1fully worked

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

1both equal $y$, so $x^2=x+2$Set the two $y$-expressions equal

2$x^2-x-2=0$Rearrange to a quadratic $=0$

3$(x-2)(x+1)=0$, so $x=2$ or $x=-1$Factorise and solve for $x$

4into $y=x+2$: $(2,4)$ and $(-1,1)$Substitute back to find each $y$

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

2you finish it

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

1both equal $y$, so $x^2+2=3x$Set the two $y$-expressions equal

2$x^2-3x+2=0$Rearrange to a quadratic $=0$

3$(x-1)(x-2)=0$, so $x=1$ or $x=2$Factorise and solve for $x$

4Substitute back to find each $y$

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

3you finish it

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

1both equal $y$, so $x^2-1=x+1$Set the two $y$-expressions equal

2$x^2-x-2=0$Rearrange to a quadratic $=0$

3Factorise and solve for $x$

4Substitute back to find each $y$

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 $-1$

4you do it all

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

1Set the two $y$-expressions equal

2Rearrange to a quadratic $=0$

3Factorise and solve for $x$

4Substitute back to find each $y$

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 $x=2$