the product of five less than three times a certain number and one less than four times the number is 7. find the number

Question

QUADRATIC EQUATIONS

Answer 1

If the number is x, (3x-5)(4x-1)=7. Without making this into a quadratic, we know the factors of 7 are 1 and 7, so either 4x-1=1 or 7. If 4x-1=7, 4x=8 and x=2. 3x-5=6-5=1, therefore x=2 is a solution.

Now quadratically: 12x^2-23x+5-7=12x^2-23x-2=0=(x-2)(12x+1). There are two solutions, x=2 and x=-1/12.

Answer 2

Let x be the required number 

3 times of x = 3x 

4 times of x = 4x 

​​​​​​According to condition 

(3x-5)(4x-1)=7

12x² -3x -20x +5 -7=0

12x² -23x -2 =0 

12x² -24x +x -2 =0 

12x(x-2) +1(x-2) =0 

(X-2) (12x+1)

​​​​so, 

Either.           Or 

​​​​x-2=0.            12x+1=0

X=2.                12x=-1

                         X= -1/12 

Hence 2 and -1/12 are required numbers.