Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve for all values of 
x.

(x-7)/(x+3)=(4)/(x)
Answer: 
x=

Solve for all values of x x .\newlinex7x+3=4x \frac{x-7}{x+3}=\frac{4}{x} \newlineAnswer: x= x=

Full solution

Q. Solve for all values of x x .\newlinex7x+3=4x \frac{x-7}{x+3}=\frac{4}{x} \newlineAnswer: x= x=
  1. Find Common Denominator: First, we need to find a common denominator to combine the fractions on both sides of the equation. The common denominator here is x(x+3)x(x+3).
  2. Multiply by Common Denominator: Next, we multiply both sides of the equation by the common denominator to eliminate the fractions: x(x+3)x7x+3=x(x+3)4xx(x+3) \cdot \frac{x-7}{x+3} = x(x+3) \cdot \frac{4}{x}
  3. Simplify Equation: Simplify the equation by canceling out the common terms on both sides: x(x7)=4(x+3)x(x-7) = 4(x+3)
  4. Distribute Terms: Now, distribute the xx on the left side and the 44 on the right side: x27x=4x+12x^2 - 7x = 4x + 12
  5. Combine Like Terms: Bring all terms to one side to set the equation to zero and combine like terms:\newlinex27x4x12=0x^2 - 7x - 4x - 12 = 0\newlinex211x12=0x^2 - 11x - 12 = 0
  6. Factor Quadratic Equation: We now have a quadratic equation. To solve for xx, we can factor the quadratic equation: (x12)(x+1)=0(x - 12)(x + 1) = 0
  7. Solve for x: Set each factor equal to zero and solve for x:\newlinex12=0x - 12 = 0 or x+1=0x + 1 = 0\newlinex=12x = 12 or x=1x = -1
  8. Check for Extraneous Solutions: We must check for extraneous solutions by plugging the values back into the original equation, because the original equation has restrictions (xx cannot be 00 or 3-3, as those would make the denominators zero).
  9. Check x=12x = 12: Check x=12x = 12:
    12712+3=412\frac{12-7}{12+3} = \frac{4}{12}
    515=412\frac{5}{15} = \frac{4}{12}
    13=13\frac{1}{3} = \frac{1}{3} (This is true, so x=12x = 12 is a valid solution.)
  10. Check x=1x = -1: Check x=1x = -1:171+3=41\frac{-1-7}{-1+3} = \frac{4}{-1}8/2=4-8/2 = -44=4-4 = -4 (This is true, so x=1x = -1 is a valid solution.)

More problems from Find the roots of factored polynomials