Solve for all values of x. (x+2)/(x-2)=(-3)/(x) Answer: x=

Cross-multiply fractions: Cross-multiply to eliminate the fractions. (x + 2) * x = (-3) * (x - 2)Distribute and simplify: Distribute to simplify both sides of the equation. x^2 + 2x = -3x + 6Move terms and set to zero: Move all terms to one side to set the equation to zero. x^2 + 2x + 3x - 6 = 0Combine like terms: Combine like terms. x^2 + 5x - 6 = 0Factor quadratic equation: Factor the quadratic equation. (x + 6)(x - 1) = 0Set factors equal and solve: Set each factor equal to zero and solve for x. x + 6 = 0 or x - 1 = 0Check for extraneous solutions: Solve each equation for x. x = -6 or x = 1Check for extraneous solutions by plugging the values back into the original equation. For x = -6: (-6+2)/(-6-2) = (-3)/(-6) -4/-8 = 1/2, which is not equal to -3/-6 (which is 1/2). Therefore, x = -6 is an extraneous solution.

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Algebra 2

Simplify variable expressions using properties

Full solution

Q. Solve for all values of

x

\newline

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

\newline

Answer:

x=

Cross-multiply fractions: Cross-multiply to eliminate the fractions. $(x + 2) \times x = (-3) \times (x - 2)$
Distribute and simplify: Distribute to simplify both sides of the equation. $x^2 + 2x = -3x + 6$
Move terms and set to zero: Move all terms to one side to set the equation to zero. $\newline$ $x^2 + 2x + 3x - 6 = 0$
Combine like terms: Combine like terms. $x^2 + 5x - 6 = 0$
Factor quadratic equation: Factor the quadratic equation. $\newline$ $(x + 6)(x - 1) = 0$
Set factors equal and solve: Set each factor equal to zero and solve for $x$ . $x + 6 = 0$ or $x - 1 = 0$
Check for extraneous solutions: Solve each equation for $x$ . $x = -6$ or $x = 1$
Check for extraneous solutions: Solve each equation for $x$ . $x = -6$ or $x = 1$ Check for extraneous solutions by plugging the values back into the original equation. For $x = -6$ : $\frac{-6+2}{-6-2} = \frac{-3}{-6}$ $-\frac{4}{-8} = \frac{1}{2}$ , which is not equal to $-\frac{3}{-6}$ (which is $\frac{1}{2}$ ). Therefore, $x = -6$ is an extraneous solution.