Simultaneous linear equations and quadratic equations combined – Question 2 - use of difference of two squares

Solve the following pair of equations

9m² – 25n² = 125 and

3m + 5n = 5

Solution

Given that

9m² – 25n² = 125 and

3m + 5n = 5

Step1 – Label the equations as follows

9m² – 25n² = 125  … equ1

3m + 5n = 5         … equ2

From equ1, we have that

9m² – 25n² = 125  … equ1

9m² = (3m)² and 25n² = (5n)²

Thus, 9m² – 25n² becomes

(3m)² – (5n)² and for this, the equation 9m² - 25n² = 125

 becomes

(3m)² – (5n)² = 125

(3m + 5n)(3m – 5n) = 125 (from difference of two squares)

Step2 – Label the factorized equation as equ3

(3m + 5n)(3m – 5n) = 125 … equ3

Note

In equ2, 3m + 5n = 5

Step3

Substitute 5 for 3m + 5n in equ3 as follows

(3m + 5n)(3m – 5n) = 125 … equ3

5(3m – 5n) = 125

Step4 – divide both sides by 5 to have

3m – 5n = 25 (we call this equ4)

Step5 – Add equ2 and equ4 to eliminate n as follows

3m + 5n = 5 … equ2

3m – 5n = 25 … equ4

6m + 0  = 30

6m = 30

m = ³⁰/₆

m = 5

Substitute 5 for m in equ4 to get the value of n as follows

3m – 5n = 25 … equ4

3*5 – 5n = 25

15 – 5n = 25

-5n = 25 – 15

-5n = 10

n = ¹⁰/_-5

n = -2

Thus, m=5, n=-2