1. Non-linear simultaneous equations
Non-linear equations are usually solved using the substitution method.
In this method one of the letters is made the subject of one of the equations. This is then “substituted” into the other equation.
There may be no solution, one solution of many solutions. Ensure that you check each solution in each equation.

2. Non-linear simultaneous equations
Examples
Solve each pair of simultaneous equations.
1) 2xy = –10
x – y = 6 …. 1 2
Check in 2(1)(–5) = –10 1
True
Rearrange equation 2
Check in – –5 = 6 2
True
x = 6 + y …. 3
Sub into 1 3
Check in 2(5)(–1) = –10 1
True
2(6 + y)y = –10
(6 + y)y = –5
y2 + 6y + 5 = 0
(y + 5)(y + 1) = 0
y = –5 or –1
Check in – –1 = 6 2
True
 (1, –5) and (5, –1) are both solutions
x = 6 – 5
x = 1
(1, –5)
x = 6 – 1
x = 5
(5, –1)

3. 2xy = –10
xy = –5
y = –5/x
x – y = 6
x = 6 + y
y = x – 6

4. Non-linear simultaneous equations
Examples
2) x2 + y2 = 29
2x + y – 1= 0 …. 1 2
y = 1 – 2(14/5)
y = –23/5
(–23/5, 14/5)
y = 1 – 2(–2)
y = 5
(–2, 5)
Rearrange equation 2
Check in (–23/5)2 + (14/5)2 = 29 1
True
y = 1 – 2x …. 3
Check in 2(14/5) + (–23/5) – 1 = 0 2
True
Sub into 1 3
Check in (–2)2 + (5)2 = 29 1
True
x2 + (1 – 2x)2 = 29
x2 + 1 – 4x + 4x2 = 29
5x2 – 4x – 28 = 0
(5x – 14)(x + 2) = 0
Check in 2(–2) + 5 –1 = 0 2
True
 (–23/5, 14/5) and (–2, 5) are both solutions
5x – 14 = 0
5x = 14
x = 14/5
x + 2 = 0
x = –2

5. x2 + y2 = 29
y2 = 29 – x2
y = √(29 – x2)
2x + y – 1= 0
y = 1 – 2x