Faded examples · Solving a line and a quadratic curve simultaneously

Non-calculator

Each worked example shows a little less than the one before – you complete the faded (blank) steps yourself, following the same four steps every time. In each example also answer the check question – it tests why the step works. No calculator.

①Example 1fully worked: read it through

Solve $y=x^2$ and $y=x+2$ simultaneously.

1Set the two $y$-expressions equal

both equal $y$, so $x^2=x+2$

2Rearrange to a quadratic $=0$

$x^2-x-2=0$

3Factorise and solve for $x$

$(x-2)(x+1)=0$, so $x=2$ or $x=-1$

4Substitute back to find each $y$

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

Check · Why do we set $x^2$ and $x+2$ equal to each other?

A at an intersection both equations share the same $x$ and $y$, so the $y$-expressions are equalB because $x^2$ is always bigger than $x+2$C to make the equation harder to solveD you should subtract one from the other instead

②Example 2you finish the last 1 step

Solve $y=x^2+2$ and $y=3x$ simultaneously.

1Set the two $y$-expressions equal

both equal $y$, so $x^2+2=3x$

2Rearrange to a quadratic $=0$

$x^2-3x+2=0$

3Factorise and solve for $x$

$(x-1)(x-2)=0$, so $x=1$ or $x=2$

4Substitute back to find each $y$

into $y=3x$: $(1,3)$ and $(2,6)$

Check · A student found $x=1$ and $x=2$ and wrote that as the final answer. What is missing?

A the two $x$-values should be squaredB nothing – the $x$-values are the full answerC the matching $y$-value for each $x$D the sum of the two $x$-values

③Example 3you finish the last 2 steps

Solve $y=x^2-1$ and $y=x+1$ simultaneously.

1Set the two $y$-expressions equal

both equal $y$, so $x^2-1=x+1$

2Rearrange to a quadratic $=0$

$x^2-x-2=0$

3Factorise and solve for $x$

$(x-2)(x+1)=0$, so $x=2$ or $x=-1$

4Substitute back to find each $y$

into $y=x+1$: $(2,3)$ and $(-1,0)$

Check · When you rearrange $x^2-1=x+1$ to "$=0$", what is the constant term?

A $0$B $-2$ (move the $+1$ across: $x^2-x-2=0$)C $+2$D $-1$

④Example 4your turn: every step

Solve $y=x^2+1$ and $y=2x+1$ simultaneously.

1Set the two $y$-expressions equal

both equal $y$, so $x^2+1=2x+1$

2Rearrange to a quadratic $=0$

$x^2-2x=0$

3Factorise and solve for $x$

$x(x-2)=0$, so $x=0$ or $x=2$

4Substitute back to find each $y$

into $y=2x+1$: $(0,1)$ and $(2,5)$

Check · The quadratic is $x^2-2x=0$. Why is it wrong to divide both sides by $x$ to solve it?

A you may divide by $x$ with no problemB because $x^2-2x$ is not really a quadraticC dividing by $x$ loses the solution $x=0$; factorise as $x(x-2)=0$ insteadD dividing by $x$ gives $x=1$ and $x=2$

Answers · Simultaneous Equations (Linear & Quadratic)

Faded examples · Solving a line and a quadratic curve simultaneously

① Example 1 both equal $y$, so $x^2=x+2$→$x^2-x-2=0$→$(x-2)(x+1)=0$, so $x=2$ or $x=-1$→into $y=x+2$: $(2,4)$ and $(-1,1)$ $(2,\,4)\ \text{and}\ (-1,\,1)$

Check: A: at an intersection both equations share the same $x$ and $y$, so the $y$-expressions are equal

② Example 2 both equal $y$, so $x^2+2=3x$→$x^2-3x+2=0$→$(x-1)(x-2)=0$, so $x=1$ or $x=2$→into $y=3x$: $(1,3)$ and $(2,6)$ $(1,\,3)\ \text{and}\ (2,\,6)$

Check: C: the matching $y$-value for each $x$

③ Example 3 both equal $y$, so $x^2-1=x+1$→$x^2-x-2=0$→$(x-2)(x+1)=0$, so $x=2$ or $x=-1$→into $y=x+1$: $(2,3)$ and $(-1,0)$ $(2,\,3)\ \text{and}\ (-1,\,0)$

Check: B: $-2$ (move the $+1$ across: $x^2-x-2=0$)

④ Example 4 both equal $y$, so $x^2+1=2x+1$→$x^2-2x=0$→$x(x-2)=0$, so $x=0$ or $x=2$→into $y=2x+1$: $(0,1)$ and $(2,5)$ $(0,\,1)\ \text{and}\ (2,\,5)$

Check: C: dividing by $x$ loses the solution $x=0$; factorise as $x(x-2)=0$ instead