Question

if \\(dfrac{2x}{x^2 + 1} = dfrac{2}{x + 2}\\), what is the value of \\(x\\)?
a \\(-dfrac{1}{4}\\)
b \\(dfrac{1}{2}\\)
c \\(0\\)
d \\(2\\)

Explanation:

Step1: Cross - multiply the equation

Given the equation \(\frac{2x}{x^{2}+1}=\frac{2}{x + 2}\), cross - multiplying (which is valid when \(x^{2}+1
eq0\) and \(x + 2
eq0\); since \(x^{2}+1=(x - 0)^{2}+1\geq1\) for all real \(x\), and \(x
eq - 2\)) gives us \(2x(x + 2)=2(x^{2}+1)\).

Step2: Expand both sides

Expand the left - hand side: \(2x(x + 2)=2x^{2}+4x\).
Expand the right - hand side: \(2(x^{2}+1)=2x^{2}+2\).

Step3: Simplify the equation

Subtract \(2x^{2}\) from both sides of the equation \(2x^{2}+4x=2x^{2}+2\).
We get \(4x=2\).

Step4: Solve for \(x\)

Divide both sides of the equation \(4x = 2\) by 4: \(x=\frac{2}{4}=\frac{1}{2}\).

Answer:

B. \(\frac{1}{2}\)