Example # 3
Now, let us consider the following equation :
 7x + 8     ₌      5x - 7
 2x + 7               10x - 6

N₁ + N₂  =  (7x + 8) + (5x - 7)  =  12x + 1
and   D₁ + D₂  =  (2x + 7) + (10x - 6)  =  12x + 1
Thus, since here we find : N₁ + N₂  =  D₁ + D_{2 ,} we just equate the sum = 0.
12x + 1 = 0  →  x = -1/12

Now, by observation, here we notice that on cross-multiplication, the coefficient of x² on both sides are not
equal and hence does not cancel out. In such a case, therefore we will have a quadratic equation to deal with,
which means that there are two values of x.
To obtain the second value of x, we check if the difference between the numerator and denominator on both
sides are equal. Thus, if N₁ - D₁ = N₂ - D₂, then we equate the difference to zero to obtain the second value
of x.

N₁ - D₁ = (7x + 8) - (2x + 7) = 5x + 1
and N₂ - D₂ = (10x - 6) - (5x - 7) = 5x + 1
Thus, since here we find : N₁ - D₁ = N₂ - D₂
We just equate this difference = 0
5x + 1 = 0  →  x = -1/5

Thus, the two values of x are :
x₁ = -1/2
x₂ = -1/5