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Solve for all values of 
x.

(x-5)/(x+3)=(-1)/(x)
Answer: 
x=

Solve for all values of x x .\newlinex5x+3=1x \frac{x-5}{x+3}=\frac{-1}{x} \newlineAnswer: x= x=

Full solution

Q. Solve for all values of x x .\newlinex5x+3=1x \frac{x-5}{x+3}=\frac{-1}{x} \newlineAnswer: x= x=
  1. Identify Equation: First, we need to identify the equation we are solving: (x5)/(x+3)=(1)/(x)(x-5)/(x+3)=(-1)/(x). We will solve for xx by finding a common denominator and then cross-multiplying to eliminate the fractions.
  2. Find Common Denominator: The common denominator between (x+3)(x+3) and xx is x(x+3)x(x+3). We will multiply both sides of the equation by this common denominator to clear the fractions.
  3. Clear Fractions: Multiplying both sides of the equation by x(x+3)x(x+3), we get:\newline(x5x+3)x(x+3)=(1x)x(x+3)(\frac{x-5}{x+3}) \cdot x(x+3) = (\frac{-1}{x}) \cdot x(x+3)\newlineThis simplifies to:\newlinex(x5)=1(x+3)x(x-5) = -1(x+3)
  4. Distribute and Simplify: Now we distribute on both sides of the equation: x25x=x3x^2 - 5x = -x - 3
  5. Combine Like Terms: Next, we bring all terms to one side of the equation to set it equal to zero: x25x+x+3=0x^2 - 5x + x + 3 = 0
  6. Factor Quadratic Equation: Combine like terms: x24x+3=0x^2 - 4x + 3 = 0
  7. Solve for x: Now we factor the quadratic equation:\newline(x3)(x1)=0(x - 3)(x - 1) = 0
  8. Check Solutions: Set each factor equal to zero and solve for xx:x3=0x - 3 = 0 or x1=0x - 1 = 0
  9. Check Solutions: Set each factor equal to zero and solve for xx:x3=0x - 3 = 0 or x1=0x - 1 = 0Solving each equation gives us the values of xx:x=3x = 3 or x=1x = 1
  10. Check Solutions: Set each factor equal to zero and solve for xx:x3=0x - 3 = 0 or x1=0x - 1 = 0Solving each equation gives us the values of xx:x=3x = 3 or x=1x = 1However, we must check these solutions against the original equation to ensure they do not make any denominator zero. The original equation has denominators (x+3)(x+3) and xx, so we must exclude any solutions where x=3x = -3 or x=0x = 0.
  11. Check Solutions: Set each factor equal to zero and solve for xx:x3=0x - 3 = 0 or x1=0x - 1 = 0Solving each equation gives us the values of xx:x=3x = 3 or x=1x = 1However, we must check these solutions against the original equation to ensure they do not make any denominator zero. The original equation has denominators (x+3)(x+3) and xx, so we must exclude any solutions where x=3x = -3 or x=0x = 0.Checking the solutions, we see that neither x=3x = 3 nor x=1x = 1 makes any denominator zero, so both are valid solutions.

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