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.