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# Algebra

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Find the number of ordered pairs $(a,b)$ of integers such that
\frac{a + 2}{a + 1} = \frac{b}{8}.

Jun 1, 2024

#1
+806
+1

First, let's cross multiply and simplfy. We get

$$8 (a + 2) = b ( a + 1) \\ 8a + 16 = ab + b$$

Now, in order to progress, we have to be able to factor it,

$$ab + b - 8a = 16 \\ ab + b - 8a - 8 = 16 - 8 \\ (a + 1) ( b - 8) = 8$$

That means the number of order pairs there are is the number of factors 8 has, including negatives.

There are 8 factors, meaning the answer is 8.

Here are the ordered pairs that work.

$$(3, 10), (1, 12), (0, 16), (7, 9), (-5, 6), (-3, 4), (-2, 0), (-9, 7)$$

Thanks! :)

Jun 1, 2024

#1
+806
+1

First, let's cross multiply and simplfy. We get

$$8 (a + 2) = b ( a + 1) \\ 8a + 16 = ab + b$$

Now, in order to progress, we have to be able to factor it,

$$ab + b - 8a = 16 \\ ab + b - 8a - 8 = 16 - 8 \\ (a + 1) ( b - 8) = 8$$

That means the number of order pairs there are is the number of factors 8 has, including negatives.

There are 8 factors, meaning the answer is 8.

Here are the ordered pairs that work.

$$(3, 10), (1, 12), (0, 16), (7, 9), (-5, 6), (-3, 4), (-2, 0), (-9, 7)$$

Thanks! :)

NotThatSmart Jun 1, 2024
#2
+850
0

Find the number of ordered pairs (a,b) of integers such that
\frac{a + 2}{a + 1} = \frac{b}{8}.

I don't know how to read that shorthand but I'm going to assume it means

(a + 2)          b

–––––––  =  –––

(a + 1)           8

I don't know how to solve this, so, as my geometry teacher used to say

"Go as far as you can, then see how far you can go."

Let's try some numbers and see if we can find a pattern

When a is 0, we get 2 / 1  so in order to preserve equality b = 16

1             3 / 2                                                          12

2             4 / 3  won't work, no integer is 1/3 of 8

3             5 / 4                                                          10

4             6 / 5  won't work, no integer is 1/5 of 8

5             7 / 6  won't work

6             8 / 7  won't work

7             9 / 8                                                            9

I think I can see intuitively that no a > 7 will result in an integer b.

Let's go the other direction from zero and see what happens.

When a is –1, we get   1 / 0  zero in denominator is a no-no

–2, we get   0 / –1  so                                              b = 0

–3, we get –1 / –2  so                                                    4

–4, we get –2 / –3  won't work

–5, we get –3 / –4  so                                                    6

–6, we get –4 / –5  won't work

–7, we get –5 / –6  won't work

–8, we get –6 / –7  won't work

–9, we get –7 / –8  so                                                    7

It looks like the equality stops working when (a+1)

is less than negative 8 and greater than positive 8.

I can't see a general pattern – other than, in the ones that work the denominator

is counting up in base 2 but I don't know if that means anything.  If if this were my

homework, I'd go out on that limb and say that the answer is  8 ordered pairs.

.

Jun 1, 2024