Hello, I am wondering if I have gone about this system of equations correctly ?
3x+5 = AX+2A+BX+B
using x = 0
5 = 2A+B
using x = 1
8 = 3A + 2B
therefore:
5 = 2A+B
8 = 3A + 2B
cancelling a term:
10 = 4A+2B
8 = 3A + 2B
equals:
A = 2
B= 1
Hello, I am wondering if I have gone about this system of equations correctly ?
3x+5 = AX+2A+BX+B
using x = 0
5 = 2A+B
using x = 1
8 = 3A + 2B
therefore:
5 = 2A+B
8 = 3A + 2B
cancelling a term:
10 = 4A+2B
8 = 3A + 2B
equals:
A = 2
B= 1
Hello, fran1942!
Hello, I am wondering if I have gone about this system of equations correctly ?
3x+5 = AX+2A+BX+B
using x = 0
5 = 2A+B
using x = 1
8 = 3A + 2B
therefore:
5 = 2A+B
8 = 3A + 2B
cancelling a term:
10 = 4A+2B
8 = 3A + 2B
equals:
A = 2
B= 1
You're absolutely correct! . . . Good work!
Here's another approach . . .
We have: .$\displaystyle 3x + 5 \:=\:(A+B)x + (2A+B) $
Two polynomials are equal if their corresponding coefficients are equal.
Hence: .$\displaystyle \begin{Bmatrix}A + B &=& 3 \\ 2A + B &=& 5 \end{Bmatrix}$
which yields the same answers: .$\displaystyle \begin{Bmatrix} A &=& 2 \\ B &=& 1 \end{Bmatrix}$