Rational Equations - Revision history

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Lila: Created page with "Rational equations are equations containing rational expressions (or expressions with fractions that contain real numbers and/or variables). Some examples of rational equation..."

2021-09-22T16:42:24Z

Created page with "Rational equations are equations containing rational expressions (or expressions with fractions that contain real numbers and/or variables). Some examples of rational equation..."

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Rational equations are equations containing rational expressions (or expressions with fractions that contain real numbers and/or variables).
Some examples of rational equations:
* <math> \frac{1}{8}x + 2 = \frac{1}{4}x </math>
* <math> \frac{1}{x+1} = \frac{2}{3} </math>
* <math> \frac{x^2}{3x + 7} = \frac{x}{6} </math>

Steps to solving rational equations:
# Note any value of the variable that would make any denominator zero.
# Find the least common denominator of all denominators in the equation.
# Clear the fractions by multiplying both sides of the equation by the LCD.
# Solve the resulting equation.
# Check: If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.

==Resources==
* [https://openstax.org/books/intermediate-algebra-2e/pages/7-4-solve-rational-equations Solve Rational Equations], OpenStax
* [https://www.youtube.com/watch?v=1fR_9ke5-n8 Solving Rational Equations (Example)], The Organic Chemistry Tutor
* [https://www.youtube.com/watch?v=MFjx52NW5Bw Solving Rational Equations with Different Denominators (Example)], The Organic Chemistry Tutor

@@ Line 16: / Line 16: @@
 # The least common denominator of all terms in the equation is <math> x^2 </math>.
 # Multiplying each side of the equation <math> 1 - \frac{1}{x} = \frac{2}{x^2} </math> with <math> x^2 </math> gives us <math> x^2 - x = 2 </math>
-# <math> x^2 - x = 2 \to x^2 - x - 2 = 0 \to (x - 2)(x + 1) = 0 \to x = -1, x = 2 </math>
+# <math> x^2 - x = 2 \;\;\;\to\;\;\; x^2 - x - 2 = 0 \;\;\;\to\;\;\; (x - 2)(x + 1) = 0 \;\;\;\to\;\;\; x = -1, x = 2 </math>
 # None of these solutions were noted in step 1, so we can check our two solutions:
-::: <math>x = -1</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \to 1 - \frac{1}{-1} = \frac{2}{(-1)^2} \to 1 - (-1) = 2 </math>
+::: <math>x = -1</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \;\;\;\to\;\;\; 1 - \frac{1}{-1} = \frac{2}{(-1)^2} \;\;\;\to\;\;\; 1 - (-1) = 2 </math>
-::: <math>x = 2</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \to 1 - \frac{1}{2} = \frac{2}{2^2} \to \frac{1}{2} = \frac{2}{4} </math>
+::: <math>x = 2</math>: <math>\;\;\, 1 - \frac{1}{x} = \frac{2}{x^2} \;\;\;\to\;\;\; 1 - \frac{1}{2} = \frac{2}{2^2} \;\;\;\to\;\;\; \frac{1}{2} = \frac{2}{4} </math>
-Thus <math>x = -1</math> and <math>x = 2</math> are solutions to our original rational equations.
+::Thus <math>x = -1</math> and <math>x = 2</math> are solutions to our original rational equations.
 ==Resources==

@@ Line 19: / Line 19: @@
 # None of these solutions were noted in step 1, so we can check our two solutions:
 ::: <math>x = -1</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \to 1 - \frac{1}{-1} = \frac{2}{(-1)^2} \to 1 - (-1) = 2 </math>
-::: <math>x = 2</math>: <math> 1 - \frac{1}{2} = \frac{2}{2^2} \to \frac{1}{2} = \frac{2}{4} </math>
+::: <math>x = 2</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \to 1 - \frac{1}{2} = \frac{2}{2^2} \to \frac{1}{2} = \frac{2}{4} </math>
 Thus <math>x = -1</math> and <math>x = 2</math> are solutions to our original rational equations.

@@ Line 18: / Line 18: @@
 # <math> x^2 - x = 2 \to x^2 - x - 2 = 0 \to (x - 2)(x + 1) = 0 \to x = -1, x = 2 </math>
 # None of these solutions were noted in step 1, so we can check our two solutions:
-: <math>x = -1</math>:
+::: <math>x = -1</math>: <math> 1 - \frac{1}{x} = \frac{2}{x^2} \to 1 - \frac{1}{-1} = \frac{2}{(-1)^2} \to 1 - (-1) = 2 </math>
 ==Resources==

@@ Line 16: / Line 16: @@
 # The least common denominator of all terms in the equation is <math> x^2 </math>.
 # Multiplying each side of the equation <math> 1 - \frac{1}{x} = \frac{2}{x^2} </math> with <math> x^2 </math> gives us <math> x^2 - x = 2 </math>
-# <math> x^2 - x = 2 &nbsp;\to &nbsp; x^2 - x - 2 = 0 &nbsp;\to &nbsp; (x - 2)(x + 1) = 0 &nbsp;\to &nbsp; x = -1, x = 2 </math>
+# <math> x^2 - x = 2 \to x^2 - x - 2 = 0 \to (x - 2)(x + 1) = 0 \to x = -1, x = 2 </math>
 # None of these solutions were noted in step 1, so we can check our two solutions:
 : <math>x = -1</math>:

@@ Line 10: / Line 10: @@
 # Clear the fractions by multiplying both sides of the equation by the LCD.
 # Solve the resulting equation.
-# Check: If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.
+# Check: If any values found in step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.
 Example problem: <math> 1 - \frac{1}{x} = \frac{2}{x^2} </math>
@@ Line 16: / Line 16: @@
 # The least common denominator of all terms in the equation is <math> x^2 </math>.
 # Multiplying each side of the equation <math> 1 - \frac{1}{x} = \frac{2}{x^2} </math> with <math> x^2 </math> gives us <math> x^2 - x = 2 </math>
-# <math> x^2 - x = 2 \to x^2 - x - 2 = 0 \to (x - 2)(x + 1) = 0 \to x = 1, x = 2 </math>
+# <math> x^2 - x = 2 &nbsp;\to &nbsp; x^2 - x - 2 = 0 &nbsp;\to &nbsp; (x - 2)(x + 1) = 0 &nbsp;\to &nbsp; x = -1, x = 2 </math>
 ==Resources==

@@ Line 3: / Line 3: @@
 * <math> \frac{1}{8}x + 2 =  \frac{1}{4}x </math>
 * <math> \frac{1}{x+1} =  \frac{2}{3} </math>
-* <math> \frac{x^2}{3x + 7} =  \frac{x}{6} </math>
+* <math> \frac{y^2}{3y + 7} =  \frac{y}{6} </math>
 Steps to solving rational equations:
@@ Line 11: / Line 11: @@
 # Solve the resulting equation.
 # Check: If any values found in Step 1 are algebraic solutions, discard them. Check any remaining solutions in the original equation.
 ==Resources==