3.3.1 System of Equations Long Questions (Question 1 & 2)

Question 1:

Solve the following simultaneous equations.

\begin{array}{l} y + 2 x = 2 \\ \frac{2}{x} + \frac{1}{y} = 5 \end{array}

Solution:

\begin{array}{l} y + 2 x = 2 - - - - (1) \\ \frac{2}{x} + \frac{1}{y} = 5 - - - - (2) \\ y = 2 - 2 x - - - - (3) \\ substitute (3) into (2), \\ \frac{2}{x} + \frac{1}{2 - 2 x} = 5 \\ \frac{2 (2 - 2 x) + x}{x (2 - 2 x)} = 5 \\ 4 - 4 x + x = 5 x (2 - 2 x) \\ 4 - 3 x = 10 x - 10 x^{2} \\ 10 x^{2} - 13 x + 4 = 0 \\ (5 x - 4) (2 x - 1) = 0 \\ 5 x - 4 = 0 or     2 x - 1 = 0 \\ x = \frac{4}{5} or x = \frac{1}{2} \\ Substitute values of x into (3), \\ When x = \frac{4}{5}, \\ y = 2 - 2 (\frac{4}{5}) = \frac{2}{5} \\ When x = \frac{1}{2} \\ y = 2 - 2 (\frac{1}{2}) = 1 \\ The solutions are x = \frac{4}{5}, y = \frac{2}{5} and x = \frac{1}{2}, y = 1 \end{array}

Question 2:

Solve the following simultaneous equations.

x – 3y + 5 = 3y + 5y²– 6 – x = 0

Solution:

x – 3y + 5 = 0

x = 3y – 5 —–(1)

3y + 5y² – 6 – x = 0 —–(2)

Substitute (1) into (2),

3y + 5y² – 6 – (3y – 5) = 0

3y + 5y² – 6 – 3y + 5 = 0

5y² – 1 = 0

5y² = 1

y = ±0.447

Substitute the values of y into (1),

When y = 0.447

x = 3 (0.447) – 5

x = –3.659

When y = – 0.447

x = 3 (–0.447) – 5

x = –6.341

The solutions are x = –3.659, y = 0.447 and x = –6.341, y = – 0.447.

sally

January 12, 2018 at 6:32 am | Reply

This site actually good but I want to suggest to provide more questions on this site such as exercise. Well, many student can learn if you provide more question . Thank you…..
Ahmad Hafizi Hamdan

August 19, 2019 at 3:03 am | Reply

2(2-x)+x = 4-3x not 4-4x
- myhometuition
  
  September 17, 2019 at 7:27 pm | Reply
  
  Thanks for pointing our mistake.
  We have done the correction accordingly.