Largest number of pairs that can be added while keeping the population at least 60% male

I'm doing problems from the AoPS Algebra Beginner's book.

There's this problem that states the following,

At her ranch, Georgia starts an animal shelter to save dogs. After the first three days, she has 34 male dogs, and 13 female dogs. She decides to only accept dogs in male-female pairs from then on. In this problem, we have to find the largest number of pairs of dogs she can accept to her shelter while still having 60% of dogs be male.

The question sort of sets some guidelines as to how things should look, easing the process.

Let d be the numbers of pairs of dogs she accepts. How many male dogs will she have total? Female dogs? Total dogs?

Write an inequality that means the percentage of dogs at her shelter that are male is at least 60%.

Solve the inequality to answer the question.

What I did:

I know that both sides of the dogs, males and females, receive the same # of new dogs.

I write an equation, (not an inequality though):

$$3((34 + 2x) / 5) = 2((13 + 2x) / 5)$$

From here on, I solve for x. I get x = -38. I know we can't have negative pairs and we're only bringing in more dogs, and not taking any away. I add 38, and try to see what percentage of dogs are male, and I get fairly close, getting to about 62.5% I think. Then, I divide 38 by 2, getting 19, and see what else I can get. I try to find a value between the 38 and lesser 19 so as to find a value which will give me a male dog percentage of 60%. I get about 57.5% with 19 plugged in, which is when I see that hey, the differences are the same. What I then do is find the average of 19 + 38, which is 57.5. I round that to 58, and I get the right answer.

I see that perhaps my solution wasn't really orthodox or anything, but with a little bit of trial and error and a bit of my creative thinking, I guess I managed to solve the problem. What worries me though is this is the first problem in the "questions" section of the inequalities section of AoPS, so I'm definitely struggling to deal with the concepts.

If someone could show how to do this question with a more standard approach, and provide their thinking and how they adjust their equation/inequality according to the information that is provided in the text, that would be great.

Thank you!

If there are $d$ pairs of dogs admitted after the initial 34 male and 13 female dogs then the total number of male dogs will be $34 + d$.

The total number of dogs will be $34+13+2d= 47+2d$

We want the number of male dogs to be $34 + d \geq .6(47+2d)$ This is because of the word "still". Currently the percentage is above 60% so I assume they want this situation to continue.

Now you can solve for $d$.

Let $d$ be the number of pairs accepted. Then there are $m(d) = d+34$ male dogs and $f(d) = d+13$ female dogs.

The constraint that at least 60% of the dogs be male corresponds to $m(d) \ge 0.6 (f(d)+m(d))$. Simplifying this expression gives $d \le 29$.

We want to solve $\max\{ d | d \le 29 \}$, which gives $d=29$.