Question #72cb2

2 Answers
Nov 6, 2017

The sum of the two integers is #19 +4= 23#

Explanation:

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#

The sum of the two integers is 23.

Explanation:

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#.