Subtract <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> from both sides of the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of <math><mstyle displaystyle="true"><mn>1</mn><mo>,</mo><mn>1</mn><mo>,</mo><mn>1</mn></mstyle></math> is the result of multiplying all prime factors the greatest number of times they occur in either number.

The factor for <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> is <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> itself.

The LCM of <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn><mo>,</mo><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> is the result of multiplying all factors the greatest number of times they occur in either term.

Multiply each term in <math><mstyle displaystyle="true"><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>x</mi></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>-</mo><mn>3</mn></mstyle></math> by <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> .

Simplify the left side.

Cancel the common factor of <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> .

Move the leading negative in <math><mstyle displaystyle="true"><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mstyle></math> into the numerator.

Cancel the common factor.

Rewrite the expression.

Simplify the right side.

Simplify each term.

Cancel the common factor of <math><mstyle displaystyle="true"><mi>x</mi><mo>+</mo><mn>1</mn></mstyle></math> .

Cancel the common factor.

Rewrite the expression.

Apply the distributive property.

Multiply <math><mstyle displaystyle="true"><mo>-</mo><mn>3</mn></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Subtract <math><mstyle displaystyle="true"><mn>3</mn><mi>x</mi></mstyle></math> from <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> .

Rewrite the equation as <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn><mi>x</mi><mo>-</mo><mn>3</mn><mo>=</mo><mo>-</mo><mn>1</mn></mstyle></math> .

Move all terms not containing <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> to the right side of the equation.

Add <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> to both sides of the equation.

Add <math><mstyle displaystyle="true"><mo>-</mo><mn>1</mn></mstyle></math> and <math><mstyle displaystyle="true"><mn>3</mn></mstyle></math> .

Divide each term in <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn><mi>x</mi><mo>=</mo><mn>2</mn></mstyle></math> by <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn></mstyle></math> and simplify.

Divide each term in <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn><mi>x</mi><mo>=</mo><mn>2</mn></mstyle></math> by <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn></mstyle></math> .

Simplify the left side.

Cancel the common factor of <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn></mstyle></math> .

Cancel the common factor.

Divide <math><mstyle displaystyle="true"><mi>x</mi></mstyle></math> by <math><mstyle displaystyle="true"><mn>1</mn></mstyle></math> .

Simplify the right side.

Divide <math><mstyle displaystyle="true"><mn>2</mn></mstyle></math> by <math><mstyle displaystyle="true"><mo>-</mo><mn>2</mn></mstyle></math> .

Exclude the solutions that do not make <math><mstyle displaystyle="true"><mn>3</mn><mo>-</mo><mfrac><mrow><mn>1</mn></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mi>x</mi></mrow><mrow><mi>x</mi><mo>+</mo><mn>1</mn></mrow></mfrac></mstyle></math> true.

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