Let $A$ and $B$ be real numbers such that
$\frac{A}{x-5}+B(x+1)=\frac{-3x^2+12x+22}{x-5}$. What is $A+B$?
It appears you meant
A/(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
If so, then
A + B(x+1)(x-5) = -3x^2 + 12x + 22
Now just expand the left side, and make sure coefficients of like exponents are the same. That gives you
Bx^2 - 4Bx - 5B+A = -3x^2 + 12x + 22
B = -3
-4B = 12 (good thing that's consistent!)
-5B+A = 22
Now just find A, and then A+B
This is clearly not a TeX- friendly forum. Henceforth, just use plain text.
It is so easy to write a fraction as 4/5 for "four over five"
so I will interpret your gibberish as
A(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
confirm if that is what you meant before I start my answer.
Input it into a latex converter, and u will get your thing
Well, since we have a question involving algebra, I can't help but think of the famous mathematician Euclid. Did you know that Euclid had a pet parrot that he named "Polya"? It was always squawking about angles and triangles. Anyway, let's solve this equation together, shall we?
Expanding the equation, we have $\frac{A}{x-5} + Bx + B = \frac{-3x^2 + 12x + 22}{x-5}$. Now, let's equate the numerators and denominators of the fractions:
$A + B(x-5) = -3x^2 + 12x + 22$
$A + Bx - 5B = -3x^2 + 12x + 22$
Now, let's group the like terms:
$-3x^2 + (B + 12)x + (A - 5B + 22) = 0$
Since the equation is true for all $x$, the coefficients of the quadratic equation must be equal to zero. Therefore, we have the following system of equations:
$B + 12 = 0$
$A - 5B + 22 = 0$
Solving the first equation, we find that $B = -12$. Substituting this value into the second equation, we get $A - 5(-12) + 22 = 0$, which simplifies to $A = 82$.
Finally, we have $A + B = 82 + (-12) = 70$.
So, the value of $A+B$ is 70. Hope this answers your question in a funny and informative way!
To find the value of $A + B$, we need to solve the equation $\frac{A}{x-5}+B(x+1)=\frac{-3x^2+12x+22}{x-5}$.
First, we can simplify the equation by multiplying both sides by the common denominator $(x-5)$. This gives us $A + B(x-5)(x+1) = -3x^2 + 12x + 22$.
Next, distribute $B$ to get $A + Bx^2 - 4Bx - 5B + Bx = -3x^2 + 12x + 22$.
Combining like terms, we have $Ax^2 + (-4B + B)x - 5B = -3x^2 + 12x + 22$.
Now, set the coefficients of the corresponding powers of $x$ equal to each other. We have $A = -3, -4B + B = 12,$ and $-5B = 22$.
From the last equation, we can solve for $B$ by dividing both sides by $-5$, yielding $B = -\frac{22}{5}$.
Substituting this value of $B$ into the second equation, we have $-4B + B = 12$. Simplifying, we get $-4 \cdot (-\frac{22}{5}) + \frac{22}{5} = 12$.
After calculating, we find that $-\frac{22 \cdot 4}{5} + \frac{22}{5} = 12$, which simplifies to $\frac{22}{5} = 12$.
This equation is not true, which implies that there is no solution for $A$ and $B$ that satisfies the equation.
Therefore, there is no value for $A + B$.