Simultaneous Equations with a Curve

A line meets a curve: substitute the line, expand carefully, pair each x with its y, and find the distance between the points (Higher).

In the exam for Edexcel IGCSE Maths 4MA1, Cambridge IGCSE Maths 0580, Edexcel Further Pure Maths 4PM1, IGCSE Additional Maths 0606, Cambridge O Level Maths 4024 · 47 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

A Line Meets a Curve

One equation is a straight line. The other has $x^{2}$ or $y^{2}$ in it, so it is a curve.

Rule

Make x or y the subject of the line, then substitute it into the curve. You get a quadratic, so there are usually two answers.

Key idea

Each answer is a point where the line crosses the curve. That is why they come in pairs.

Key idea

Sometimes the quadratic is a perfect square, like $x^{2} - 2 x + 1 = 0$ , which is $(x - 1)^{2} = 0$ . Then there is only one x. The line just touches the curve at one point. That is not a mistake.

Substitute the Line

This is the method for every line-and-curve question. Here the curve is a circle.

Worked example

1
$x^{2} + y^{2} = 41, 2 x + y = 3$ — The question. The line has no squares, so start with the line.
2
$y = 3 - 2 x$ — Take 2x from both sides of the line, so y is alone.
3
$x^{2} + (3 - 2 x)^{2} = 41$ — Replace y in the curve. Keep the bracket, because the whole line gets squared.
4
$x^{2} + 9 - 12 x + 4 x^{2} = 41$ — Expand the bracket: $(3 - 2 x)^{2}$ = 9 − 12x + 4x². Do not forget the middle term.
5
$5 x^{2} - 12 x + 9 = 41$ — Collect the x² terms: x² + 4x² = 5x².
6
$5 x^{2} - 12 x - 32 = 0$ — Take 41 from both sides. A quadratic must equal 0 before you solve it.
7
$(x - 4) (5 x + 8) = 0$ — Factorise. Check by expanding: 5x² + 8x − 20x − 32 = 5x² − 12x − 32.
8
$x = 4 or x = - \frac{8}{5}$ — Each bracket can be 0: x − 4 = 0 or 5x + 8 = 0.
9
$y = 3 - 2 (4) = - 5$ — Put x = 4 into the line. The line is easier than the curve.
10
$y = 3 - 2 (- \frac{8}{5}) = \frac{31}{5}$ — Now x = −8/5. Minus times minus: −2 × (−8/5) = +16/5, and 3 + 16/5 = 31/5.
11
$(4, - 5), (- \frac{8}{5}, \frac{31}{5})$ — Write each x with its own y, as two points.

Common mistake

Mixing the pairs up. Write each x with its own y, as two points.

Key idea

Show every line. Most of the marks are for the working: replacing y, getting the quadratic = 0, and solving it. Answers with no algebra score 1 mark at most.

Two y = Equations: Set Them Equal

Many questions give both equations as “y = …”. Then there is nothing to rearrange: set them equal.

Worked example

1
$y = x^{2} + 3 x - 8, y = 2 x + 4$ — The question. Both equations start with y =.
2
$x^{2} + 3 x - 8 = 2 x + 4$ — Both equal y, so they equal each other. This line is usually the first mark.
3
$x^{2} + x - 8 = 4$ — Take 2x from both sides: 3x − 2x = x.
4
$x^{2} + x - 12 = 0$ — Take 4 from both sides, so it equals 0: −8 − 4 = −12.
5
$(x + 4) (x - 3) = 0$ — Factorise. +4 × (−3) = −12 and +4 + (−3) = +1.
6
$x = - 4 or x = 3$ — Each bracket can be 0.
7
$y = 2 (3) + 4 = 10$ — Put x = 3 into the line, because it is simpler than the curve.
8
$y = 2 (- 4) + 4 = - 4$ — Now x = −4: 2 × (−4) = −8, and −8 + 4 = −4.
9
$(3, 10), (- 4, - 4)$ — Each x with its own y. Check (3, 10) in the curve: 9 + 9 − 8 = 10.

Common mistake

Sometimes the numbers cancel and you get $x^{2} + 3 x = 0$ . Do not divide by x: you lose the answer x = 0. Factorise instead: $x (x + 3) = 0$ , so x = 0 or x = −3.

Careful with the Brackets

Most marks are lost in the expanding. Put the whole line in a bracket every time it appears.

Worked example

1
$3 x^{2} + y^{2} - x y = 5, y = 2 x - 3$ — y appears twice in the curve
2
$3 x^{2} + (2 x - 3)^{2} - x (2 x - 3) = 5$ — a bracket each time
3
$3 x^{2} + 4 x^{2} - 12 x + 9 - 2 x^{2} + 3 x = 5$ — expand each part
4
$5 x^{2} - 9 x + 4 = 0 \Rightarrow x = 1, 0.8$ — tidy and solve
5
$(1, - 1), (0.8, - 1.4)$ — y = 2x − 3 for each

Common mistake

Writing (2x − 3)² as 4x² + 9. It is $4 x^{2} - 12 x + 9$ : the middle term is easy to lose.

Replace x Instead

If x has no number in front in the line, make x the subject instead. It keeps fractions away.

Worked example

1
$x - 3 y = 1, x y - y^{2} = 6$ — The question. In the line, x stands on its own.
2
$x = 3 y + 1$ — Add 3y to both sides. Making y the subject would give $\frac{x - 1}{3}$ , a fraction.
3
$(3 y + 1) y - y^{2} = 6$ — Replace x in the curve, with a bracket.
4
$3 y^{2} + y - y^{2} = 6$ — Expand: y times each term in the bracket.
5
$2 y^{2} + y = 6$ — Collect the squares: 3y² − y² = 2y².
6
$2 y^{2} + y - 6 = 0$ — Take 6 from both sides, so it equals 0.
7
$(2 y - 3) (y + 2) = 0$ — Factorise. Check: 4y − 3y = y and −3 × 2 = −6.
8
$y = 1.5 or y = - 2$ — Each bracket can be 0. This time the answers are y values.
9
$x = 3 (1.5) + 1 = 5.5$ — Put y = 1.5 into x = 3y + 1.
10
$x = 3 (- 2) + 1 = - 5$ — Now y = −2: 3 × (−2) = −6, and −6 + 1 = −5.
11
$(5.5, 1.5), (- 5, - 2)$ — Write x first, even though you found y first.

Common mistake

Writing (1.5, 5.5). A point is always (x, y), so x goes first.

When It Won’t Factorise

If the question says “give your answers to 2 decimal places”, the quadratic will not factorise. Use the quadratic formula (see the lesson The Quadratic Formula and Completing the Square).

Worked example

1
$y = x^{2} + x - 3, y = 5 x - 2$ — The question. Both start with y =, so set them equal.
2
$x^{2} + x - 3 = 5 x - 2$ — Both equal y.
3
$x^{2} - 4 x - 3 = - 2$ — Take 5x from both sides: x − 5x = −4x.
4
$x^{2} - 4 x - 1 = 0$ — Add 2 to both sides, so it equals 0.
5
$a = 1, b = - 4, c = - 1$ — Read off a, b and c. Keep the minus signs.
6
$b^{2} - 4 a c = (- 4)^{2} - 4 \times 1 \times (- 1)$ — The part under the square root. Put the negatives in brackets.
7
$= 16 + 4 = 20$ — (−4)² = 16, and −4 × 1 × (−1) = +4.
8
$x = \frac{- ( - 4 ) \pm 20}{2} = \frac{4 \pm 20}{2}$ — −b means −(−4) = 4. The bottom is 2a = 2.
9
$x = 4.2360 \dots or x = - 0.2360 \dots$ — Work out + and − separately. Keep all the digits for now.
10
$y = 5 (4.2360 \dots) - 2 = 19.180 \dots$ — Put the unrounded x into the line.
11
$y = 5 (- 0.2360 \dots) - 2 = - 3.180 \dots$ — Same for the other x.
12
$(4.24, 19.18), (- 0.24, - 3.18)$ — Round only at the end, to 2 decimal places.

Common mistake

Rounding x first. 5 × 4.24 − 2 = 19.20, not 19.18, so the answer loses its accuracy mark.

The Distance Between Them (Higher)

A 6-mark version asks for the length AB after you find the points. Use Pythagoras.

Worked example

1
$A (4, - 5), B (- 1.6, 6.2)$ — the points from the earlier card
2
$5.6, 11.2$ — the across and up distances
3
$A B = 5. 6^{2} + 11. 2^{2} = 156.8$ — Pythagoras
4
$A B = \frac{28 5}{5} \approx 12.5$ — exact form if asked, otherwise a decimal

Which Method? Read the Question

Look for the squared letter.

You see	You do
Two straight lines	Use the lesson Simultaneous Equations by Elimination
A line and a curve	Substitute the line into the curve
Both start with y =	Set them equal, then make it = 0
y alone in the line	Replace y; use brackets
x alone in the line	Replace x instead
“2 decimal places”	Quadratic formula; round at the end
“Length AB”	Find A and B, then Pythagoras
“Midpoint of AB”	Find A and B, then average the x's and the y's

Key idea

Check both points in the line and the curve. Both equations must work.

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