To find the value of [tex]\( a \)[/tex] that satisfies the equation [tex]\[\frac{a}{2x+1} + \frac{1}{x+2} = \frac{4x+5}{2x^2 + 5x + 2},\][/tex]
we can proceed by equating the numerators since the denominators on both sides are equivalent.
First, factorize the denominator on the right-hand side:
[tex]\[2x^2 + 5x + 2\][/tex]
We can rewrite this quadratic expression as:
[tex]\[2x^2 + 5x + 2 = (2x+1)(x+2)\][/tex]
Now our equation is:
[tex]\[\frac{a}{2x+1} + \frac{1}{x+2} = \frac{4x+5}{(2x+1)(x+2)}\][/tex]
Since the denominators are the same, we equate the numerators. Hence,
[tex]\[ a(x+2) + 1(2x+1) = 4x + 5\][/tex]
Now, expand and simplify the left-hand side:
[tex]\[ a(x + 2) + 2x + 1 = ax + 2a + 2x + 1 = (a+2)x + 2a + 1\][/tex]
So we have:
[tex]\[(a+2)x + 2a + 1 = 4x + 5\][/tex]
By comparing the coefficients on both sides of the equation:
1. The coefficients of [tex]\( x \)[/tex]:
[tex]\[ a + 2 = 4\][/tex]
2. The constant terms:
[tex]\[ 2a + 1 = 5\][/tex]
Solve these two equations:
From the first equation:
[tex]\[ a + 2 = 4\][/tex]
[tex]\[ a = 4 - 2\][/tex]
[tex]\[ a = 2\][/tex]
Verify with the second equation:
[tex]\[ 2a + 1 = 5\][/tex]
Substitute [tex]\( a = 2 \)[/tex]:
[tex]\[ 2(2) + 1 = 5\][/tex]
[tex]\[ 4 + 1 = 5\][/tex]
[tex]\[ 5 = 5\][/tex]
Both equations are satisfied with [tex]\( a = 2 \)[/tex].
Thus, the value of [tex]\( a \)[/tex] is [tex]\(\boxed{2}\)[/tex].
