iii)

Method 1: Elimination

Multiply the equation for the triangle by 3.

$\begin{array}{rcl} 3 a + 5 b & = & 170 \times 3 \\ 9 a + 15 b & = & 510 \end{array}$

Subtract the equation for the quadrilateral from the adapted triangle equation.

$9 a + 15 b = 510 - 9 a + 7 b = 310 8 b = 200$

Divide both sides by 8.

$b = 25$

Substitute $b = 25$ into either equation and solve for $a$ .

$\begin{array}{rcl} 3 a + 5 \times 25 & = & 170 \\ 3 a + 125 & = & 170 \end{array}$

Subtract 125 from both sides.

$3 a = 45$

Divide both sides by 3.

$a = 15$

Correct method of elimination shown [1]

$\begin{array}{rcl} a & = & 15 \end{array}$ [1]
$\begin{array}{rcl} b & = & 25 \end{array}$ [1]

Method 2: Substitution

Rearrange the equation for the triangle to make $a$ the subject.

$\begin{array}{rcl} 3 a + 5 b & = & 170 \end{array}$

Subtract $5 b$ from both sides.

$3 a = 170 - 5 b$

Divide both sides by 3.

$a = \frac{170 - 5 b}{3}$

Substitute the rearranged equation into the quadrilateral equation in place of $a$ .

$9 (\frac{170 - 5 b}{3}) + 7 b = 310$

Multiply both sides by 3.

$9 (170 - 5 b) + 21 b = 930$

Expand the brackets on the left hand side.

$\begin{array}{rcl} 1530 - 45 b + 21 b & = & 930 \end{array}$

Simplify the left hand side by collecting like terms.

$1530 - 24 b = 930$

Add $24 b$ to both sides.

$1530 = 930 + 24 b$

Subtract 930 from both sides.

$\begin{array}{rcl} 600 & = & 24 b \end{array}$

Divide both sides by 24.

$25 = b$

Substitute $b = 25$ into either equation and solve for $a$ .

$\begin{array}{rcl} 3 a + 5 \times 25 & = & 170 \\ 3 a + 125 & = & 170 \end{array}$

Subtract 125 from both sides.

$3 a = 45$

Divide both sides by 3.

$a = 15$

Correct method of substitution shown [1]

$\begin{array}{rcl} a & = & 15 \end{array}$ [1]
$\begin{array}{rcl} b & = & 25 \end{array}$ [1]

Angles in Polygons & Parallel Lines (CIE IGCSE Maths: Extended)

Exam Questions