The values of a and b can be found by substituting terms in from the the first few patterns.
Substitute the terms in from diagram 1 (n = 1, T = 1), call this equation (1).

$\begin{array}{rcl} n & = & 1, T = 1 \\ 1 & = & a {(1)}^{3} + b {(1)}^{2} \\ 1 & = & a + b \end{array}$

[1]

Substitute the terms in from diagram 2 (n = 2, T = 1 + 5 = 6), call this equation (2).

$\begin{array}{rcl} n & = & 2, T = 6 \\ 6 & = & a {(2)}^{3} + b {(2)}^{2} \\ 6 & = & 8 a + 4 b \end{array}$

[1]

Write the two equations out next to each other and number them clearly.

$\begin{array}{rcl} a + b & = & 1 (1) \\ 8 a + 4 b & = & 6 (2) \end{array}$

Make the $b$ terms equal by multiplying all parts of equation (1) by 4.

$\begin{array}{rcl} 4 a + 4 b & = & 4 (3) \\ 8 a + 4 b & = & 6 (2) \end{array}$

Eliminate the $b$ terms by subtracting equation (3) from equation (2). Be careful to subtract every term.

$8 a + 4 b = 6 (2) - (4 a + 4 b = 4) (3) 4 a = 2$

[1]

Solve the equation to find $a$ by dividing both sides by 4.

$\begin{array}{rcl} a & = & \frac{2}{4} = \frac{1}{2} \end{array}$

[1]

Substitute $a = \frac{1}{2}$ into either of the two original equations.

$\begin{array}{rcl} (1) a + b & = & 1 \\ (\frac{1}{2}) + b & = & 1 \end{array}$

Solve this equation to find $b$ .

$\begin{array}{rcl} b & = & 1 - \frac{1}{2} \\ = & \frac{1}{2} \end{array}$

Substitute $a = \frac{1}{2}$ and $b = \frac{1}{2}$ into the other equation to check that they are correct

$\begin{array}{r} (2) 8 a + 4 b = 6 \\ 8 (\frac{1}{2}) + 4 (\frac{1}{2}) = 6 \\ 4 + 2 = 6 \\ 6 = 6 \end{array}$

$a = \frac{1}{2}, b = \frac{1}{2}$ [1]

Simultaneous Equations (CIE IGCSE Maths: Extended)

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