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

 May 2, 2022
 #1
avatar+2440 
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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: \(b+a=\color{brown}\boxed{-3a+5+3x \over x}\)

 May 2, 2022
 #2
avatar+9459 
+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.

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

 May 2, 2022

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