Question

the least common denominator is:
(x - 4)(x + 1)(x - 2)
what is the numerator in the sum?
2(x - 2) + 3(x + 1)
2(x - 2)(x - 4) + 3(x + 1)(x - 4)
2(x + 1)(x - 2)(x - 4) + 3(x + 1)(x - 2)(x - 4)

Explanation:

Step1: Recall LCD for fractions

When adding fractions, we use the least common denominator (LCD). Here, LCD is \((x - 4)(x + 1)(x - 2)\).

Step2: Adjust numerators for LCD

Suppose the fractions are \(\frac{2}{(x + 1)(x - 2)}\) and \(\frac{3}{(x - 4)(x + 1)}\). To get LCD, first fraction needs to multiply numerator and denominator by \((x - 4)\), so numerator becomes \(2(x - 4)\)? Wait, no, wait the LCD given is \((x - 4)(x + 1)(x - 2)\). Wait, the first term's denominator is \((x + 1)(x - 2)\), so to get LCD, multiply numerator and denominator by \((x - 4)\): numerator \(2(x - 4)\)? No, wait the options show \(2(x - 2)+3(x + 1)\)? Wait, maybe the original fractions are \(\frac{2}{(x + 1)(x - 4)}\) and \(\frac{3}{(x - 2)(x - 4)}\)? Wait, no, the LCD is \((x - 4)(x + 1)(x - 2)\). Wait, let's re-express: if we have two fractions, say \(\frac{2}{(x + 1)(x - 2)}\) and \(\frac{3}{(x - 4)(x + 1)}\). Then, to add them, we need to make denominators equal to LCD \((x - 4)(x + 1)(x - 2)\). So first fraction: multiply numerator and denominator by \((x - 4)\), so numerator is \(2(x - 4)\)? No, that's not matching. Wait, the first option is \(2(x - 2)+3(x + 1)\). Wait, maybe the original fractions are \(\frac{2}{(x + 1)(x - 4)}\) and \(\frac{3}{(x - 2)(x - 4)}\), but no. Wait, the LCD is \((x - 4)(x + 1)(x - 2)\). Wait, maybe the fractions are \(\frac{2}{(x + 1)(x - 2)}\) and \(\frac{3}{(x - 4)(x + 1)}\). Then, first fraction: denominator is \((x + 1)(x - 2)\), so to get LCD, multiply numerator and denominator by \((x - 4)\): numerator \(2(x - 4)\). Second fraction: denominator is \((x - 4)(x + 1)\), so multiply numerator and denominator by \((x - 2)\): numerator \(3(x - 2)\). Then sum numerators: \(2(x - 4)+3(x - 2)\)? No, that's not matching. Wait, the first option is \(2(x - 2)+3(x + 1)\). Wait, maybe the original fractions are \(\frac{2}{(x + 1)(x - 4)}\) and \(\frac{3}{(x - 2)(x - 4)}\), but then LCD is \((x - 4)(x + 1)(x - 2)\). So first fraction: multiply numerator and denominator by \((x - 2)\): numerator \(2(x - 2)\). Second fraction: multiply numerator and denominator by \((x + 1)\): numerator \(3(x + 1)\). Then sum numerators: \(2(x - 2)+3(x + 1)\). Ah! That makes sense. So the two fractions are \(\frac{2}{(x + 1)(x - 4)}\) and \(\frac{3}{(x - 2)(x - 4)}\), but wait, no, their denominators are \((x + 1)(x - 4)\) and \((x - 2)(x - 4)\), so LCD is \((x - 4)(x + 1)(x - 2)\). So to add them, first fraction: denominator is \((x + 1)(x - 4)\), so multiply numerator and denominator by \((x - 2)\): numerator \(2(x - 2)\). Second fraction: denominator is \((x - 2)(x - 4)\), multiply numerator and denominator by \((x + 1)\): numerator \(3(x + 1)\). Then, when we add, the numerators are \(2(x - 2)+3(x + 1)\), because the denominators are now both \((x - 4)(x + 1)(x - 2)\), so we can add the numerators. So the correct numerator is \(2(x - 2)+3(x + 1)\).

Answer:

\(2(x - 2)+3(x + 1)\) (the first option)