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Solve for all values of 
x.

(x+2)/(x+6)=(1)/(x)
Answer: 
x=

Solve for all values of x x .\newlinex+2x+6=1x \frac{x+2}{x+6}=\frac{1}{x} \newlineAnswer: x= x=

Full solution

Q. Solve for all values of x x .\newlinex+2x+6=1x \frac{x+2}{x+6}=\frac{1}{x} \newlineAnswer: x= x=
  1. Find Common Denominator: First, we need to find a common denominator to combine the fractions. The common denominator for (x+6)(x+6) and xx is x(x+6)x(x+6). We will multiply both sides of the equation by this common denominator to eliminate the fractions.\newlineCalculation: x+2x+6×x(x+6)=1x×x(x+6)\frac{x+2}{x+6} \times x(x+6) = \frac{1}{x} \times x(x+6)
  2. Multiply and Simplify: After multiplying both sides by the common denominator, we can simplify the equation.\newlineCalculation: (x+2)×x=1×(x+6)(x+2) \times x = 1 \times (x+6)
  3. Distribute and Simplify: Now, we distribute the xx on the left side and simplify the right side.\newlineCalculation: x2+2x=x+6x^2 + 2x = x + 6
  4. Bring Terms Together: Next, we bring all terms to one side to set the equation to zero and solve for xx.\newlineCalculation: x2+2xx6=0x^2 + 2x - x - 6 = 0
  5. Factor the Equation: Simplify the equation by combining like terms.\newlineCalculation: x2+x6=0x^2 + x - 6 = 0
  6. Set Factors to Zero: Now, we need to factor the quadratic equation.\newlineCalculation: (x+3)(x2)=0(x + 3)(x - 2) = 0
  7. Solve for x: To find the values of x, we set each factor equal to zero and solve for x.\newlineCalculation: x+3=0x + 3 = 0 or x2=0x - 2 = 0
  8. Check Valid Solutions: Solving each equation gives us the values of xx.\newlineCalculation: x=3x = -3 or x=2x = 2\newlineHowever, we must check these solutions in the original equation to ensure they do not make the denominator zero.
  9. Check Valid Solutions: Solving each equation gives us the values of xx.\newlineCalculation: x=3x = -3 or x=2x = 2\newlineHowever, we must check these solutions in the original equation to ensure they do not make the denominator zero.Checking x=3x = -3 in the original equation's denominators:\newline(x+6)(x+6) becomes (3+6)(-3+6) which is 33, not zero.\newlinexx becomes 3-3, which is not zero.\newlineSo, x=3x = -3 is a valid solution.
  10. Check Valid Solutions: Solving each equation gives us the values of xx.\newlineCalculation: x=3x = -3 or x=2x = 2\newlineHowever, we must check these solutions in the original equation to ensure they do not make the denominator zero.Checking x=3x = -3 in the original equation's denominators:\newline(x+6)(x+6) becomes (3+6)(-3+6) which is 33, not zero.\newlinexx becomes 3-3, which is not zero.\newlineSo, x=3x = -3 is a valid solution.Checking x=2x = 2 in the original equation's denominators:\newline(x+6)(x+6) becomes x=3x = -322 which is x=3x = -333, not zero.\newlinexx becomes x=3x = -355, which is not zero.\newlineSo, x=2x = 2 is also a valid solution.

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