To solve a pair where both equations are nonlinear, choose the equation from which one variable is easiest to express, substitute it into the other, and solve the quadratic that appears, often in x². Questions of this kind ask for every pair (x, y) that satisfies both equations.
This lesson builds on substituting a line into a circle and feeds the word problems in forming a system from two constraints.
Which equation do you start from?
Look for the equation that lets you write one variable on its own with the least effort. Three patterns cover most questions.
- xy = k. Write y = k/x. After substitution you multiply through by x², so you meet an x⁴ term.
- y = x² (or similar). Replace x² in the other equation with y. You then solve for y, which is often the easier unknown.
- Two equations in x² and y². Add or subtract them to eliminate one squared term directly.
Whichever route you take, the job is to reach a single equation in a single unknown.
How do you handle the quadratic in x²?
When the result is ax⁴ + bx² + c = 0, treat x² as one unknown. Write u = x² and solve the quadratic in u. Then go back: x = ±√u for each positive u, while a negative u gives no real x.
This step is where most solutions lose roots. One positive u produces two x-values, and each x has its own partner y.
Worked example
Solve xy = 6 and x² + y² = 13.
Eliminate y: y = 6/x, so x² + 36/x² = 13.
Clear the fraction: multiply by x²: x⁴ + 36 = 13x², so x⁴ − 13x² + 36 = 0.
Let u = x²: u² − 13u + 36 = 0, so (u − 4)(u − 9) = 0 and u = 4 or u = 9.
Return to x: x² = 4 gives x = ±2, and x² = 9 gives x = ±3.
Find each partner with y = 6/x:
| x | 2 | −2 | 3 | −3 |
|---|---|---|---|---|
| y | 3 | −3 | 2 | −2 |
Answer: (2, 3), (−2, −3), (3, 2), (−3, −2).
Check: in every pair xy = 6, and x² + y² = 4 + 9 = 13 or 9 + 4 = 13.
The mistake to watch for
The usual slip is to take only the positive square roots, or to pair values carelessly.
Mistaken answer: x = 2, y = 3 and x = 3, y = 2.
The student solved for u correctly but wrote only x = √u, forgetting that x = −√u also satisfies x² = u.
The correction is to write ± at once whenever you take a square root, and to test the negative pair in both original equations. (−2)(−3) = 6 and 4 + 9 = 13, so the negative pairs are genuine solutions. A quick sketch shows why: the curve xy = 6 has a branch in the lower left as well as the upper right, and the circle cuts both.
Check yourself
Work these on paper, then open each answer.
1. Solve xy = 10 and x² + y² = 29.
Show answer
y = 10/x, so x² + 100/x² = 29, giving x⁴ − 29x² + 100 = 0. Then (x² − 25)(x² − 4) = 0, so x = ±5 or x = ±2.
Partners: y = 2, −2, 5, −5 respectively.
(5, 2), (−5, −2), (2, 5), (−2, −5). Check: 25 + 4 = 29 and 4 + 25 = 29.
2. Solve y = x² and x² + y² = 20.
Show answer
Replace x² with y: y + y² = 20, so y² + y − 20 = 0 and (y + 5)(y − 4) = 0. Then y = 4 or y = −5.
y = −5 would need x² = −5, which has no real solution, so reject it. When y = 4, x² = 4 and x = ±2.
(2, 4) and (−2, 4). Check: 4 + 16 = 20.
3. Solve x² − y² = 5 and x² + y² = 13.
Show answer
Add the equations: 2x² = 18, so x² = 9 and x = ±3. Subtract them: 2y² = 8, so y² = 4 and y = ±2.
All four combinations work because only the squares appear: (3, 2), (3, −2), (−3, 2), (−3, −2). Check (3, −2): 9 − 4 = 5 and 9 + 4 = 13.
Where this leads next
When the elimination produces a double root, the picture changes, as interpreting a repeated solution geometrically explains. The non-calculator working trainer helps with exact arithmetic along the way, and the quadratic structure explorer shows how a quadratic in u behaves.
If you can follow these steps but stall when choosing the first substitution, our teachers can practise that choice with you in online one-to-one Additional Mathematics tuition.