How is (3x+11)/((x-3)(x+2)) = (4/(x-3))-(1/(x+2))?? Missing logic in algebra step..?

Question

How is (3x+11)/((x-3)(x+2)) = (4/(x-3))-(1/(x+2))??

**I don't see where the 4 and 1 came from**

And

(x^2 + x)/(x-1) = (x2+x+2 + (2/(x-1)))??

What am I missing the logic here, cuz I don't see how they got from one to the other?

Thanks!

Answer 1

(3x+11)/(x-3)(x+2)=a/(x-3)+b/(x+2) .The dominator is (X-3)(x+2),then we have: 3x+11=a(x+2)+b(x-3) (*)
In (*) put x=3 then 3*3+11=5a => a=20/5=4
In(*) put x=-2 then -6+11=-5b =>b=5/-5=-1 . OK!

Answer 2

Actually this is a partial fractions problem. So the way you solve it is to set the original equation

(3x+11)/((x-3)(x+2)) = A/(x-3) + B/(x+2).

We then multiply the equations out and we get

A(x+2) + B(x-3) = 3x+11

Now you set x= to a number that will eliminate one of the terms either A or B. So lets elimate A first so x=-2. Plug this in and solve for B.

A(-2+2) + B(-2-3) = 3(-2)+11
A(0) +B(-5) = 5
B=-1

Now do the same thing but eliminate B (x=3)

A(3+2) +B(3-3) = 3(3) +11
A(5) + B(0) = 20
A=4

Now you have A/(x-3) + B/(x+2) or

4/(x-3) - 1/(x+2)

Answer 3

It's easy to get from the RHS to the LHS:
(4/(x-3))-(1/(x+2))
= 4(x+2)/(x-3)(x+2) - (x-3)/(x-3)(x+2) common denominator
= (4x + 8 - x + 3) / (x-3)(x+2)
= (3x + 11) / (x-3)(x+2).

But the other way around (without prior knowledge of what the RHS is supposed to look like)... that's a little more convoluted. basically you're looking for numbers a and b such that
(3x+11)/((x-3)(x+2)) = (a/(x-3)) + (b/(x+2))
This means
3x + 11 = a(x+2) + b(x-3) after you've put everything to the LCD.
3x+11 = (a+b)x + (2a-3b)

The x coefficient and constant term have to be identical so you have two equations from this:
a+b = 3 (x coefficient)
2a-3b = 11 (constant term)

Solving gives you a=4 and b=-1.

Answer 4

You are being asked to solve the equation.
To make it easier to see what is going on, rename the denominators. Let A = (x-3), B = (x+2).
The equation becomes:
(3x+11)/AB = 4/A - 1/B
Multiply the eq. by AB:
3x + 11 = 4B - A
3x + 11 = 4(x+2) - (x-3)
3x + 11 = 4x + 8 - x + 3
11 = 11.

Answer 5

Get the RIGHT side to the Least Common Denominator:
(x-3)(x+2). You get {4(x+2) - 1(x-3)} / {(x-3)(x+2)}; now just simplify the numerator of this fraction.

The x2 on the right side of your second equation looks suspiciously like a typo. Whether the first term on the right is (x times 2) or (x squared) the left side of this equation isn't always equal to the right side. (Just do a quick test with simple whole numbers. Notice that such a simple test in your first equation will work for ALL numbers EXCEPT x=3 or x= -2. You know that division by zero isn't allowed because it provokes inconsistencies.)

Answer 6

Start with the right hand side of the equation, you want to have a common denominator of (x-3)(x-2) for each fraction so that you may add them togther.

So multiply
[4/(x-3)]*[(x+2)/(x+2)]

this is the same as multiplying by 1 so you only need to do this to the one fraction instead of all fractions on both sides of the equation

you get:
(4x+8)/[(x-3)(x+2)]

multiply the other fraction by (x-3)/(x-3)

you get:
(3-x)/[(x-3)(x+2)]

now that these fraction have a common denominator you may add the numerators and have them divided by the common denominator:

(4x + 8 + 3 - x)/[(x-3)(x+2)]

simplify the numerator and you get:

(3x - 11)/[(x-3)(x+2)]

which is equal to the left hand side of the equation and your proof is complete!

Hope that helps.

Answer 7

Multiply out.

ie.

Left hand side:

(3x+11)/((x-3)(x+2)) = (3x+11)/(x^2-3x+2x-6)

= (3x+11)/(x^2-x-6)

Right hand side:

Multiply each term by the denominator of the other ie.

(4/(x-3))-(1/(x+2)) = (4(x+2)-(x-3)/((x-3)(x+2))

Multiply out

= (4x+8-x+3)/(x^2+2x-3x-6)

= (3x+11)/(x^2-x-6)

Voila!

Can't be bothered to do the 2nd ;-)

Answer 8

(3x+11)/((x-3)(x+2)) = (4(x + 2) - (x - 3))/((x-3)(x+2))
= (4/(x-3))-(1/(x+2))

(x^2 + x)/(x-1) = (x2+x+2 + (2/(x-1))) : it must be something wrong here