Factor out the common term \(2\).
\[3-\frac{1}{2x}=6+\frac{2(3+5x)}{2x+4}-2\]
Factor out the common term \(2\).
\[3-\frac{1}{2x}=6+\frac{2(3+5x)}{2(x+2)}-2\]
Cancel \(2\).
\[3-\frac{1}{2x}=6+\frac{3+5x}{x+2}-2\]
Simplify \(6+\frac{3+5x}{x+2}-2\) to \(4+\frac{3+5x}{x+2}\).
\[3-\frac{1}{2x}=4+\frac{3+5x}{x+2}\]
Multiply both sides by the Least Common Denominator: \(2x(x+2)\).
\[6x(x+2)-x-2=8x(x+2)+2x(3+5x)\]
Simplify.
\[6{x}^{2}+11x-2=18{x}^{2}+22x\]
Move all terms to one side.
\[6{x}^{2}+11x-2-18{x}^{2}-22x=0\]
Simplify \(6{x}^{2}+11x-2-18{x}^{2}-22x\) to \(-12{x}^{2}-11x-2\).
\[-12{x}^{2}-11x-2=0\]
Multiply both sides by \(-1\).
\[12{x}^{2}+11x+2=0\]
Split the second term in \(12{x}^{2}+11x+2\) into two terms.
Multiply the coefficient of the first term by the constant term.
\[12\times 2=24\]
Ask: Which two numbers add up to \(11\) and multiply to \(24\)?
Split \(11x\) as the sum of \(8x\) and \(3x\).
\[12{x}^{2}+8x+3x+2\]
\[12{x}^{2}+8x+3x+2=0\]
Factor out common terms in the first two terms, then in the last two terms.
\[4x(3x+2)+(3x+2)=0\]
Factor out the common term \(3x+2\).
\[(3x+2)(4x+1)=0\]
Solve for \(x\).
Ask: When will \((3x+2)(4x+1)\) equal zero?
When \(3x+2=0\) or \(4x+1=0\)
Solve each of the 2 equations above.
\[x=-\frac{2}{3},-\frac{1}{4}\]
\[x=-\frac{2}{3},-\frac{1}{4}\]
Decimal Form: -0.666667, -0.25
x=-2/3,-1/4
