How many pairs of integers $(b,c)$ satisfy the equation
\[\frac{b + 7}{b + 4} = \frac{c}{9}?\]
Don't knock BASIC. I knew a girl in college who was using it to program a gyroscopically stable table in orbit in 1972.
Well, let's first cross multiply and simplify the equation. We have $(b + 7)(9) = (b + 4)(c)$.
Expanding both sides, we get $9b + 63 = cb + 4c$. Rearranging terms, we have $cb - 9b + 4c - 63 = 0$.
Now, let's try to apply some comedy to lighten the mood.
This equation looks a bit complex, but hey, isn't life complex too? I mean, some people might argue, "I'm not arguing, I'm simply explaining why I'm right!" So maybe this equation is just trying to explain why it's right...or why we're wrong? It's hard to say.
Now, let's focus back on the problem. To make things less complicated, let's use Simon's Favorite Factoring Trick.
We notice that $cb - 9b + 4c - 63$ looks a bit lonely, so let's group the terms and rearrange:
$(cb + 4c) - (9b + 63) = 0$.
Ah, much better! Now the equation has some friends to keep it company.
Let's try a little trickery to factor this expression. We want to find two numbers that multiply to give $(cb + 4c)$ and subtract to give $(9b + 63)$.
Hmm, let me think...oh, I know! We can combine the b's and c's to make things easier. How about we write them as $(c - 9)(b + 4)$?
And ta-da! Our equation becomes $(c - 9)(b + 4) = 0$.
So, we have two cases to consider. Either $(c - 9) = 0$ or $(b + 4) = 0$.
In the first case, we have $c = 9$.
In the second case, we have $b = -4$.
Therefore, we have two pairs of integers $(b,c)$ that satisfy the equation: $\boxed{(-4, 9)}$.
But remember, math can be funny too! Just like a comedian once said, "Why was the math book sad? Because it had too many problems!" So let's just solve one problem at a time and find joy in the process.
To solve the equation, we start by cross-multiplying.
This gives us $(b + 7) \cdot 9 = c \cdot (b + 4)$.
Expanding both sides, we have $9b + 63 = bc + 4c$.
Rearranging the terms, we get $bc - 9b - 4c + 63 = 0$.
We now attempt to factor the left side of the equation.
We can rewrite $bc - 9b - 4c + 63$ as $(b-4)(c-9) - 4$.
Setting this equal to zero, we have $(b-4)(c-9) - 4 = 0$.
Adding 4 to both sides of the equation, we get $(b-4)(c-9) = 4$.
The pairs of integers that satisfy this equation are $(b-4,c-9) = (1, 4), (2, 2), (-1, -4), (-2, -2)$.
Adding 4 to both sides of each pair gives $(b,c) = (5,13), (6,11), (3,5), (2,7)$.
Therefore, there are $\boxed{4}$ pairs of integers $(b,c)$ that satisfy the equation.
To begin, let's cross-multiply to get rid of the fractions. We have:
\[9(b + 7) = (b + 4)c.\]
Expanding both sides, we obtain:
\[9b + 63 = bc + 4c.\]
Rearranging terms, we have:
\[bc - 9b - 4c + 63 = 0.\]
Now, we apply Simon's Favorite Factoring Trick. We add 1 to both sides to obtain:
\[bc - 9b - 4c + 64 = 1.\]
Rearranging the terms once again, we have:
\[bc - 9b - 4c + 16 \cdot 4 = 1.\]
Now, we notice that the left-hand side of the equation can be factored as:
\[(b - 4)(c - 9) = 1.\]
Since 1 is the only positive integer that has no positive factors other than 1 and itself, the equation $(b - 4)(c - 9) = 1$ only has two possible pairs of values for $(b, c)$: $(b - 4, c - 9) = (1, 1)$ or $(b - 4, c - 9) = (-1, -1)$. Solving for $b$ and $c$ in each case, we obtain the pairs $(b, c) = \boxed{(5, 10), (3, 8)}$.
your notation clearly did not show up as you intended.
I will guess you meant:
(b+7)/(b+4) = c/9
c = 9(b+7)/(b+4)
I have this very old fashioned computer language called BASIC.
I ran a stupid little program for b = -5000 to b = 5000 and it found the following pairs (b,c)
>list
10 for b = -5000 to 5000
20 c = 9*(b+7)/(b+4)
30 if c = int(c) then print b,c
40 next b
>run
-31 8
-13 6
-7 0
-5 -18
-3 36
-1 18
5 12
23 10
So I found 8 of them