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
[tex]x \times y =76[/tex]
substituting the values of x from eq(1)
[tex]( 3+4y) \times y =76[/tex]
[tex]3y + 4y^2 = 76[/tex]
[tex]3y + 4y^2-76= 0[/tex]
[tex] 4y^2 + 3y -76= 0[/tex]
Solving the quadratic equation equation we get
[tex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]
here
a = 4
b= 3
c = -76
susbtituting the above values in the formula
[tex]y=\frac{-3 \pm \sqrt{3^2-4(4)(-76)}}{2(4)}[/tex]
[tex]y=\frac{-3 \pm \sqrt{9- 4(4)(-76)}}{8}[/tex]
[tex]y=\frac{-3 \pm \sqrt{9- (16)(-76)}}{8}[/tex]
[tex]y=\frac{-3 \pm \sqrt{9- (-1216)}}{8}[/tex]
[tex]y=\frac{-3 \pm \sqrt{9 + 1216}}{8}[/tex]
[tex]y=\frac{-3 \pm \sqrt{1225}}{8}[/tex]
[tex]y=\frac{-3 \pm 35}{8}[/tex]
[tex]y=\frac{-3 +35}{8}[/tex] [tex]y=\frac{-3 -35}{8}[/tex]
[tex]y=\frac{32}{8}[/tex] [tex]y=\frac{-38}{8}[/tex]
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
