What Level is Your GCSE Maths Algebra At?

MME
Maths Made Easy Team 26 March 2018
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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


Level 4 (Low Grade C)

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} 


Level 5 (High Grade C)

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^2and the height is twice as long as the width.

Calculate the value of x given that the area of the rectangle is 12.5m^2and 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}


Level 7 (Grade A)

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}

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