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Express (7)/(x-3) - (8)/(1-2x) as a single fraction in its simplest form.

Mathematics

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Answer

a

22x−312x2−7x+3\frac{22x - 31}{2x^2 - 7x + 3}

Step-by-step solution

8 steps

Part a· 5 steps

To express the given expression as a single fraction, we find a common denominator.

  1. Find the common denominator.

    The common denominator for 7x−3\frac{7}{x-3} and 81−2x\frac{8}{1-2x} is (x−3)(1−2x)(x-3)(1-2x).

  2. Rewrite each fraction with the common denominator.

    7x−3−81−2x=7(1−2x)(x−3)(1−2x)−8(x−3)(x−3)(1−2x)\frac{7}{x-3} - \frac{8}{1-2x} = \frac{7(1-2x)}{(x-3)(1-2x)} - \frac{8(x-3)}{(x-3)(1-2x)}

  3. Combine the fractions and simplify the numerator.

    7(1−2x)−8(x−3)(x−3)(1−2x)=7−14x−(8x−24)(x−3)(1−2x)\frac{7(1-2x) - 8(x-3)}{(x-3)(1-2x)} = \frac{7 - 14x - (8x - 24)}{(x-3)(1-2x)} =7−14x−8x+24(x−3)(1−2x)= \frac{7 - 14x - 8x + 24}{(x-3)(1-2x)} =31−22x(x−3)(1−2x)= \frac{31 - 22x}{(x-3)(1-2x)}

  4. Expand the denominator.

    (x−3)(1−2x)=x(1)+x(−2x)−3(1)−3(−2x)(x-3)(1-2x) = x(1) + x(-2x) - 3(1) - 3(-2x) =x−2x2−3+6x= x - 2x^2 - 3 + 6x =−2x2+7x−3= -2x^2 + 7x - 3

  5. Write the final simplified fraction.

    31−22x−2x2+7x−3\frac{31 - 22x}{-2x^2 + 7x - 3}

    To make the leading term in the denominator positive, we can multiply the numerator and denominator by −1-1:

    −(22x−31)−(2x2−7x+3)=22x−312x2−7x+3\frac{-(22x - 31)}{-(2x^2 - 7x + 3)} = \frac{22x - 31}{2x^2 - 7x + 3}

    The final answer is 22x−312x2−7x+3\boxed{\frac{22x - 31}{2x^2 - 7x + 3}}.

Part b· 3 steps

(i) To draw a tree diagram, we consider the probabilities at each stage of selection.

Initially, there are 2 green (G) balls and 3 blue (B) balls, making a total of 5 balls.

  1. First selection probabilities.

    • Probability of selecting a Green ball first (G1): P(G1)=25P(G1) = \frac{2}{5}
    • Probability of selecting a Blue ball first (B1): P(B1)=35P(B1) = \frac{3}{5}

  2. Second selection probabilities (without replacement).

    • If the first ball was Green (G1): Remaining balls: 1 Green, 3 Blue (total 4 balls).

      • Probability of selecting a Green ball second (G2 | G1): P(G2∣G1)=14P(G2 | G1) = \frac{1}{4}
      • Probability of selecting a Blue ball second (B2 | G1): P(B2∣G1)=34P(B2 | G1) = \frac{3}{4}

    • If the first ball was Blue (B1): Remaining balls: 2 Green, 2 Blue (total 4 balls).

      • Probability of selecting a Green ball second (G2 | B1): P(G2∣B1)=24=12P(G2 | B1) = \frac{2}{4} = \frac{1}{2}
      • Probability of selecting a Blue ball second (B2 | B1): P(B2∣B1)=24=12P(B2 | B1) = \frac{2}{4} = \frac{1}{2}

  3. Tree Diagram Illustration.

    Here is a textual representation of the tree diagram:

    • First Selection
      • Branch 1: Green (G)
        • Probability: 25\frac{2}{5}
        • Second Selection (after G)
          • Branch 1.1: Green (G)
            • Probability: 14\frac{1}{4}
            • Outcome: GG
          • Branch 1.2: Blue (B)
            • Probability: 34\frac{3}{4}
            • Outcome: GB
      • Branch 2: Blue (B)
        • Probability: 35\frac{3}{5}
        • Second Selection (after B)
          • Branch 2.1: Green (G)
            • Probability: 12\frac{1}{2}
            • Outcome: BG
          • Branch 2.2: Blue (B)
            • Probability: 12\frac{1}{2}
            • Outcome: BB
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