There are values A and B such that (Bx-11) / (x^2-7x+10) = A/(x-2) - 3/(x-5).

Find A+B

Guest May 2, 2022

#1**0 **

We have the equation: \({{bx-11} \over {x^2-7x+10}} = {a \over {x -2}} - {3 \over {x-5}}\)

We begin by multiplying \(a \over {x-2}\) by \({x - 5} \over {x - 5}\). This gives us \({ax-5a } \over {x^2-7x+10}\)

We can then multiply \({3 \over{x-5}}\) by \({ x-2} \over {x -2}\), which gives us \({3x-6} \over {x^2-7x+10}\)

Note: We don't have to do anything to the other side of the equation because we were multiplying the equation by 1

We now disregard the denominator (because they're the same) and have the equation: \(bx-11 = ax-5a+3x-6\)

We can now combine like terms: \(bx-ax=-5a+5+3x\)

Now, we can factor out the x: \(x(b-a)=-5a+5+3x\)

Dividing by x, we have: \({b-a= {{-5a+5+3x} \over x}}\)

Adding 2a to both sides, we get: \(b+a={{-5a +5 +3x+2a} \over x}\)

We can then simplify further: \(b+a=\color{brown}\boxed{-3a+5+3x \over x}\)

BuilderBoi May 2, 2022

#2**+1 **

Multiply \(x^2 - 7x + 10 = (x -2)(x - 5)\) on both sides:

\(Bx - 11 = A(x - 5) - 3(x - 2)\)

Since this is true for any real number x, we can substitute whatever real number we want.

Substitute x = 5.

\(5B - 11 = -3(5 - 2)\\ 5B - 11 = -9\\ 5B = 2\\ B = \dfrac25\)

Substitute B = 2/5 and x = 2.

\(\dfrac45 - 11 = A(2 - 5)\\ A = \dfrac{17}5\)

MaxWong May 2, 2022