Question 277166
find the value of a, b, and c so that i can decompose it to partial functions
{{{(3x -13)/(6x^2-x-12)}}} = {{{(3x - 13)/((2x-3)(3x+4))}}}
:
Write the given expression as a sum using A & B
{{{(3x - 13)/((2x-3)(3x+4))}}} = {{{A/(2x-3)}}} + {{{B/(3x+4)}}}
:
Multiply by the LCD, (2x-3)(3x+4)
(2x-3)(3x+4){{{(3x - 13)/((2x-3)(3x+4))}}} = (2x-3)(3x+4){{{A/(2x-3)}}} + (2x-3)(3x+4){{{B/(3x+4)}}}
results
3x - 13 = A(3x+4) + B(2x-3)
:
3x - 13 = 3xA + 4A + 2xB - 3B
:
3x - 13 - (3A + 2B)x + 4A - 3B
Coefficients of x are equal
3A + 2B = 3
Constants are equal
4A - 3B = -13
;
Solve for two unknowns using elimination, multiply 1st eq by 3, 2nd eq by 2
9A + 6B = 9
8A - 6B =-26
----------------adding eliminates B, find A
17A = -17
A = -1
:
Find B
3A + 2B = 3
3(-1) + 2B = 3
-3 + 2B = 3
2B = 3 + 3
2B = 6
B = 3
:
We have
{{{(3x - 13)/(6x^2-x-12)}}} = {{{-1/(2x-3)}}} + {{{3/(3x+4)}}}
:
You should check this by adding the two fractions on the right to see if they equal the expression on the left