Start with the equation: We start with the equation: y = (-2)/(2y + 3) Our goal is to solve for y.Multiply to eliminate fraction: To eliminate the fraction, we multiply both sides of the equation by (2y + 3) to get: y(2y + 3) = -2Distribute y on left side: Distribute y on the left side of the equation: 2y^2 + 3y = -2Set equation to zero: We want to set the equation to zero to solve for y, so we add 2 to both sides: 2y^2 + 3y + 2 = 0Factor the quadratic equation: Now we have a quadratic equation. We can attempt to factor it, or use the quadratic formula. Let's try factoring first: (2y + 1)(y + 2) = 0Solve for y in first equation: Set each factor equal to zero and solve for y: 2y + 1 = 0 or y + 2 = 0Solve for y in second equation: Solve the first equation for y: 2y = -1 y = -1/2Solve the second equation for y: y = -2

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$y=\frac{-2}{2 y+3}$

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Algebra 1

Evaluate integers raised to rational exponents

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y=\frac{-2}{2 y+3}

Start with the equation: We start with the equation: $y = \frac{-2}{2y + 3}$ Our goal is to solve for $y$ .
Multiply to eliminate fraction: To eliminate the fraction, we multiply both sides of the equation by $(2y + 3)$ to get: $\newline$ $y(2y + 3) = -2$
Distribute $y$ on left side: Distribute $y$ on the left side of the equation: $2y^2 + 3y = -2$
Set equation to zero: We want to set the equation to zero to solve for $y$ , so we add $2$ to both sides: $2y^2 + 3y + 2 = 0$
Factor the quadratic equation: Now we have a quadratic equation. We can attempt to factor it, or use the quadratic formula. Let's try factoring first: $\newline$ $(2y + 1)(y + 2) = 0$
Solve for $y$ in first equation: Set each factor equal to zero and solve for $y$ : $2y + 1 = 0$ or $y + 2 = 0$
Solve for $y$ in second equation: Solve the first equation for $y$ : $2y = -1$ $y = -\frac{1}{2}$
Solve for y in second equation: Solve the first equation for y: $\newline$ $2y = -1$ $\newline$ $y = -\frac{1}{2}$ Solve the second equation for y: $\newline$ $y = -2$