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)

Factor out common terms: Factor out the common terms in the numerator and the denominator. The numerator 56y^2 - 28y can be factored by taking out the common factor of 28y, which gives us 28y(2y - 1). The denominator 35y^2 + 21y can be factored by taking out the common factor of 7y, which gives us 7y(5y + 3). So, the expression becomes (28y(2y - 1)) / (7y(5y + 3)).Simplify by canceling out: Simplify the expression by canceling out the common terms. The term 28y in the numerator and 7y in the denominator can be simplified because 28y / 7y = 4. The expression now is 4(2y - 1) / (5y + 3).Distribute the 4: Distribute the 4 in the numerator. Multiplying 4 by each term in the parentheses gives us 8y - 4. The expression now is (8y - 4) / (5y + 3).Match with answer choices: Match the simplified expression with the given answer choices. The expression (8y - 4) / (5y + 3) matches with choice (D).

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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 $y \neq 0$ , which of the following expressions is equivalent to $\frac{56 y^{2}-28 y}{35 y^{2}+21 y}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{8 y-4}{5 y-3}$ $\newline$ (B) $\frac{8 y-4 y}{5 y+3 y}$ $\newline$ (C) $\frac{8 y-28}{5 y+21}$ $\newline$ (D) $\frac{8 y-4}{5 y+3}$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

Q. For

y \neq 0

, which of the following expressions is equivalent to

\frac{56 y^{2}-28 y}{35 y^{2}+21 y}

\newline

Choose

1

answer:

\newline

(A)

\frac{8 y-4}{5 y-3}

\newline

(B)

\frac{8 y-4 y}{5 y+3 y}

\newline

(C)

\frac{8 y-28}{5 y+21}

\newline

(D)

\frac{8 y-4}{5 y+3}

Factor out common terms: Factor out the common terms in the numerator and the denominator. $\newline$ The numerator $56y^2 - 28y$ can be factored by taking out the common factor of $28y$ , which gives us $28y(2y - 1)$ . $\newline$ The denominator $35y^2 + 21y$ can be factored by taking out the common factor of $7y$ , which gives us $7y(5y + 3)$ . $\newline$ So, the expression becomes $\frac{28y(2y - 1)}{7y(5y + 3)}$ .
Simplify by canceling out: Simplify the expression by canceling out the common terms. $\newline$ The term $28y$ in the numerator and $7y$ in the denominator can be simplified because $\frac{28y}{7y} = 4$ . $\newline$ The expression now is $\frac{4(2y - 1)}{5y + 3}$ .
Distribute the $4$ : Distribute the $4$ in the numerator. $\newline$ Multiplying $4$ by each term in the parentheses gives us $8y - 4$ . $\newline$ The expression now is $\frac{8y - 4}{5y + 3}$ .
Match with answer choices: Match the simplified expression with the given answer choices. $\newline$ The expression $(8y - 4) / (5y + 3)$ matches with choice (D).