Is there a rigorous approach to solving this, instead of trial and error?

This is a question from a 4th grade math book. The solution in the teacher's guide suggests that one might not be able to model the question easily and the best approach may be to solve by trial and error.

Indeed, by trial and error, the student found 7 to be a correct answer. But is that the only solution?

Harriet bought some crackers for her favorite birds.
She wanted to give each of them the same number of crackers.
If she gave each bird 5 crackers, she would have 3 left.
If she gave each bird 6 crackers, she would be short 4 crackers.
How many favorite birds does she have?


Is there any algebra that would help solve this type of problem instead of solving by trial and error?

• Of course it can be done with algebra, but it can also be done with just clear fourth-grade thinking. After giving each bird 5 crackers, Harriet has 3 remaining. In order to give each bird 6 crackers, she needs to give each exactly one more. She can give each of the three remaining crackers to a bird, but she still needs 4 more, so there must be 3+4 birds altogether. I think this could be a gentle way of motivating the concept of negative numbers. Sep 3, 2015 at 14:54
• I follow the logic up to ..'so there must be 3 + 4 birds'. Why is it 3+4? Sep 3, 2015 at 16:07
• I do understand now. No explanation is necesssary. Sep 3, 2015 at 16:26

If $c$ is the number of crackers, and $b$ is the number of birds, \left.\begin{align*} c&=5b+3\\ c&=6b-4 \end{align*}\right\}\implies5b+3=6b-4\implies 3+4=6b-5b\implies b=7