Martin-Gay, Beginning Algebra, 5ed 22 33 Solve the following rational equation.EXAMPLE Because no variable appears in the denominator, no restrictions.

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Presentation transcript:

Martin-Gay, Beginning Algebra, 5ed 22

33 Solve the following rational equation.EXAMPLE Because no variable appears in the denominator, no restrictions exist. The LCM of 5, 2, and 4 is 20, so we multiply both sides by 20: Using the multiplication principle to multiply both sides by the LCM. Parentheses are important! Using the distributive law. Be sure to multiply EACH term by the LCM.

Martin-Gay, Beginning Algebra, 5ed 44 Simplifying and solving for x. If fractions remain, we have either made a mistake or have not used the LCM of the denominators. We should check our solution, but no need to since we never make a mistake

Martin-Gay, Beginning Algebra, 5ed 55 SolveEXAMPLE Restrictions: x cannot equal 0 LCD = 6x

Martin-Gay, Beginning Algebra, 5ed 66 Restrictions: x cannot equal  3 LCD = 5(p + 3)

Martin-Gay, Beginning Algebra, 5ed 77 Restrictions: x cannot equal  LCD = x

Martin-Gay, Beginning Algebra, 5ed 88 Restrictions: x cannot equal 6 LCD = x  6

Martin-Gay, Beginning Algebra, 5ed 99 EXAMPLE Solve Restrictions: x cannot equal 1 or  1 LCD = (x  1)(x + 1) Because of the restriction above, 1 must be rejected as a solution. This equation has no solution.

Martin-Gay, Beginning Algebra, 5ed 10 = (x  2)(x + 2) Restrictions: x cannot equal 2 or  2 LCD = (x  2)(x + 2)

Martin-Gay, Beginning Algebra, 5ed 11 Restrictions: x cannot equal  1 or 3

Martin-Gay, Beginning Algebra, 5ed 12