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Thursday 21st of November



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	The Square of a Binomial
	Properties of Negative Exponents
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	Combining Operations
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	Graphing Inequalities in Two Variables
	Solving Trigonometric Equations
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	Simple Trinomials as Products of Binomials
	Ratios and Proportions
	Solving Equations
	Multiplying and Dividing Fractions 2
	Rational Numbers
	Difference of Two Squares
	Factoring Polynomials by Grouping
	Solving Equations That Contain Rational Expressions
	Solving Quadratic Equations
	Dividing and Subtracting Rational Expressions
	Square Roots and Real Numbers
	Order of Operations
	Solving Nonlinear Equations by Substitution
	The Distance and Midpoint Formulas
	Linear Equations
	Graphing Using x- and y- Intercepts
	Properties of Exponents
	Solving Quadratic Equations
	Solving One-Step Equations Using Algebra
	Relatively Prime Numbers
	Solving a Quadratic Inequality with Two Solutions
	Quadratics
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	Factoring a Difference of Two Squares
	Straight Lines
	Solving Quadratic Equations by Factoring
	Graphing Logarithmic Functions
	Simplifying Expressions Involving Variables
	Adding Integers
	Decimals
	Factoring Completely General Quadratic Trinomials
	Using Patterns to Multiply Two Binomials
	Adding and Subtracting Rational Expressions With Unlike Denominators
	Rational Exponents
	Horizontal and Vertical Lines

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Solving Equations That Contain Rational Expressions

Letâ€™s consider the case where some of the rational expressions have a variable in the denominator.

Example

Solve:

Solution
Multiply each side of the equation by 4x, the LCD of the rational expressions.
Distribute 4x to each term on the left side.
Reduce by canceling common factors.
	8 + x(x - 2)			= 23
Distribute x.	8 + x² - 2x			= 23
The fractions have been eliminated and we are left with the quadratic equation 8 + x² - 2x = 23.
To solve this equation, first write it in standard form, ax² + bx + c = 0.	x² - 2x - 15			= 0
To factor x² - 2x - 15, find two integers whose product is -15 and whose sum is -2. They are -5 and 3.
Use the Zero Product Property. Solve each equation.	x - 5 x	= 0 = 5	or or	x + 3 = 0 x = -3

So, the equation has two solutions: 5 and -3.

To check the solutions, we substitute each value of x in the original equation and simplify.