# 4.1.1 Simultaneous Equations Example 1 & 2

Example 1:
Solve the simultaneous equations.

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

Substitute (3) into (2),
$\begin{array}{l}{y}^{2}-8=4\left(1-\frac{1}{4}y\right)\\ {y}^{2}-8=4-\frac{4}{4}y\end{array}$
y2 + y – 12 = 0
(y + 4)(y – 3) = 0
y = –4 or y = 3

Substitute the values of y into (3),

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

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 – 2+ 4x2 + x – 5 = 0
4x2 – 3x – 4 = 0

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.

### 1 thought on “4.1.1 Simultaneous Equations Example 1 & 2”

1. The answer for the first example is wrong, by substituting -4 into 3 we get 2