# Algebra Examples

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Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.

Since does not contain the variable to solve for, move it to the right side of the equation by adding to both sides.

Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.

Since does not contain the variable to solve for, move it to the right side of the equation by subtracting from both sides.

The Intermediate Value Theorem states that, if is a real-valued continuous function on the interval , and is a number between and , then there is a contained in the interval such that .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Simplify each term.

Raise to the power of to get .

Multiply by to get .

Raise to the power of to get .

Multiply by to get .

Multiply by to get .

Simplify by adding and subtracting.

Add and to get .

Add and to get .

Subtract from to get .

Simplify each term.

Remove parentheses around .

Raising to any positive power yields .

Multiply by to get .

Remove parentheses around .

Raising to any positive power yields .

Multiply by to get .

Multiply by to get .

Simplify by adding zeros.

Add and to get .

Add and to get .

Subtract from to get .

Rewrite the equation as .

The roots of this equation could not be found algebraically, so the roots were determined numerically.

The Intermediate Value Theorem states that there is a root on the interval because is a continuous function on .

The roots on the interval are located at .