Question

One positive integer is 3 greater than 4 times another positive integer. If the product of the two integers is 76, then what is the sum of the two integers?
14
23
21
19

Answer 1

Answer:

The sum of the two integers is 23

Step-by-step explanation:

Let one integer be x and the other integer be y

Then according to the statement "One positive integer is 3 greater than 4 times another positive integer.."

x be the integer that is One positive integer is 3 greater than 4 times another positive integer.

Then

x= 3+4y----------------------------------(1)

Product of the two integer is 76, this can  written as

$x \times y =76$

substituting the values of x from eq(1)

$( 3+4y) \times y =76$

$3y + 4y^2 = 76$

$3y + 4y^2-76= 0$

$4y^2 + 3y -76= 0$

Solving the quadratic equation equation we get

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

here

a = 4

b= 3

c = -76

susbtituting the above values in the formula

$y=\frac{-3 \pm \sqrt{3^2-4(4)(-76)}}{2(4)}$

$y=\frac{-3 \pm \sqrt{9- 4(4)(-76)}}{8}$

$y=\frac{-3 \pm \sqrt{9- (16)(-76)}}{8}$

$y=\frac{-3 \pm \sqrt{9- (-1216)}}{8}$

$y=\frac{-3 \pm \sqrt{9 + 1216}}{8}$

$y=\frac{-3 \pm \sqrt{1225}}{8}$

$y=\frac{-3 \pm 35}{8}$

$y=\frac{-3 +35}{8}$                       $y=\frac{-3 -35}{8}$    

$y=\frac{32}{8}$                             $y=\frac{-38}{8}$

y= 4                                                           y =  −4.75

Since in the  question it is given that it is a positive integer

so  y = 4

substituting y=4 in eq (1) we get,

x= 3+4(4)

x= 3+16

x= 19

The sum of the two integers

=> x + y

=> 19+4

=>23