Simultaneous Equations (Example 1 & 2)

Posted on April 18, 2020 by user

Example 1:

Solve the simultaneous equations.

x + \frac{1}{4} y = 1 and y^{2} - 8 = 4 x .

Solution:

\begin{array}{l} x + \frac{1}{4} y = 1 \to (1) \\ y^{2} - 8 = 4 x \to (2) \\ x = 1 - \frac{1}{4} y \to (3) \end{array}

Substitute (3) into (2),

\begin{array}{l} y^{2} - 8 = 4 (1 - \frac{1}{4} y) \\ y^{2} - 8 = 4 - \frac{4}{4} y \end{array}

y² + y – 12 = 0

(y + 4)(y – 3) = 0

y = –4 or y = 3

Substitute the values of y into (3),

\begin{array}{l} when y = - 4, \\ x = 1 - \frac{1}{4} (- 4) = 2 \\ when y = 3, \\ x = 1 - \frac{1}{4} (3) = \frac{1}{4} \\ The solutions are x = 2, y = - 4 and x = \frac{1}{4}, y = 3 \end{array}

Example 2:

Solve the simultaneous equations 2x + y = 1 and 2x²+ y² + xy = 5.

Correct your answer to three decimal places.

Solution:

2x + y = 1-----(1)

2x² + y²+ xy = 5-----(2)

From (1),

y = 1 – 2x-----(3)

Substitute (3) into (2).

2x² + (1 – 2x)² + x(1 – 2x) = 5

2x² + (1 – 2x)(1 – 2x) + x – 2x² = 5

1 – 2x – 2x + 4x² + x – 5 = 0

4x² – 3x – 4 = 0

\begin{array}{l} From x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ a = 4, b = - 3, c = - 4 \\ x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (4) (- 4)}}{2 (4)} \\ x = \frac{3 \pm \sqrt{73}}{8} \\ x = - 0.693 or 1.443 \end{array}

Substitute the values of x into (3).

When x = –0.693,

y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places)

When x = 1.443,

y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places)

The solutions are x = –0.693, y = 2.386 and x = 1.443, y = –1.886.