 and x⁴ − 17x² + 16 = 0. Then (x² − 1)(x² − 16) = 0, so x = ±1 or x = ±4.
Partners from y = 4/x: x = 1 gives 4, x = −1 gives −4, x = 4 gives 1, x = −4 gives −1.
(1, 4), (−1, −4), (4, 1), (−4, −1). Check: 1 + 16 = 17.
6. Show that y = 2x − 5 is a tangent to x² + y² = 5, and find the point of contact.
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x² + (2x − 5)² = 5 gives 5x² − 20x + 25 = 5, so 5x² − 20x + 20 = 0 and x² − 4x + 4 = 0. That is (x − 2)² = 0, a repeated root, so the line is a tangent.
Then y = 2(2) − 5 = −1. Point of contact: (2, −1). Check: 4 + 1 = 5.
7. How many points do y = x + 3 and x² + y² = 4 have in common?
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x² + (x + 3)² = 4 gives 2x² + 6x + 9 = 4, so 2x² + 6x + 5 = 0. The discriminant is 36 − 4(2)(5) = 36 − 40 = −4, which is negative.
None. The line misses the circle. Check by distance: 3/√2 is about 2.12, which is more than the radius 2.
8. A rectangle has perimeter 46 cm and diagonal 17 cm. Find its length and width.
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Let the sides be x and y. Then x + y = 23 and x² + y² = 289. Put y = 23 − x: x² + 529 − 46x + x² = 289, so 2x² − 46x + 240 = 0 and x² − 23x + 120 = 0. Then (x − 8)(x − 15) = 0.
The rectangle is 15 cm by 8 cm. Check: 2(15 + 8) = 46 and 225 + 64 = 289.
9. Two positive numbers differ by 3, and the sum of their squares is 117. Find them.
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Let the numbers be x and y with x − y = 3. Then x = y + 3 and (y + 3)² + y² = 117, so 2y² + 6y + 9 = 117 and y² + 3y − 54 = 0. Then (y + 9)(y − 6) = 0.
Both numbers are positive, so reject y = −9. Then y = 6 and x = 9.
The numbers are 9 and 6. Check: 9 − 6 = 3 and 81 + 36 = 117.
10. Find the values of k for which y = 2x + k meets x² + y² = 20 at two distinct points.
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x² + (2x + k)² = 20 gives 5x² + 4kx + k² − 20 = 0. The discriminant is 16k² − 20(k² − 20) = 400 − 4k². For two points, 400 − 4k² > 0, so k² < 100.
−10 < k < 10. Check by distance: |k|/√5 = √20 gives |k| = 10 at the tangent position.
11. Find the values of k for which y = x + k does not meet the curve y = x².
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Set x² = x + k, so x² − x − k = 0. The discriminant is 1 + 4k. No intersection needs 1 + 4k < 0.
k < −1/4. Check with k = −1: x² − x + 1 = 0 has discriminant 1 − 4 = −3, which is negative, so no real roots.
12. The line y = 2x + k is a tangent to the curve y = x² − 4x + 7. Find k and the point of contact.
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Set x² − 4x + 7 = 2x + k, so x² − 6x + (7 − k) = 0. The discriminant is 36 − 4(7 − k) = 8 + 4k. For a tangent, 8 + 4k = 0.
k = −2. Then x² − 6x + 9 = 0, so x = 3 and y = 2(3) − 2 = 4. Point of contact: (3, 4). Check: 9 − 12 + 7 = 4.
If you got these wrong
- The bracket expanded wrongly, or a middle term was missing (questions 1, 3, 4): return to substituting a line into a circle.
- Lost roots, negative pairs, or an impossible root kept (questions 2, 5): return to eliminating a variable from a nonlinear pair.
- A single answer treated as a mistake, or the tangent point not found (questions 6, 7, 12): return to interpreting a repeated solution geometrically.
- The equations came out wrong from the words (questions 8, 9): return to forming a system from two constraints.
- The range for k was wrong or one-sided (questions 10, 11, 12): return to checking parameter values for a requested number of intersections.
Log each error in the mistake log and retest queue and retry a fresh question a few days later. Your work on the discriminant depends on quadratic structure and discriminants.
If the same type of error keeps appearing in your written solutions, a teacher can study it with you in online one-to-one Additional Mathematics tuition.