Question

1. solve the equation \\(\frac{5}{x} - 2 = \frac{2}{x + 3}\\) by using the lcd. select the solution(s).

Explanation:

Step1: Identify the LCD

The denominators are \( x \) and \( x + 3 \), so the least common denominator (LCD) is \( x(x + 3) \).

Step2: Multiply each term by the LCD

Multiply each term in the equation \( \frac{5}{x}-2=\frac{2}{x + 3} \) by \( x(x + 3) \):
\[ \frac{5}{x}\times x(x + 3)-2\times x(x + 3)=\frac{2}{x + 3}\times x(x + 3) \]
Simplify each term:
- For \( \frac{5}{x}\times x(x + 3) \), the \( x \) cancels out, giving \( 5(x + 3) \).
- For \( -2\times x(x + 3) \), we get \( -2x(x + 3) \).
- For \( \frac{2}{x + 3}\times x(x + 3) \), the \( x + 3 \) cancels out, giving \( 2x \).

So the equation becomes:
\[ 5(x + 3)-2x(x + 3)=2x \]

Step3: Expand and simplify

Expand \( 5(x + 3) \) to \( 5x + 15 \) and \( -2x(x + 3) \) to \( -2x^{2}-6x \):
\[ 5x + 15-2x^{2}-6x = 2x \]
Combine like terms:
\[ -2x^{2}-x + 15 = 2x \]
Subtract \( 2x \) from both sides to set the equation to zero:
\[ -2x^{2}-3x + 15 = 0 \]
Multiply both sides by \( -1 \) to make the quadratic coefficient positive:
\[ 2x^{2}+3x - 15 = 0 \]

Step4: Solve the quadratic equation

Use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for the quadratic equation \( ax^{2}+bx + c = 0 \). Here, \( a = 2 \), \( b = 3 \), and \( c=-15 \).

First, calculate the discriminant \( \Delta=b^{2}-4ac \):
\[ \Delta = 3^{2}-4\times2\times(-15)=9 + 120 = 129 \]
Then, find the solutions:
\[ x=\frac{-3\pm\sqrt{129}}{4} \]
Wait, this seems wrong. Let's check the expansion again.

Wait, going back to Step 2:

Original equation: \( \frac{5}{x}-2=\frac{2}{x + 3} \)

Multiply by \( x(x + 3) \):

\( 5(x + 3)-2x(x + 3)=2x \)

Expand \( 5(x + 3)=5x + 15 \), \( -2x(x + 3)=-2x^{2}-6x \)

So left side: \( 5x + 15-2x^{2}-6x=-2x^{2}-x + 15 \)

Set equal to \( 2x \):

\( -2x^{2}-x + 15 = 2x \)

Bring all terms to left:

\( -2x^{2}-3x + 15 = 0 \)

Multiply by -1: \( 2x^{2}+3x - 15 = 0 \)

Wait, maybe I made a mistake in the sign when moving terms. Let's do it again:

\( -2x^{2}-x + 15-2x = 0 \)

\( -2x^{2}-3x + 15 = 0 \)

Yes, that's correct. Then quadratic formula:

\( x=\frac{3\pm\sqrt{9 + 120}}{-4}=\frac{3\pm\sqrt{129}}{-4}=\frac{-3\pm\sqrt{129}}{4} \)

Wait, but let's check the original equation for extraneous solutions. Wait, maybe I made a mistake in the multiplication step.

Wait, let's start over.

Original equation: \( \frac{5}{x}-2=\frac{2}{x + 3} \)

LCD is \( x(x + 3) \), \( x eq0 \) and \( x eq - 3 \)

Multiply each term by \( x(x + 3) \):

\( 5(x + 3)-2x(x + 3)=2x \)

Expand:

\( 5x + 15-2x^{2}-6x = 2x \)

Combine like terms:

\( -2x^{2}-x + 15 = 2x \)

Bring \( 2x \) to left:

\( -2x^{2}-3x + 15 = 0 \)

Multiply by -1:

\( 2x^{2}+3x - 15 = 0 \)

Discriminant \( D = 9 + 120 = 129 \)

Solutions:

\( x=\frac{-3\pm\sqrt{129}}{4} \)

Wait, but maybe I made a mistake in the problem. Wait, the original equation is \( \frac{5}{x}-2=\frac{2}{x + 3} \). Let's check with \( x = 5 \):

Left side: \( 5/5 - 2=1 - 2=-1 \)

Right side: \( 2/(5 + 3)=2/8 = 0.25 \). Not equal.

Wait, maybe the equation is \( \frac{5}{x}-2=\frac{2}{x + 3} \) with a typo? Or maybe I made a mistake.

Wait, let's try another approach. Let's move all terms to left:

\( \frac{5}{x}-2-\frac{2}{x + 3}=0 \)

Combine fractions:

\( \frac{5(x + 3)-2x(x + 3)-2x}{x(x + 3)}=0 \)

Numerator must be zero:

\( 5(x + 3)-2x(x + 3)-2x = 0 \)

Expand:

\( 5x + 15-2x^{2}-6x - 2x = 0 \)

Combine like terms:

\( -2x^{2}-3x + 15 = 0 \)

Which is the same as before. So the solutions are \( x=\frac{-3\pm\sqrt{129}}{4} \)

Wait, but maybe the original equation is \( \frac{5}{x}-2=\frac{2}{x + 3} \) with a different number. Wait, maybe I misread the equation. Let me check again.

The equation is \( \frac{5}{x}-2=\frac{2}{x + 3} \). Yes. So the solutions are \( x=\frac{-3+\sqrt{129}}{4} \) and \( x=\frac{-3-\sqrt{129}}{4} \)

But let's check if these are valid. Since \( \sqrt{129}\approx11.36 \), so \( -3 + 11.36\approx8.36 \), divided by 4 is approx 2.09, which is not zero or -3. \( -3 - 11.36\approx -14.36 \), divided by 4 is approx -3.59, which is not -3. So both are valid.

Wait, but maybe I made a mistake in the calculation. Let's re-express the quadratic equation:

From \( -2x^{2}-3x + 15 = 0 \), multiply by -1: \( 2x^{2}+3x - 15 = 0 \)

Quadratic formula: \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-3\pm\sqrt{9 + 120}}{4}=\frac{-3\pm\sqrt{129}}{4} \)

Yes, that's correct.

Answer:

The solutions are \( x = \frac{-3+\sqrt{129}}{4} \) and \( x = \frac{-3-\sqrt{129}}{4} \)