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Question
9 what is the solution to the equation \\(\frac{2x - 3}{x - 1} = \frac{8x + 1}{4x + 5}\\)? a \\(-\frac{14}{5}\\) c \\(\frac{14}{9}\\) b \\(-\frac{14}{9}\\) d \\(\frac{14}{5}\\) \\(\bigcirc\\) d \\(\bigcirc\\) c \\(\bigcirc\\) b \\(\bigcirc\\) a
Step1: Cross - multiply the fractions
To solve the equation \(\frac{2x - 3}{x - 1}=\frac{8x + 1}{4x + 5}\), we cross - multiply. Cross - multiplying gives us \((2x - 3)(4x + 5)=(8x + 1)(x - 1)\).
Step2: Expand both sides
First, expand the left - hand side:
Using the FOIL method \((a + b)(c + d)=ac+ad+bc+bd\), for \((2x - 3)(4x + 5)\), we have:
\(2x\times4x+2x\times5-3\times4x - 3\times5=8x^{2}+10x-12x - 15=8x^{2}-2x - 15\)
Then, expand the right - hand side:
For \((8x + 1)(x - 1)\), we have:
\(8x\times x-8x\times1 + 1\times x-1\times1=8x^{2}-8x+x - 1=8x^{2}-7x - 1\)
Step3: Set up the resulting equation
Now we have the equation \(8x^{2}-2x - 15 = 8x^{2}-7x - 1\)
Subtract \(8x^{2}\) from both sides of the equation:
\(8x^{2}-2x - 15-8x^{2}=8x^{2}-7x - 1-8x^{2}\)
This simplifies to \(-2x - 15=-7x - 1\)
Step4: Solve for x
Add \(7x\) to both sides:
\(-2x+7x - 15=-7x + 7x-1\)
\(5x - 15=-1\)
Add 15 to both sides:
\(5x-15 + 15=-1 + 15\)
\(5x=14\)
Divide both sides by 5:
\(x=\frac{14}{5}\)
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D. \(\frac{14}{5}\)