2+x=3/x

Amanita, let's solve the equation 2+x=3x2 + x = \frac{3}{x} step by step. Here’s our plan:

Step 1: Move all terms to one side to set the equation to zero.
Step 2: Multiply both sides by xx to eliminate the fraction.
Step 3: Rearrange the equation into standard quadratic form.
Step 4: Solve the quadratic equation for xx.
Step 5: Check the solutions to make sure they are valid (since xx cannot be zero).

Let's begin!

Step 1:
Start with the equation:

2+x=3x2 + x = \frac{3}{x}

Subtract 3x\frac{3}{x} from both sides:

2+x3x=02 + x - \frac{3}{x} = 0

Step 2:
Multiply both sides by xx to get rid of the fraction:

x(2+x3x)=0x(2 + x - \frac{3}{x}) = 0

This simplifies to:

2x+x23=02x + x^2 - 3 = 0

Step 3:
Rearrange the equation into standard quadratic form:

x2+2x3=0x^2 + 2x - 3 = 0

Step 4:
Now, let's solve the quadratic equation x2+2x3=0x^2 + 2x - 3 = 0.

Amanita, after solving the quadratic equation, we find two possible solutions: x=3x = -3 and x=1x = 1.

Step 5:
Let's check if both solutions are valid by substituting them back into the original equation 2+x=3x2 + x = \frac{3}{x}:

  • For x=3x = -3: 2+(3)=12 + (-3) = -1 and 33=1\frac{3}{-3} = -1, so this works.
  • For x=1x = 1: 2+1=32 + 1 = 3 and 31=3\frac{3}{1} = 3, so this also works.

So, the solutions to the equation 2+x=3x2 + x = \frac{3}{x} are:

x=3 and x=1\boxed{x = -3 \text{ and } x = 1}