Question

question #2 solve the equation. remember to check for extraneous solutions. $\frac{9}{x}=\frac{3x - 9}{x}$
$o x=\frac{2}{3}$
$o x = 6$
$o x = 1$
$o x = 3$
question #3 solve the equation. remember to check for extraneous solutions $\frac{5x + 30}{x^{2}}=\frac{2x + 3}{x^{2}}$
$o x=-4$ and $x=-3$
$o x=-4$
$o x = 1$
$o x=-9$

Explanation:

Question #2

Step1: Cross - multiply (since denominators are the same)

Since the denominators of $\frac{9}{x}$ and $\frac{3x - 9}{x}$ are the same ($x
eq0$), we can set the numerators equal: $9=3x - 9$.

Step2: Solve for $x$

Add 9 to both sides of the equation: $9 + 9=3x-9 + 9$, which simplifies to $18 = 3x$. Then divide both sides by 3: $\frac{18}{3}=\frac{3x}{3}$, so $x = 6$.

Step3: Check for extraneous solutions

The original equation $\frac{9}{x}=\frac{3x - 9}{x}$ has a denominator of $x$. When $x = 6$, the denominators are non - zero.

Step1: Cross - multiply (since denominators are the same)

Since the denominators of $\frac{5x + 30}{x^{2}}$ and $\frac{2x+3}{x^{2}}$ are the same ($x
eq0$), we set the numerators equal: $5x + 30=2x + 3$.

Step2: Solve for $x$

Subtract $2x$ from both sides: $5x-2x + 30=2x-2x + 3$, which gives $3x+30 = 3$. Then subtract 30 from both sides: $3x+30 - 30=3 - 30$, so $3x=-27$. Divide both sides by 3: $\frac{3x}{3}=\frac{-27}{3}$, and $x=-9$.

Step3: Check for extraneous solutions

For the original equation $\frac{5x + 30}{x^{2}}=\frac{2x + 3}{x^{2}}$, when $x=-9$, the denominator $x^{2}=(-9)^{2}=81
eq0$.

Answer:

$x = 6$

Question #3