Question

*Will fan and medal*
Model each problem as an equation, and then match it to its solution.

1) When the sum of the squares of two positive integers is 185. If one integer is 3 less than the other, find the larger integer.

2) The numerator of a fraction is 1 more than twice its denominator. If 4 is added to both the numerator and the denominator, the fraction reduces to 5/3. Find the denominator.

3) The difference of a positive integer and its inverse is 15/4. Find the integer.

*Solutions*

A) 11

B) 4

C) 5

Answer 1

@netlopes1 help me por favor

Answer 2

3) The difference of a positive integer and its inverse is 15/4. Find the integer.
\(a-\dfrac{1}{a}=\dfrac{15}{4}\iff \dfrac{a}{1}-\dfrac{1}{a}=\dfrac{15}{4}\iff \dfrac{a^2}{a}-\dfrac{1}{a}=\dfrac{15}{4}\\\implies \dfrac{a^2-1}{a}= \dfrac{15}{4}\)

from there you have 2 things you can solve for \(a^2-1=15\)
or \(a=4\)

one of them is obvious

Answer 3

The first one is a= 4,-4?

Answer 4

c is 4, just 4

Answer 5

so 3) is B)?

Answer 6

1) When the sum of the squares of two positive integers is 185. If one integer is 3 less than the other, find the larger integer.
\(a^2+b^2=185
\\a=b-3
\\(b-3)^2+b^2=185
\\(b-3)(b-3)=b^2-6b+9
\\b^2-6b+9+b^2=185
\\2b^2-6b+9=185
\\2b^2-6b\cancel{+9-9}=185-9
\\2b^2-6b=176
\\2(b^2-3b)=176
\\b^2-3b-88=0
\\(b+8)(b-11)=0
\\b=-8,11\)

Answer 7

since we want only positive integers we can discard -8

Answer 8

ok I see

Answer 9

Thanks so much I appreciate it a lot, I got it now! :)

Answer 10

oh there's another one

Answer 11

I have no idea how to do it lol