Question #72cb2

One positive integer is 3 greater than 4 times another positive integer.

Let the second positive integer be `x`.

So, the first positive integer will be :

`3+4x`

Now product of theses two is given to be 76, so:

`=>(3+4x) xx x= 76`

`=> 3x +4x^2 = 76`

`=> 4x^2 +3x - 76 = 0`

Solving this quadratic equation:

We need to find two such numbers whose sum is equal to the coefficient of the middle term (i.e. `3)` and the ir product is equal to the product of the coefficients of first and last term (i.e.`4 xx -76 = -304)` :

Two such numbers are : 19 and -16

`=> 4x^2 -16x +19x -76 = 0`

`=>4x(x- 4) +19(x- 4) =0`

`=> (4x+19)(x- 4) =0`

`=> 4x+19 =0 or x- 4=0`

`therefore x= -19/4 or x=4`

But as mentioned in the question statement, the integers are positive . so we will consider `x=4`

So. the second integer is `x=4` and the first integer is `3+4x= 3+ 16 =19`

Cross check `=>`product of the two integers `= 19 xx 4 =76`

And, the sum 4 and 19 = 23

Answer : Sum of the two integers is `19 +4= 23`

This can be written as `y = 4x + 3`.

If `xy = 76`, you can rearrange the equation to solve for either x or y. In my example, I will use x.

Divide both sides by x, and you will get `y=76/x`.

Now, you can substitute `76/x` for y in the original equation.

`76/x = 4x + 3`

Solve for x.

`x = 4`.

Plug 4 back into the original equation.

`y = 4(4) + 3`.
`y = 16 + 3`.
`y = 19`.

`x + y = ?`
`4 + 19 = 23`.