To find the value of [tex]\( k \)[/tex] in the given equation:
[tex]\[ \frac{10x + 1}{(x-2)(x+1)} = \frac{k}{x-2} + \frac{3}{x+1}, \][/tex]
we follow these steps:
1. Rewrite the Right-Hand Side with a Common Denominator:
[tex]\[
\frac{k}{x-2} + \frac{3}{x+1}
\][/tex]
To combine these fractions, we need a common denominator. The common denominator for [tex]\((x-2)\)[/tex] and [tex]\((x+1)\)[/tex] is [tex]\((x-2)(x+1)\)[/tex]. Rewrite both fractions with this common denominator:
[tex]\[
\frac{k(x + 1)}{(x-2)(x+1)} + \frac{3(x - 2)}{(x-2)(x+1)}
\][/tex]
2. Combine the Fractions:
[tex]\[
\frac{k(x + 1) + 3(x - 2)}{(x-2)(x+1)}
\][/tex]
Now, the equation becomes:
[tex]\[
\frac{10x + 1}{(x-2)(x+1)} = \frac{k(x + 1) + 3(x - 2)}{(x-2)(x+1)}
\][/tex]
3. Equate the Numerators:
Since the denominators are the same, we can equate the numerators:
[tex]\[
10x + 1 = k(x + 1) + 3(x - 2)
\][/tex]
4. Expand and Simplify:
Expand the right-hand side:
[tex]\[
10x + 1 = kx + k + 3x - 6
\][/tex]
Combine like terms:
[tex]\[
10x + 1 = (k + 3)x + (k - 6)
\][/tex]
5. Match Coefficients:
For the equation to be true for all [tex]\( x \)[/tex], the coefficients of [tex]\( x \)[/tex] and the constant terms on both sides must be equal:
[tex]\[
10 = k + 3
\][/tex]
[tex]\[
1 = k - 6
\][/tex]
6. Solve for [tex]\( k \)[/tex]:
From the first equation:
[tex]\[
k + 3 = 10 \implies k = 10 - 3 \implies k = 7
\][/tex]
From the second equation (as a consistency check):
[tex]\[
k - 6 = 1 \implies k = 1 + 6 \implies k = 7
\][/tex]
Both equations give us [tex]\( k = 7 \)[/tex].
Hence, the value of [tex]\( k \)[/tex] is:
[tex]\[ k = 7 \][/tex]
