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# What Level is Your GCSE Maths Algebra At?

What Level is Your GCSE Maths Algebra At?

Try our Maths questions below to see what your Algebra level is at. Can you work your way up the Algebra questions? Could you get a GCSE Grade C or Level 4?

Level 1

$2x = 10$ what is $x$?

Solve

$2x = 10$

$x=5$

Level 2

$3x + 1 = 10$ what is $x$?

Solve

$3x + 1 = 10$

\begin{aligned}3x &=9 \\ x=3\end{aligned}

Level 3

Simplify the following:

$2y + y + y- 10y$

Simplify $2y + y + y- 10y$

$-6y$

Solve the following:

$\dfrac{3x^2-10}{2}=1$

Solve

$\dfrac{3x^2-10}{2}=1$

\begin{aligned}3x^2-10 &= 2 \\ 3x^2 &= 12 \\ x^2 &= 4 \\ x &= \sqrt{4} \\ x &=2\end{aligned}

Make $y$ the subject of the following equation

$\dfrac{5y+2}{3} = 6x+2y$

Make $y$ the subject of the following equation

$\dfrac{5y+2}{3} = 6x+2y$

\begin{aligned}5y+2 &= 18x+6y \\ 5y+2 -18x &= 6y \\ 2-18x &= y\end{aligned}

Level 6

Calculate the value of $x$ given that the area of the rectangle is $12.5m^2$and the height is twice as long as the width.

Calculate the value of $x$ given that the area of the rectangle is $12.5m^2$and the height is twice as long as the width.

\begin{aligned}(2x+1)(x+\dfrac{1}{2})&=12.5 \\ 2x^2 + x + x + \dfrac{1}{2} &= 12.5 \\ 4x^2 + 4x + 1 &= 25 \\ 4x^2 + 4x - 24 &= 0 \\ x^2 + x -6 &= 0 \\ (x+3)(x-2) &= 0 \\ \text{so }x=-3&\text{ or }2\text{. }x\text{ is a length so only }x=2\text{ is an acceptable answer.}\end{aligned}

Prove that the product of any two odd integers is always odd.

Prove that the product of any two odd integers is always odd.

$(2x+1)(2y+1)$

$4xy + 2x + 2y + 1$

$2(2xy + x + y) +1$

This is written in the form of an odd number so is always odd.

Level 8 and 9 (Grade A*)

The functions $f$ and $g$ are as follows:

$f(x)=x+4$

$g(x)=2 - \dfrac{1}{x}$

(Level 8 Gade A*): Calculate fg(2)

$fg(2)=2- \dfrac{1}{2} + 4 = 5.5$

$fg(2)=2- \dfrac{1}{2} + 4 = 5.5$

(Level 9 Gade A**): Show that $g^{-1}(x) +f(x) = \dfrac{9 - 2x -x^2}{2-x}$

\begin{aligned}g(x) &= 2 - \dfrac{1}{x} \\ g(x) - 2 &= -\dfrac{1}{x} \\ 2-g(x) &= \dfrac{1}{x} \\ x &= \dfrac{1}{2-g(x)} \\ \text{The inverse is:} \\ g^{-1}(x) &= \dfrac{1}{2-x} \end{aligned}

$\begin{gathered}g^{-1}(x) + f(x) = \dfrac{1}{2-x} + (x+4) \\ \dfrac{1}{2-x} +\dfrac{(2-x)(x+4)}{(2-x)} \\ \dfrac{1+2x-x^2+8-4x}{2-x} \\ \dfrac{9-2x-x^2}{2-x}\end{gathered}$