Simplify (2x + 1)/(x + 5) - (3x - 2)/(x + 4) into a single fraction.

Step-by-step solution

1. 1

   Build the common denominator. The two denominators $x+5$ and $x+4$ share no factor, so the least common denominator is their product $(x+5)(x+4)$. Rewrite each fraction by multiplying top and bottom by whatever the other denominator is:

   $\frac{2x+1}{x+5}=\frac{(2x+1)(x+4)}{(x+5)(x+4)},\qquad \frac{3x-2}{x+4}=\frac{(3x-2)(x+5)}{(x+5)(x+4)}.$

   Note also the restrictions $x\ne-5$ and $x\ne-4$, which carry over to the final answer.

2. 2

   Expand the first numerator. Using FOIL:

   $(2x+1)(x+4)=2x^{2}+8x+x+4=2x^{2}+9x+4.$

   The middle coefficient is $8+1=9$; combining those two like terms before moving on avoids carrying an unwieldy four-term expression.

3. 3

   Expand the second numerator. Likewise,

   $(3x-2)(x+5)=3x^{2}+15x-2x-10=3x^{2}+13x-10.$

   Watch the constant: $(-2)(+5)=-10$, not $+10$.

4. 4

   Subtract, distributing the minus sign over all three terms. This is the step where errors cluster:

   $\left(2x^{2}+9x+4\right)-\left(3x^{2}+13x-10\right)=2x^{2}+9x+4-3x^{2}-13x+10.$

   Every sign inside the second bracket flips, including $-10$ becoming $+10$. Combining like terms:

   $-x^{2}-4x+14.$

5. 5

   Write the result and check numerically. The simplified expression is

   $\frac{-x^{2}-4x+14}{(x+5)(x+4)}=-\frac{x^{2}+4x-14}{(x+5)(x+4)},\qquad x\ne-5,\,-4.$

   The numerator does not factor over the rationals (its discriminant is $16+56=72$), so nothing cancels. Numerical check at $x=2$: the original gives $\tfrac57-\tfrac46=0.7143-0.6667=0.04762$, and the answer gives $\tfrac{-4-8+14}{7\cdot 6}=\tfrac{2}{42}=0.04762$ ✓.