Simultaneous linear equations and quadratic equations combined – the use of difference of two squares

Sometimes, we may be given a pair of equations. Being a pair make it correct to call them simultaneous equation. And the fact that the power of the unknowns in such equation(s) is/are 2, makes it correct to call it/them quadraticequation. Where this happens, we say we have simultaneous linear and quadratic equations that are combined in one question.

To solve such questions, we need to factorize the quadratic part of the equations in such a way that a difference of two squares principle is used.

Example.

If x² – y² = 3 and x + y = 1, find the values of x and y

Solution

Given that

x² – y² = 3 and

x + y = 1

Step1

We label the two equations as follows

x² – y² = 3 … equ1

x + y = 1 … equ2

From equ1, we have that

x² – y² = 3 which gives

(x + y)(x – y) = 3 (by difference of two squares)

Thus, we call this new expression

(x + y)(x – y) = 3 … equ3

And recall that from equ2, x + y = 1.

Substitute 1 for x + y in equ3 as follows

(x + y)(x – y) = 3

1(x – y) = 3

x – y = 3 (by opening bracket)

Make x the subject of the equation as follows

x = 3 + y … equ4

Substitute 3 + y for x in equ2 as follows

x + y = 1 … equ2

3 + y + y = 1

3 + 2y = 1

2y = 1 – 3 (by collecting like terms)

2y = -2

y = -2÷2 = -1

y = -1

Substitute -1 for y in equ4 to get x as follows

x = 3 + y … equ4

x = 3 + (-1)

x = 3 – 1

x = 2

Thus, x=2, y=-1