Even easy problems like this one can show up on the SAT.

Simple factorization gives us : (a-b)x + (b-a)y = (a-b)(x-y) =3*5 =15

a-b =3 therefor b-a=-3, by substituting (a-b)x+(b-a)y=3x-3y , by taking 3 as a common factor : 3x-3y=3(x-y)=3*(5) = 15

$\Rightarrow (a-b)x+(b-a)y\quad =\quad x(a-b)+y(b-a)\\$

Since, $b-a=-(a-b)$

$\\ \Rightarrow x(a-b)-y(a-b)\\ \Rightarrow (a-b)(x-y)\\ \Rightarrow 3\times 5\\ \Rightarrow 15$

$a-b=3$ and $b-a=-3$

Then

$3x-3y=3(x-y)=3(5)=15$

lol, i just guess... a-b=3 therefore b-a=(-3) i don't really know the formula how to solve the value of x and y so i just replace them with 8-3=5 (from x-y=5) then...

3(x)-3(y)

3(8)-3(3)= 24-9= 15

am I just that lucky to get the answer? or am I somehow correct with my solution?

a-b=3,x-y=5, (a-b)*(x-y)=15, ax-ay-bx+by=15, (a-b)x+(b-a)y=15.

First you need to find the value of ( $b-a$ )

This can be done easily by factorising it

$b-a = -a + b = -(a-b)$

Since we already know that

$a-b = 3$

We get $b-a = -(3) = -3$

These two equations can be substituted into the question to get

$(a-b)x+(b-a)y = (3)x+(-3)y = 3x-3y$

Factorising this gives

$3(x-y)$

Substituting $x-y = 5$ into the question gives

$3(x-y) = 3(5) = 15$

So the answer is $15$