idk what topic this is problem supposed to be

homeschool girl

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Suppose [MATH]a[/MATH] and [MATH]b[/MATH] are integers. How many solutions are there to the equation [MATH]ab = 2a + 3b?[/MATH]
tried

[MATH]ab = 2a + 3b[/MATH]
[MATH]ab -2a - 3b=0[/MATH]
then I added 5 to both sides and factored

[MATH]ab -2a - 3b+5=5[/MATH]
[MATH](a-3)(b -2)=5[/MATH]
then I got (a,b)=(8,3),(4,7) but when I plug it in it doesn't work.
 
That's not how you do this.

\(\displaystyle ab=2a+3b\\

ab-3b = 2a\\

b(a-3) = 2a\\

b = \dfrac{2a}{a-3},~a \neq 3\\
\)

Now the question is how many pairs \(\displaystyle (a,b)\) produce integers for \(\displaystyle \left(a, \dfrac{2a}{a-3}\right)\)

and indeed there are only 8 of them. I leave you to discover what the 8 pairs are.

Hint: [MATH]-3 \leq a \leq 9[/MATH]
 
Suppose [MATH]a[/MATH] and [MATH]b[/MATH] are integers. How many solutions are there to the equation [MATH]ab = 2a + 3b?[/MATH]
tried

[MATH]ab = 2a + 3b[/MATH]
[MATH]ab -2a - 3b=0[/MATH]
then I added 5 to both sides and factored

[MATH]ab -2a - 3b+5=5[/MATH]
[MATH](a-3)(b -2)=5[/MATH]

then I got (a,b)=(8,3),(4,7) but when I plug it in it doesn't work.
(a-3)(b-2) = ab-2a -3b + 6 not + 5.
Why add 5 to both sides?
 
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That's not how you do this.

\(\displaystyle ab=2a+3b\\

ab-3b = 2a\\

b(a-3) = 2a\\

b = \dfrac{2a}{a-3},~a \neq 3\\
\)

Now the question is how many pairs \(\displaystyle (a,b)\) produce integers for \(\displaystyle \left(a, \dfrac{2a}{a-3}\right)\)

and indeed there are only 8 of them. I leave you to discover what the 8 pairs are.

Hint: [MATH]-3 \leq a \leq 9[/MATH]
you gave me the answer but didn't tell me how to do the problem...
 
you gave me the answer but didn't tell me how to do the problem...
You want to find all integer values for a such that b= 2a/(a-3) is an integer.

I will give you a hint. Do the long division for 2a/(a-3) and pick values for a so the remainder is an integer.

You can also think of \(\displaystyle \dfrac{2a}{a-3}\ \ \ as\ \dfrac{2(a-3) + 6}{a-3} = ...\)
 
Suppose [MATH]a[/MATH] and [MATH]b[/MATH] are integers. How many solutions are there to the equation [MATH]ab = 2a + 3b?[/MATH]
tried

[MATH]ab = 2a + 3b[/MATH]
[MATH]ab -2a - 3b=0[/MATH]
then I added 5 to both sides and factored

[MATH]ab -2a - 3b+5=5[/MATH]
[MATH](a-3)(b -2)=5[/MATH]
then I got (a,b)=(8,3),(4,7) but when I plug it in it doesn't work.
oh I see it now I should have added 6 to both sides
Yes. Try again using 5 instead of 6 (and factoring correctly), and show your work to get your answer. I think you have a good method (clearly there are others!) but you have to carry it out correctly!

And don't forget to include negative numbers ...
 
Suppose [MATH]a[/MATH] and [MATH]b[/MATH] are integers. How many solutions are there to the equation [MATH]ab = 2a + 3b?[/MATH]
tried

[MATH]ab = 2a + 3b[/MATH]
[MATH]ab -2a - 3b=0[/MATH]
then I added 5 to both sides and factored

[MATH]ab -2a - 3b+5=5[/MATH]
[MATH](a-3)(b -2)=5[/MATH]
then I got (a,b)=(8,3),(4,7) but when I plug it in it doesn't work.
This elaborates on romsek's and jomo's answers.

Did you notice that you had two unknowns and only one equation?

Thus, if the equation is valid, an infinite number of pairs of real numbers solve the equation. But not all real numbers solve the equation. So let's first see how we can limit the possible real numbers that work.

[MATH]ab = 2a - 3b \implies ab + 3b = 2a \implies b(a + 3) = 2a.[/MATH]
[MATH]\text {If } a = -\ 3 \implies b(-\ 3 + 3) = 2 * (-\ 3) \implies b * 0 = -\ 6 \implies 0 = -\ 6.[/MATH]
That's absurd so we can conclude that none of the pairs that solve this equation include any where a = minus 3.

[MATH]\therefore b(a + 3) = 2a \implies b = \dfrac{2a}{a + 3} = \dfrac{2a + 6 - 6}{a + 3} = \dfrac{2(a + 3) - 6}{a + 3} = 2 - \dfrac{6}{a + 3}.[/MATH]
How in the world does that help you? Well, you do not have a second equation, but you are looking only for numbers that are integers. There may be a finite number of pairs of integers that work. Let's look at that algebraic fraction because fractions and integers do not usually play nicely together.

If 6/(a + 3) is not an integer, then b is not an integer. That means that (a + 3) can have what possible values?

By the way, this one common technique for solving equations where answers must be restricted to integers, so-called Diophantine equations, which is a more advanced field than elementary algebra.
 
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Yes. Try again using 6 instead of 5 (and factoring correctly), and show your work to get your answer.
Oops. This is what I meant.

6 will work nicely; and you may notice a strong relationship between your work with your method, and the other method you've been shown (which is also worth learning).
 
My previous answer started with an incorrect statement of the problem. I have corrected it below.

Find all integer solutions if any for

[MATH]ab = 2a + 3b.[/MATH]
[MATH]ab = 2a + 3b \implies ab - 3b = 2a \implies b(a - 3) = 2a.[/MATH]
[MATH]\text {If } a = 3 \implies b(3 - 3) = 2 * 3 \implies b * 0 = 6 \implies 0 = 6.[/MATH]
That's absurd so we can conclude that none of the pairs that solve this equation include any where a = 3. Thus we can divide by a-3.

[MATH]\therefore b(a - 3) = 2a \implies b = \dfrac{2a}{a - 3} = \dfrac{2a - 6 + 6}{a - 3} = \dfrac{2(a - 3) + 6}{a - 3} = 2 + \dfrac{6}{a - 3}.[/MATH]
How in the world does that help you? Well, you do not have a second equation, but you are looking only for numbers that are integers. There may be a finite number of pairs of integers that work. Let's look at that algebraic fraction because fractions and integers do not usually play nicely together.

If 6/(a - 3) is not an integer, then b is not an integer. That means that (a - 3) can have what possible values?

By the way, this is one common technique for solving equations where answers must be restricted to integers, so-called Diophantine equations, which is a more advanced field than elementary algebra.
 
BIG hint as you have not replied.
6/1 is an integer. 6/2 is an integer. 6/3 is an integer. 6/6 is an integer. 6/-1, 6/-2 6/-3 and 6/-6 are also integers.
So what values can a-3 equal? What values can then a be?
 
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