Long Questions (Question 3 & 4)

Posted on April 18, 2020 by user

Question 3:

Solve the following simultaneous equations.

3y – 2x= – 4

y² + 4x² = 2

Solution:

3y – 2x= – 4 -----(1)

y² + 4x²= 2 -----(2)

\begin{array}{l} From (1), \\ y = \frac{2 x - 4}{3} ------- (3) \\ Substitute (3) into (2), \\ {(\frac{2 x - 4}{3})}^{2} + 4 x^{2} = 2 \\ (\frac{4 x^{2} - 16 x + 16}{9}) + 4 x^{2} = 2 \\ 4 x^{2} - 16 x + 16 + 36 x^{2} = 18 (\times 9) \\ 40 x^{2} - 16 x - 2 = 0 \\ 20 x^{2} - 8 x - 1 = 0 \\ (10 x + 1) (2 x - 1) = 0 \\ x = - \frac{1}{10} or x = \frac{1}{2} \\ Substitute the values of x into (3), \\ When x = - \frac{1}{10}, \\ y = \frac{2 (- \frac{1}{10}) - 4}{3} = - 1 \frac{2}{5} \\ When x = \frac{1}{2}, \\ y = \frac{2 (\frac{1}{2}) - 4}{3} = - \frac{3}{3} = 1 \\ The solutions are x = - \frac{1}{10}, y = - 1 \frac{2}{5} and x = \frac{1}{2}, y = 1. \end{array}

Question 4:

Solve the simultaneous equations x – 3y = –1 and y + yx – 2x = 0.

Give your answers correct to three decimal places.

Solution:

x – 3y = –1 -----(1)

y + yx – 2x = 0 -----(2)

From (1),

x = 3y – 1 -----(3)

Substitute (3) into (2),

y + y (3y – 1) – 2(3y – 1) = 0

y + 3y² – y – 6y+ 2 = 0

3y² – 6y + 2 = 0

\begin{array}{l} a = 3, b = - 6, c = 2 \\ y = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ y = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 (3) (2)}}{2 (3)} \\ y = \frac{6 \pm \sqrt{12}}{6} \\ y = 1.577 or 0 .423 \end{array}

Substitute the values of y into (3).

When y = 1.577,

x = 3 (1.577) – 1 = 3.731 (correct to 3 decimal places)

When y = 0.423,

x = 3 (0.423) – 1 = 0.269 (correct to 3 decimal places)

The solutions are x = 3.731, y = 1.577 and x = 0.269, y = 0.423.