George can cut and split a cord of firewood in $3$ fewer hours than SkylerSkyler can. When they work together, it takes them $2$ hours. How long would it take each of them to do the job alone?

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Q. George can cut and split a cord of firewood in

3

fewer hours than SkylerSkyler can. When they work together, it takes them

2

hours. How long would it take each of them to do the job alone?

Denote Time for Skyler: Let's denote the time it takes Skyler to cut and split a cord of firewood alone as $S$ hours. Since George can do the job in $3$ fewer hours than Skyler, George's time would be $S - 3$ hours.
Find Rates for George and Skyler: We need to find the rate at which George and Skyler work. The rate is the reciprocal of the time. So, Skyler's rate is $\frac{1}{S}$ and George's rate is $\frac{1}{S - 3}$ .
Set Up Combined Rates Equation: When George and Skyler work together, their rates add up. The combined rate is $\frac{1}{S} + \frac{1}{S - 3}$ . We know that together they take $2$ hours to complete the job, so their combined rate is $\frac{1}{2}$ cord per hour.
Solve Combined Rates Equation: Now we set up the equation based on their combined rates: $\frac{1}{S} + \frac{1}{S - 3} = \frac{1}{2}$ .
Expand and Rearrange Equation: To solve the equation, we need a common denominator. The common denominator for $S$ and $S - 3$ is $S(S - 3)$ . We rewrite the equation as $(S - 3 + S) / [S(S - 3)] = \frac{1}{2}$ .
Factor Quadratic Equation: Simplify the numerator: $(2S - 3) / [S(S - 3)] = \frac{1}{2}$ .
Final Solution: Cross-multiply to solve for $S$ : $2(2S - 3) = S(S - 3)$ .
Final Solution: Cross-multiply to solve for $S$ : $2(2S - 3) = S(S - 3)$ .Expand both sides: $4S - 6 = S^2 - 3S$ .
Final Solution: Cross-multiply to solve for $S$ : $2(2S - 3) = S(S - 3)$ . Expand both sides: $4S - 6 = S^2 - 3S$ . Rearrange the equation to form a quadratic equation: $S^2 - 3S - 4S + 6 = 0$ , which simplifies to $S^2 - 7S + 6 = 0$ .
Final Solution: Cross-multiply to solve for $S$ : $2(2S - 3) = S(S - 3)$ . Expand both sides: $4S - 6 = S^2 - 3S$ . Rearrange the equation to form a quadratic equation: $S^2 - 3S - 4S + 6 = 0$ , which simplifies to $S^2 - 7S + 6 = 0$ . Factor the quadratic equation: $(S - 6)(S - 1) = 0$ .
Final Solution: Cross-multiply to solve for $S$ : $2(2S - 3) = S(S - 3)$ . Expand both sides: $4S - 6 = S^2 - 3S$ . Rearrange the equation to form a quadratic equation: $S^2 - 3S - 4S + 6 = 0$ , which simplifies to $S^2 - 7S + 6 = 0$ . Factor the quadratic equation: $(S - 6)(S - 1) = 0$ . Solve for $S$ : $S = 6$ or $S = 1$ . Since George takes $3$ fewer hours than Skyler, and it cannot be negative or zero, $S$ must be $2(2S - 3) = S(S - 3)$ $1$ . Therefore, Skyler takes $2(2S - 3) = S(S - 3)$ $1$ hours and George takes $2(2S - 3) = S(S - 3)$ $3$ hours.