Simultaneous (linear/quadratic) Worked Solutions — GCSE Maths

Fully worked, step-by-step solutions to GCSE Simultaneous (linear/quadratic) questions. See exactly how to solve problems on substitution, quadratic and horizontal line, factorising by taking out x, common factor.

substitutionquadratic and horizontal linefactorising by taking out xcommon factorcirclevertical line

GCSE Higher70 questionsStep-by-step solutions

Question 1

1 markeasy

Solve the simultaneous equations

y = x^2

and

y = 9

Worked solution

Substitute the line into the curve
$x^{2} = 9$
Both equations give $y$ , so where the graphs meet the two expressions for $y$ are equal. That leaves one equation in $x$ alone.
Rearrange and solve for x
$x^{2} - 9 = 0 \ \Rightarrow \ (x + 3)(x - 3) = 0 \ \Rightarrow \ x = -3 \ \text{or} \ x = 3$
Rearrange so the quadratic equals zero, factorise, then set each factor to zero. Never divide through by $x$ — that would throw away the root $x = 0$ .
State the two solutions
$x = -3,\ y = 9 \quad \text{or} \quad x = 3,\ y = 9$
The line meets the curve at (-3, 9) and (3, 9).

Answer

x = -3,\ y = 9 \quad \text{or} \quad x = 3,\ y = 9

Question 2

2 markseasy

Solve the simultaneous equations

y = x^2

and

y = 2x

Worked solution

Substitute the line into the curve
$x^{2} = 2 x$
Both equations give $y$ , so where the graphs meet the two expressions for $y$ are equal. That leaves one equation in $x$ alone.
Rearrange and solve for x
$x^{2} - 2 x = 0 \ \Rightarrow \ x(x - 2) = 0 \ \Rightarrow \ x = 0 \ \text{or} \ x = 2$
Rearrange so the quadratic equals zero, factorise, then set each factor to zero. Never divide through by $x$ — that would throw away the root $x = 0$ .
State the two solutions
$x = 0,\ y = 0 \quad \text{or} \quad x = 2,\ y = 4$
The line meets the curve at (0, 0) and (2, 4).

Answer

x = 0,\ y = 0 \quad \text{or} \quad x = 2,\ y = 4

Question 3

2 markseasy

Solve the simultaneous equations

y = x^2

and

y = x

Worked solution

Substitute the line into the curve
$x^{2} = x$
Both equations give $y$ , so where the graphs meet the two expressions for $y$ are equal. That leaves one equation in $x$ alone.
Rearrange and solve for x
$x^{2} - x = 0 \ \Rightarrow \ x(x - 1) = 0 \ \Rightarrow \ x = 0 \ \text{or} \ x = 1$
Rearrange so the quadratic equals zero, factorise, then set each factor to zero. Never divide through by $x$ — that would throw away the root $x = 0$ .
State the two solutions
$x = 0,\ y = 0 \quad \text{or} \quad x = 1,\ y = 1$
The line meets the curve at (0, 0) and (1, 1).

Answer

x = 0,\ y = 0 \quad \text{or} \quad x = 1,\ y = 1

Question 4

2 markseasy

The line

x = 4

crosses the circle

x^2 + y^2 = 25

at two points. Work out the coordinates of both points.

Worked solution

Substitute the x-value into the circle equation
$\left(4\right)^2 + y^2 = 25$
The line fixes $x$ , so put that value straight into the circle equation.
Solve for y
$y^2 = 25 - 16 = 9 \ \Rightarrow \ y = \pm 3$
Square-rooting gives a positive AND a negative value — that is why the line cuts the circle twice.
State the two points
$\left(4,\ 3\right) \quad \text{and} \quad \left(4,\ -3\right)$
Both points have $x = 4$ ; they are the top and bottom of the chord.

Answer

\left(4,\ 3\right) \quad \text{and} \quad \left(4,\ -3\right)

Question 5

2 markseasy

The line

y = 5

meets the circle

x^2 + y^2 = 25

at exactly one point. Write down the coordinates of that point.

Worked solution

Substitute the line into the circle equation
$x^2 + \left(5\right)^2 = 25$
Replace $y$ in the circle equation by the expression from the line. Every $y$ disappears, leaving one equation in $x$ alone.
Rearrange so the quadratic equals zero
$x^{2} = 0$
Collect every term on one side. A quadratic can only be factorised once it is equal to zero.
State the point of contact
$\left(0,\ 5\right)$
The line touches the curve at exactly one point, (0, 5), so it is a tangent.

Answer

\left(0,\ 5\right)

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