6.4.7 Linear Equations, PT3 Focus Practice

Question 19:

Calculate the value of x and of y that satisfy the following simultaneous linear equations:

\begin{array}{l} x + 6 y = 12 \\ \frac{2}{3} x + 2 y = 6 \end{array}

[4 marks]

Solution:

\begin{array}{l} x + 6 y = 12 \\ x = 12 - 6 y - - - (1) \\ \frac{2}{3} x + 2 y = 6 \\ (\times 3) 2 x + 6 y = 18 \\ (\div 2) x + 3 y = 9 - - - (2) \end{array}

Substitute (1) into (2),
12 – 6y + 3y = 9
–3y = –3
y = 1

Substitute y = 1 into (1),

x = 12 – 6 (1)
x = 6