Topics

A line and a curve: non-linear systems

Covers AS91269

What makes a system non-linear

A non-linear equation contains a term that is not simply a multiple of $x$ or $y$ — an $x^2$ , a $y^2$ , an $xy$ , or a $\frac{1}{x}$ .
Its graph is a curve, not a line: a parabola, a circle, a hyperbola.
This standard pairs one non-linear equation with one linear equation — a curve and a straight line.
The solutions are the points where the line meets the curve, and a line can meet a curve twice, once, or not at all.

The method — always substitution

Elimination cannot work, because adding or subtracting will never cancel a squared term.
The steps:
- Rearrange the LINEAR equation to make one variable the subject. Always the linear one — rearranging the curve creates square roots.
- Substitute into the non-linear equation.
- Rearrange to the form $ax^2 + bx + c = 0$ .
- Solve by factorising, the quadratic formula, or completing the square.
- Find the matching value for each solution by substituting back into the linear equation — it is easier and cannot introduce extra roots.
- Write the solutions as points, and check both in the non-linear equation.

Worked through — a line and a parabola

Solve $y = x^2 - 2x + 3$ and $y = 2x + 3$ :
- Both give $y$ , so set them equal: $x^2 - 2x + 3 = 2x + 3$
- Rearrange: $x^2 - 4x = 0$
- Factorise: $x(x - 4) = 0$
- So $x = 0$ or $x = 4$
- From the line, $y = 2(0) + 3 = 3$ and $y = 2(4) + 3 = 11$
- Solutions: $(0, 3)$ and $(4, 11)$
Note the factorising. $x^2 - 4x = 0$ has a common factor of $x$ ; dividing both sides by $x$ would lose the solution $x = 0$ . Never divide by a variable.

Worked through — a line and a circle

Solve $x^2 + y^2 = 25$ $x^{2} + y^{2} = 25$ and $y = x + 1$ $y = x + 1$ :
- Substitute the line into the circle: $x^2 + (x + 1)^2 = 25$
- Expand carefully: $x^2 + x^2 + 2x + 1 = 25$
- Simplify: $2x^2 + 2x - 24 = 0$
- Divide by 2: $x^2 + x - 12 = 0$
- Factorise: $(x + 4)(x - 3) = 0$ , so $x = -4$ or $x = 3$
- From the line: $y = -3$ and $y = 4$
- Solutions: $(-4, -3)$ and $(3, 4)$
$(x+1)^2$ is $x^2 + 2x + 1$ , not $x^2 + 1$ . That expansion error is the single most common mistake on this page.

Expecting two answers

A quadratic normally has two roots, so a non-linear system normally has two solution points.
Both must be given, each as a complete pair of values.
Then check both against the context. A negative length or a negative price is rejected — with a stated reason, not silently.

The number of solutions

After substituting you have a quadratic $ax^2 + bx + c = 0$ . The discriminant decides how many times the line meets the curve:

$\Delta = b^2 - 4ac$

Discriminant	Roots	Geometry
$\Delta > 0$	Two	The line cuts the curve at two points
$\Delta = 0$	One (repeated)	The line is a tangent — it touches at exactly one point
$\Delta < 0$	None (real)	The line misses the curve entirely

$\Delta = 0$ is the tangent condition, and it is how questions about "just touching" are answered without any calculus.
A repeated root is one point, not two. Saying "two solutions, both the same" describes the algebra but misreads the geometry.

Finding an unknown constant

A standard question gives a line with an unknown, such as $y = 2x + k$ , and asks for the value that makes it a tangent.
The method:
- Substitute to get a quadratic in $x$ whose coefficients contain $k$ .
- Set the discriminant to zero and solve for $k$ .
- Substitute $k$ back to find the point of contact.
Worked through — for what $k$ $k$ is $y = 2x + k$ $y = 2 x + k$ a tangent to $y = x^2 - 4x + 7$ $y = x^{2} - 4 x + 7$ ?
- $x^2 - 4x + 7 = 2x + k$ , so $x^2 - 6x + (7 - k) = 0$
- $\Delta = (-6)^2 - 4(1)(7 - k) = 36 - 28 + 4k = 8 + 4k$
- Tangent when $\Delta = 0$ : $4k = -8$ , so $k = -2$
- Then $x^2 - 6x + 9 = 0$ , $(x - 3)^2 = 0$ , $x = 3$ and $y = 2(3) - 2 = 4$
- The tangent is $y = 2x - 2$ , touching at $(3, 4)$

Tips and Tricks

Rearrange the linear equation, never the curve. Making $y$ the subject of a circle introduces a square root and a $\pm$ that you will lose track of.

Expand brackets in full. $(x + 1)^2 = x^2 + 2x + 1$ . Writing $x^2 + 1$ turns a solvable question into a wrong answer that still looks tidy.

Never divide both sides by $x$ . $x^2 = 4x$ has solutions $x = 0$ and $x = 4$ ; dividing by $x$ silently deletes the first one. Factorise instead.

Give both solutions as points. Two $x$ values with one $y$ value, or a list of four numbers with no pairing, does not answer the question.

Substitute back into the linear equation to find the second coordinate. The curve will give you the right answer too, but with more work and more chances to slip.

$\Delta = 0$ means tangent. Whenever a question says "just touches", "meets at exactly one point" or "is a tangent to", set the discriminant to zero — no calculus is needed.

Check both answers against the context before writing them down. A rectangle with side $-4$ m is a valid root of the equation and an invalid answer to the question.

Worked ExampleA rectangular paddock

A rectangular paddock is fenced on all four sides using 60 m of fencing, and encloses an area of 216 m². Find its dimensions.

Step 1 — Define the variables

Step 2 — Form one equation from each condition

The fencing gives the perimeter:

$2x + 2y = 60$

Divide by 2 immediately — smaller numbers, fewer errors:

$x + y = 30 \qquad (1) \quad \textbf{linear}$

The area gives:

$xy = 216 \qquad (2) \quad \textbf{non-linear}$

Step 3 — Rearrange the LINEAR equation

$y = 30 - x$

Step 4 — Substitute into the non-linear equation

$x(30 - x) = 216$

Expand:

$30x - x^2 = 216$

Step 5 — Rearrange into standard quadratic form

Move everything to one side, keeping the $x^2$ term positive:

$0 = x^2 - 30x + 216$

Step 6 — Solve the quadratic

Try factorising. We need two numbers multiplying to 216 and adding to 30:

$12 \times 18 = 216 \quad \text{and} \quad 12 + 18 = 30 \ \checkmark$

$(x - 12)(x - 18) = 0$

$x = 12 \quad \text{or} \quad x = 18$

Step 7 — Find the matching $y$ for each, from the LINEAR equation

If $x = 12$ : $y = 30 - 12 = 18$
If $x = 18$ : $y = 30 - 18 = 12$

Step 8 — Check both in the non-linear equation

$12 \times 18 = 216$ ✓
$18 \times 12 = 216$ ✓

And the perimeter: $2(12) + 2(18) = 60$ ✓

Step 9 — Interpret the two solutions in context

The algebra gives two solution points, $(12, 18)$ and $(18, 12)$ — but they describe the same paddock, once with the long side called $x$ and once with it called $y$ .

Both roots are positive, so neither is rejected; they are simply the same rectangle labelled two ways.

Step 10 — A useful extension check

Would 60 m of fencing enclose an area of, say, 250 m²? The same working gives

$x^2 - 30x + 250 = 0, \qquad \Delta = 900 - 1000 = -100 < 0$

No real solutions — it is impossible. The largest area 60 m of fence can enclose is the square case $x = y = 15$ , giving $225$ m², and the discriminant is what detects that limit.

A line and a curve: non-linear systems

Covers AS91269

What makes a system non-linear

A non-linear equation contains a term that is not simply a multiple of $x$ or $y$ — an $x^2$ , a $y^2$ , an $xy$ , or a $\frac{1}{x}$ .
Its graph is a curve, not a line: a parabola, a circle, a hyperbola.
This standard pairs one non-linear equation with one linear equation — a curve and a straight line.
The solutions are the points where the line meets the curve, and a line can meet a curve twice, once, or not at all.

The method — always substitution

Elimination cannot work, because adding or subtracting will never cancel a squared term.
The steps:
- Rearrange the LINEAR equation to make one variable the subject. Always the linear one — rearranging the curve creates square roots.
- Substitute into the non-linear equation.
- Rearrange to the form $ax^2 + bx + c = 0$ .
- Solve by factorising, the quadratic formula, or completing the square.
- Find the matching value for each solution by substituting back into the linear equation — it is easier and cannot introduce extra roots.
- Write the solutions as points, and check both in the non-linear equation.

Worked through — a line and a parabola

Solve $y = x^2 - 2x + 3$ and $y = 2x + 3$ :
- Both give $y$ , so set them equal: $x^2 - 2x + 3 = 2x + 3$
- Rearrange: $x^2 - 4x = 0$
- Factorise: $x(x - 4) = 0$
- So $x = 0$ or $x = 4$
- From the line, $y = 2(0) + 3 = 3$ and $y = 2(4) + 3 = 11$
- Solutions: $(0, 3)$ and $(4, 11)$
Note the factorising. $x^2 - 4x = 0$ has a common factor of $x$ ; dividing both sides by $x$ would lose the solution $x = 0$ . Never divide by a variable.

Worked through — a line and a circle

Solve $x^2 + y^2 = 25$ $x^{2} + y^{2} = 25$ and $y = x + 1$ $y = x + 1$ :
- Substitute the line into the circle: $x^2 + (x + 1)^2 = 25$
- Expand carefully: $x^2 + x^2 + 2x + 1 = 25$
- Simplify: $2x^2 + 2x - 24 = 0$
- Divide by 2: $x^2 + x - 12 = 0$
- Factorise: $(x + 4)(x - 3) = 0$ , so $x = -4$ or $x = 3$
- From the line: $y = -3$ and $y = 4$
- Solutions: $(-4, -3)$ and $(3, 4)$
$(x+1)^2$ is $x^2 + 2x + 1$ , not $x^2 + 1$ . That expansion error is the single most common mistake on this page.

Expecting two answers

A quadratic normally has two roots, so a non-linear system normally has two solution points.
Both must be given, each as a complete pair of values.
Then check both against the context. A negative length or a negative price is rejected — with a stated reason, not silently.

The number of solutions

After substituting you have a quadratic $ax^2 + bx + c = 0$ . The discriminant decides how many times the line meets the curve:

$\Delta = b^2 - 4ac$

Discriminant	Roots	Geometry
$\Delta > 0$	Two	The line cuts the curve at two points
$\Delta = 0$	One (repeated)	The line is a tangent — it touches at exactly one point
$\Delta < 0$	None (real)	The line misses the curve entirely

$\Delta = 0$ is the tangent condition, and it is how questions about "just touching" are answered without any calculus.
A repeated root is one point, not two. Saying "two solutions, both the same" describes the algebra but misreads the geometry.

Finding an unknown constant

A standard question gives a line with an unknown, such as $y = 2x + k$ , and asks for the value that makes it a tangent.
The method:
- Substitute to get a quadratic in $x$ whose coefficients contain $k$ .
- Set the discriminant to zero and solve for $k$ .
- Substitute $k$ back to find the point of contact.
Worked through — for what $k$ $k$ is $y = 2x + k$ $y = 2 x + k$ a tangent to $y = x^2 - 4x + 7$ $y = x^{2} - 4 x + 7$ ?
- $x^2 - 4x + 7 = 2x + k$ , so $x^2 - 6x + (7 - k) = 0$
- $\Delta = (-6)^2 - 4(1)(7 - k) = 36 - 28 + 4k = 8 + 4k$
- Tangent when $\Delta = 0$ : $4k = -8$ , so $k = -2$
- Then $x^2 - 6x + 9 = 0$ , $(x - 3)^2 = 0$ , $x = 3$ and $y = 2(3) - 2 = 4$
- The tangent is $y = 2x - 2$ , touching at $(3, 4)$

Tips and Tricks

Rearrange the linear equation, never the curve. Making $y$ the subject of a circle introduces a square root and a $\pm$ that you will lose track of.

Expand brackets in full. $(x + 1)^2 = x^2 + 2x + 1$ . Writing $x^2 + 1$ turns a solvable question into a wrong answer that still looks tidy.

Never divide both sides by $x$ . $x^2 = 4x$ has solutions $x = 0$ and $x = 4$ ; dividing by $x$ silently deletes the first one. Factorise instead.

Give both solutions as points. Two $x$ values with one $y$ value, or a list of four numbers with no pairing, does not answer the question.

Substitute back into the linear equation to find the second coordinate. The curve will give you the right answer too, but with more work and more chances to slip.

$\Delta = 0$ means tangent. Whenever a question says "just touches", "meets at exactly one point" or "is a tangent to", set the discriminant to zero — no calculus is needed.

Check both answers against the context before writing them down. A rectangle with side $-4$ m is a valid root of the equation and an invalid answer to the question.

Worked ExampleA rectangular paddock

A rectangular paddock is fenced on all four sides using 60 m of fencing, and encloses an area of 216 m². Find its dimensions.

Step 1 — Define the variables

Step 2 — Form one equation from each condition

The fencing gives the perimeter:

$2x + 2y = 60$

Divide by 2 immediately — smaller numbers, fewer errors:

$x + y = 30 \qquad (1) \quad \textbf{linear}$

The area gives:

$xy = 216 \qquad (2) \quad \textbf{non-linear}$

Step 3 — Rearrange the LINEAR equation

$y = 30 - x$

Step 4 — Substitute into the non-linear equation

$x(30 - x) = 216$

Expand:

$30x - x^2 = 216$

Step 5 — Rearrange into standard quadratic form

Move everything to one side, keeping the $x^2$ term positive:

$0 = x^2 - 30x + 216$

Step 6 — Solve the quadratic

Try factorising. We need two numbers multiplying to 216 and adding to 30:

$12 \times 18 = 216 \quad \text{and} \quad 12 + 18 = 30 \ \checkmark$

$(x - 12)(x - 18) = 0$

$x = 12 \quad \text{or} \quad x = 18$

Step 7 — Find the matching $y$ for each, from the LINEAR equation

If $x = 12$ : $y = 30 - 12 = 18$
If $x = 18$ : $y = 30 - 18 = 12$

Step 8 — Check both in the non-linear equation

$12 \times 18 = 216$ ✓
$18 \times 12 = 216$ ✓

And the perimeter: $2(12) + 2(18) = 60$ ✓

Step 9 — Interpret the two solutions in context

The algebra gives two solution points, $(12, 18)$ and $(18, 12)$ — but they describe the same paddock, once with the long side called $x$ and once with it called $y$ .

Both roots are positive, so neither is rejected; they are simply the same rectangle labelled two ways.

Step 10 — A useful extension check

Would 60 m of fencing enclose an area of, say, 250 m²? The same working gives

$x^2 - 30x + 250 = 0, \qquad \Delta = 900 - 1000 = -100 < 0$

No real solutions — it is impossible. The largest area 60 m of fence can enclose is the square case $x = y = 15$ , giving $225$ m², and the discriminant is what detects that limit.

A line and a curve: non-linear systems

What makes a system non-linear

The method — always substitution

Worked through — a line and a parabola

Worked through — a line and a circle

Expecting two answers

The number of solutions

Finding an unknown constant

Step 1 — Define the variables

Step 2 — Form one equation from each condition

Step 3 — Rearrange the LINEAR equation

Step 4 — Substitute into the non-linear equation

Step 5 — Rearrange into standard quadratic form

Step 6 — Solve the quadratic

Step 7 — Find the matching yyy for each, from the LINEAR equation

Step 8 — Check both in the non-linear equation

Step 9 — Interpret the two solutions in context

Step 10 — A useful extension check

A line and a curve: non-linear systems

What makes a system non-linear

The method — always substitution

Worked through — a line and a parabola

Worked through — a line and a circle

Expecting two answers

The number of solutions

Finding an unknown constant

Step 1 — Define the variables

Step 2 — Form one equation from each condition

Step 3 — Rearrange the LINEAR equation

Step 4 — Substitute into the non-linear equation

Step 5 — Rearrange into standard quadratic form

Step 6 — Solve the quadratic

Step 7 — Find the matching yyy for each, from the LINEAR equation

Step 8 — Check both in the non-linear equation

Step 9 — Interpret the two solutions in context

Step 10 — A useful extension check

Step 7 — Find the matching $y$ for each, from the LINEAR equation

Step 7 — Find the matching $y$ for each, from the LINEAR equation