Advanced Simultaneous Equations

Solving one Linear and one Quadratic Equation

1. The Visual Meaning: Why Two Answers?

When you solved simple linear simultaneous equations, you were finding where two straight lines crossed (which happens at exactly one point).

Now, one of your equations will have an $x^2$ or $y^2$ in it. This means it is a curve (like a parabola or a circle). When a straight line cuts through a curve, it usually crosses it in two different places.

Because the line hits the curve twice, you will get two pairs of answers (two $x$ values and two $y$ values).

2. The Foolproof 5-Step Method

CRITICAL RULE: You cannot use the "Elimination" method (adding or subtracting equations) when one is a quadratic. You must use the Substitution method.

Rearrange the LINEAR equation so that either $x$ or $y$ is on its own (e.g., $y = x + 3$).
Substitute this into the QUADRATIC equation.
Expand and Simplify to make a standard quadratic equation equal to zero ($ax^2 + bx + c = 0$).
Solve the quadratic (by factorising or using the quadratic formula) to find your two values for the first letter.
Substitute BACK into the LINEAR equation to find the two matching values for the other letter. Pair them up clearly!

3. Worked Examples

Example 1: A Line and a Parabola
Solve the simultaneous equations:
(1) $y = x^2 - 3x + 4$
(2) $y = x + 1$
Step 1 & 2: Substitute the linear into the quadratic Since both equal $y$, we can set them equal to each other:
$$x^2 - 3x + 4 = x + 1$$ Step 3: Rearrange to make it equal zero Subtract $x$ and subtract 1 from both sides:
$$x^2 - 4x + 3 = 0$$ Step 4: Solve the quadratic Factorise (find numbers that multiply to 3 and add to -4):
$$(x - 3)(x - 1) = 0$$ So, $x = 3$ OR $x = 1$. Step 5: Substitute back into the LINEAR equation Always use the linear one ($y = x + 1$) because it's easier and avoids extra squares!
When $x = 3 \Rightarrow y = 3 + 1 \Rightarrow y = 4$
When $x = 1 \Rightarrow y = 1 + 1 \Rightarrow y = 2$
Solution: $(x=3, y=4)$ and $(x=1, y=2)$

Example 2: A Line and a Circle (Harder)
Solve the simultaneous equations:
(1) $x^2 + y^2 = 25$
(2) $x + y = 7$
Step 1: Rearrange the linear equation Make $y$ the subject of eq (2):
$$y = 7 - x$$ Step 2: Substitute into the quadratic Replace the $y$ in eq (1) with $(7 - x)$. Use brackets!
$$x^2 + (7 - x)^2 = 25$$ Step 3: Expand and Simplify Expand $(7 - x)(7 - x) = 49 - 7x - 7x + x^2 = 49 - 14x + x^2$
Put it back in:
$$x^2 + 49 - 14x + x^2 = 25$$ Combine the $x^2$s: $$2x^2 - 14x + 49 = 25$$ Subtract 25 to equal zero: $$2x^2 - 14x + 24 = 0$$ Step 4: Solve the quadratic Divide everything by 2 to make it easier:
$$x^2 - 7x + 12 = 0$$ Factorise: $(x - 3)(x - 4) = 0$
So, $x = 3$ OR $x = 4$. Step 5: Substitute back into the rearranged LINEAR equation Use $y = 7 - x$:
When $x = 3 \Rightarrow y = 7 - 3 \Rightarrow y = 4$
When $x = 4 \Rightarrow y = 7 - 4 \Rightarrow y = 3$
Solution: $(x=3, y=4)$ and $(x=4, y=3)$

10 Practice Questions & Solutions

Try these on paper. Show all your workings clearly. Click the question to check your steps!

Q1: Solve $y = x^2$ and $y = x + 2$

Set them equal: $x^2 = x + 2$
Make equal to zero: $x^2 - x - 2 = 0$
Factorise: $(x - 2)(x + 1) = 0$
Solutions for x: $x = 2$ or $x = -1$

Substitute into $y = x + 2$:
If $x = 2$, $y = 2 + 2 = 4$
If $x = -1$, $y = -1 + 2 = 1$
Answer: $(x=2, y=4)$ and $(x=-1, y=1)$

Q2: Solve $y = x^2 - 2x - 3$ and $y = 2x + 2$

Set them equal: $x^2 - 2x - 3 = 2x + 2$
Subtract $2x$ and subtract $2$: $x^2 - 4x - 5 = 0$
Factorise: $(x - 5)(x + 1) = 0$
Solutions for x: $x = 5$ or $x = -1$

Substitute into $y = 2x + 2$:
If $x = 5$, $y = 10 + 2 = 12$
If $x = -1$, $y = -2 + 2 = 0$
Answer: $(x=5, y=12)$ and $(x=-1, y=0)$

Q3: Solve $x^2 + y^2 = 13$ and $y = x + 1$

Substitute $y$ into the circle equation: $x^2 + (x + 1)^2 = 13$
Expand bracket: $x^2 + (x^2 + 2x + 1) = 13$
Simplify: $2x^2 + 2x + 1 = 13$
Make equal to zero: $2x^2 + 2x - 12 = 0$
Divide by 2: $x^2 + x - 6 = 0$
Factorise: $(x + 3)(x - 2) = 0$
$x = -3$ or $x = 2$

Sub into $y = x + 1$:
If $x = -3$, $y = -2$
If $x = 2$, $y = 3$
Answer: $(x=-3, y=-2)$ and $(x=2, y=3)$

Q4: Solve $xy = 12$ and $y = x - 4$

Substitute the linear into the curve: $x(x - 4) = 12$
Expand: $x^2 - 4x = 12$
Equal to zero: $x^2 - 4x - 12 = 0$
Factorise: $(x - 6)(x + 2) = 0$
$x = 6$ or $x = -2$

Sub into $y = x - 4$:
If $x = 6$, $y = 6 - 4 = 2$
If $x = -2$, $y = -2 - 4 = -6$
Answer: $(x=6, y=2)$ and $(x=-2, y=-6)$

Q5: Solve $x^2 + y^2 = 20$ and $y = 2x$

Substitute $y$ into the circle: $x^2 + (2x)^2 = 20$
Be careful! $(2x)^2 = 4x^2$.
So, $x^2 + 4x^2 = 20$
$5x^2 = 20$
Divide by 5: $x^2 = 4$
Square root: $x = 2$ or $x = -2$

Sub into $y = 2x$:
If $x = 2$, $y = 4$
If $x = -2$, $y = -4$
Answer: $(x=2, y=4)$ and $(x=-2, y=-4)$

Q6: Solve $y = x^2 - 5x$ and $y = 2x - 12$

Set equal: $x^2 - 5x = 2x - 12$
Rearrange: $x^2 - 7x + 12 = 0$
Factorise: $(x - 3)(x - 4) = 0$
$x = 3$ or $x = 4$

Sub into $y = 2x - 12$:
If $x = 3$, $y = 6 - 12 = -6$
If $x = 4$, $y = 8 - 12 = -4$
Answer: $(x=3, y=-6)$ and $(x=4, y=-4)$

Q7: Solve $x^2 + y^2 = 10$ and $x + y = 2$

Rearrange linear: $y = 2 - x$
Sub into circle: $x^2 + (2 - x)^2 = 10$
Expand bracket: $x^2 + (4 - 4x + x^2) = 10$
Simplify: $2x^2 - 4x + 4 = 10$
Equal zero: $2x^2 - 4x - 6 = 0$
Divide by 2: $x^2 - 2x - 3 = 0$
Factorise: $(x - 3)(x + 1) = 0$
$x = 3$ or $x = -1$

Sub into $y = 2 - x$:
If $x = 3$, $y = 2 - 3 = -1$
If $x = -1$, $y = 2 - (-1) = 3$
Answer: $(x=3, y=-1)$ and $(x=-1, y=3)$

Q8: Solve $y = 2x^2 + 3x - 1$ and $y = x + 3$

Set equal: $2x^2 + 3x - 1 = x + 3$
Rearrange: $2x^2 + 2x - 4 = 0$
Divide by 2: $x^2 + x - 2 = 0$
Factorise: $(x + 2)(x - 1) = 0$
$x = -2$ or $x = 1$

Sub into $y = x + 3$:
If $x = -2$, $y = -2 + 3 = 1$
If $x = 1$, $y = 1 + 3 = 4$
Answer: $(x=-2, y=1)$ and $(x=1, y=4)$

Q9: Solve $x^2 - y^2 = 9$ and $x - y = 1$

Rearrange linear: $x = y + 1$
Sub into quadratic: $(y + 1)^2 - y^2 = 9$
Expand: $(y^2 + 2y + 1) - y^2 = 9$
The $y^2$s cancel out!: $2y + 1 = 9$
$2y = 8 \Rightarrow y = 4$

Notice there's only one $y$ value here! Let's find $x$.
Sub into $x = y + 1$:
$x = 4 + 1 = 5$
Answer: $(x=5, y=4)$

Q10: Challenge! Solve $y = x^2 + 4x + 4$ and $y = 2x + 3$

Set equal: $x^2 + 4x + 4 = 2x + 3$
Rearrange: $x^2 + 2x + 1 = 0$
Factorise: $(x + 1)(x + 1) = 0$
$x = -1$

Because we get a repeated root, there is only one solution! This means the straight line is a tangent to the curve; it just brushes against it.
Sub into $y = 2x + 3$:
$y = 2(-1) + 3 = 1$
Answer: $(x=-1, y=1)$