Let A and B be real numbers such that A/(x-5)+B(x+1)=(-3x^2+12x+22)/(x-5)
What is A+B
see my suggestions of yesterday
https://www.jiskha.com/questions/1799358/Let-A-and-B-be-real-numbers-such-that-frac-A-x-5-B-x-1-frac-3x-2-12x-22-x-5#1884830
To find the value of A + B, we need to equate the coefficients of like terms on both sides of the equation.
The given equation is:
A/(x-5) + B(x+1) = (-3x^2 + 12x + 22)/(x-5)
First, let's simplify the right side of the equation by dividing each term by (x-5):
(-3x^2 + 12x + 22)/(x-5) = (-(3x^2 - 12x - 22))/(x-5)
Now, let's put the equation in a common denominator, which is (x-5):
A(x+1) + B(x-5) = -(3x^2 - 12x - 22)/(x-5)
Multiplying both sides of the equation by (x-5) to eliminate the denominator, we get:
A(x+1)(x-5) + B(x-5)(x-5) = -(3x^2 - 12x - 22)
Expanding and simplifying both sides, we get:
A(x^2 - 4x - 5) + B(x^2 - 10x + 25) = -3x^2 + 12x + 22
Distributing, we have:
Ax^2 - 4Ax - 5A + Bx^2 - 10Bx + 25B = -3x^2 + 12x + 22
Now, let's group the like terms:
(A + B)x^2 + (-4A - 10B)x + (25B - 5A) = -3x^2 + 12x + 22
Equate the coefficients of the like terms on both sides:
A + B = -3 ...(1)
-4A - 10B = 12 ...(2)
25B - 5A = 22 ...(3)
Now, we have a system of linear equations with three equations (1), (2), and (3) and two unknowns (A and B). We can solve this system of equations to find the values of A and B.
To solve the system of equations, we can use the method of substitution or elimination. Solving the equations, we find that A = -5 and B = 2.
Therefore, A + B = -5 + 2 = -3.
To find the values of A and B that satisfy the given equation, we can start by multiplying both sides of the equation by (x-5) to eliminate the fraction:
A + B(x-5) = -3x^2 + 12x + 22
Next, we can distribute B to simplify the equation:
A + Bx - 5B = -3x^2 + 12x + 22
Since both sides of the equation are polynomials, the equation must hold for all values of x. This means that the coefficients of corresponding powers of x on both sides of the equation must be equal.
Comparing the coefficients of x^2, we have:
-3 = 0
This implies that there is no x^2 term on the right-hand side of the equation, which is contradictory. Therefore, there are no real numbers A and B that satisfy the given equation.
Hence, A + B is undefined in this case.