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For y!=0, which of the following expressions is equivalent to (56y^(2)-28 y)/(35y^(2)+21 y) ? Choose 1 answer: (A) (8y-4)/(5y-3) (B) (8y-4y)/(5y+3y) (C) (8y-28)/(5y+21) (D) (8y-4)/(5y+3)

For y≠0 y \neq 0 , which of the following expressions is equivalent to 56y2−28y35y2+21y \frac{56 y^{2}-28 y}{35 y^{2}+21 y} ?\newlineChoose 11 answer:\newline(A) 8y−45y−3 \frac{8 y-4}{5 y-3} \newline(B) 8y−4y5y+3y \frac{8 y-4 y}{5 y+3 y} \newline(C) 8y−285y+21 \frac{8 y-28}{5 y+21} \newline(D) 8y−45y+3 \frac{8 y-4}{5 y+3}

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Full solution

Q. For y≠0 y \neq 0 , which of the following expressions is equivalent to 56y2−28y35y2+21y \frac{56 y^{2}-28 y}{35 y^{2}+21 y} ?\newlineChoose 11 answer:\newline(A) 8y−45y−3 \frac{8 y-4}{5 y-3} \newline(B) 8y−4y5y+3y \frac{8 y-4 y}{5 y+3 y} \newline(C) 8y−285y+21 \frac{8 y-28}{5 y+21} \newline(D) 8y−45y+3 \frac{8 y-4}{5 y+3}
  1. Factor out common terms: Factor out the common terms in the numerator and the denominator.\newlineThe numerator 56y2−28y56y^2 - 28y can be factored by taking out the common factor of 28y28y, which gives us 28y(2y−1)28y(2y - 1).\newlineThe denominator 35y2+21y35y^2 + 21y can be factored by taking out the common factor of 7y7y, which gives us 7y(5y+3)7y(5y + 3).\newlineSo, the expression becomes 28y(2y−1)7y(5y+3)\frac{28y(2y - 1)}{7y(5y + 3)}.
  2. Simplify by canceling out: Simplify the expression by canceling out the common terms.\newlineThe term 28y28y in the numerator and 7y7y in the denominator can be simplified because 28y7y=4\frac{28y}{7y} = 4.\newlineThe expression now is 4(2y−1)5y+3\frac{4(2y - 1)}{5y + 3}.
  3. Distribute the 44: Distribute the 44 in the numerator.\newlineMultiplying 44 by each term in the parentheses gives us 8y−48y - 4.\newlineThe expression now is 8y−45y+3\frac{8y - 4}{5y + 3}.
  4. Match with answer choices: Match the simplified expression with the given answer choices.\newlineThe expression (8y−4)/(5y+3)(8y - 4) / (5y + 3) matches with choice (D).

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