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Objective Standard 15.0 I will solve a rational equation by multiplying the LCM of the denominators to both sides.

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Presentation on theme: "Objective Standard 15.0 I will solve a rational equation by multiplying the LCM of the denominators to both sides."β€” Presentation transcript:

1 Objective Standard 15.0 I will solve a rational equation by multiplying the LCM of the denominators to both sides

2 Review 1 Solving Linear Equation
Solve for x for the following equation. 2π‘₯+10=3π‘₯+4

3 Review 2 Solving Quadratic Equation
Solve for x for the following equation. 2π‘₯ π‘₯βˆ’6 +3 π‘₯+3 =27

4 Review 3 Solving Equation with x in Denominator
Solve for x in the following equation. 3 2 = 1 π‘₯ + 7 6

5 Solving a Rational Equation
Example 1 Solve the equation: 3 π‘₯βˆ’2 = 1 π‘₯βˆ’1 + 7 π‘₯βˆ’1 (π‘₯βˆ’2) Step 1: State the excluded values for x. Step 2: Find the LCM of all the denominators. Step 3: Multiply both sides by the LCM. Step 4: Simplify the denominators. Step 5: Solve for π‘₯. Reject some of the answers.

6 Solving a Rational Equation
Example 2 Solve the equation: 3 π‘₯ + 2 π‘₯+1 = 23 π‘₯ 2 +π‘₯ Step 1: Factorize the denominator Step 2: State the excluded values for x. Step 3: Find the LCM of all the denominators. Step 4: Multiply both sides by the LCM. Step 5: Simplify the denominators. Step 6: Solve for π‘₯.

7 Solving a Rational Equation
Example 3 Solve the equation: 4 π‘₯βˆ’2 βˆ’ 2 π‘₯ = 14 π‘₯ 2 βˆ’2π‘₯ Step 1: Factorize the denominator Step 2: State the excluded values for x. Step 3: Find the LCM of all the denominators. Step 4: Multiply both sides by the LCM. Step 5: Simplify the denominators. Step 6: Solve for π‘₯.

8 Solving a Rational Equation
Example 4 Solve the equation: 4 π‘₯ 2 βˆ’6π‘₯+8 = 3π‘₯ π‘₯βˆ’2 + 2 π‘₯βˆ’4 Step 1: Factorize the denominator Step 2: State the excluded values for x. Step 3: Find the LCM of all the denominators. Step 4: Multiply both sides by the LCM. Step 5: Simplify the denominators. Step 6: Solve for π‘₯.

9 Solving a Rational Equation
Example 4 Solve the equation: 4 π‘₯ 2 βˆ’6π‘₯+8 = 3π‘₯ π‘₯βˆ’2 + 2 π‘₯βˆ’4 4 (π‘₯βˆ’2)(π‘₯βˆ’4) = 3π‘₯ π‘₯βˆ’2 + 2 π‘₯βˆ’4 𝐿𝐢𝑀=(π‘₯βˆ’2)(π‘₯βˆ’4) Multiply (π‘₯βˆ’2)(π‘₯βˆ’4) to both sides (π‘₯βˆ’2)(π‘₯βˆ’4) βˆ™ 4 π‘₯βˆ’2 π‘₯βˆ’4 = 3π‘₯ π‘₯βˆ’2 βˆ™ π‘₯βˆ’2 π‘₯βˆ’4 + 2 π‘₯βˆ’4 βˆ™ π‘₯βˆ’2 π‘₯βˆ’4 4=3π‘₯ π‘₯βˆ’4 +2(π‘₯βˆ’2) 4=3π‘₯ π‘₯βˆ’4 +2 π‘₯βˆ’2 4=3 π‘₯ 2 βˆ’12π‘₯+2π‘₯βˆ’4 3 π‘₯ 2 βˆ’10π‘₯βˆ’8=0 3π‘₯+2 π‘₯βˆ’4 =0 π‘₯=βˆ’ π‘œπ‘Ÿ π‘₯=4 (π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘–π‘’π‘‘) π‘₯β‰ 2,4

10 Solving a Rational Equation
Example 5 Solve the equation: 2π‘₯ π‘₯ π‘₯βˆ’6 = 27 π‘₯ 2 βˆ’3π‘₯βˆ’18 Step 1: Factorize the denominator Step 2: State the excluded values for x. Step 3: Find the LCM of all the denominators. Step 4: Multiply both sides by the LCM. Step 5: Simplify the denominators. Step 6: Solve for π‘₯.

11 Solving a Rational Equation
Example 5 Solve the equation: 2π‘₯ π‘₯ π‘₯βˆ’6 = 27 π‘₯ 2 βˆ’3π‘₯βˆ’18 2π‘₯ π‘₯ π‘₯βˆ’6 = 27 (π‘₯βˆ’6)(π‘₯+3) 𝐿𝐢𝑀=(π‘₯βˆ’6)(π‘₯+3) Multiply (π‘₯βˆ’6)(π‘₯+3) to both sides (π‘₯βˆ’6)(π‘₯+3)βˆ™ 2π‘₯ π‘₯ π‘₯βˆ’6 βˆ™(π‘₯βˆ’6)(π‘₯+3)= 27 (π‘₯βˆ’6)(π‘₯+3) βˆ™(π‘₯βˆ’6)(π‘₯+3) 2π‘₯ π‘₯βˆ’6 +3 π‘₯+3 =27 2 π‘₯ 2 βˆ’12π‘₯+3π‘₯+9=27 2 π‘₯ 2 βˆ’9π‘₯βˆ’18=0 2π‘₯+3 π‘₯βˆ’6 =0 π‘₯=βˆ’ π‘œπ‘Ÿ π‘₯=6 (π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘) π‘₯β‰ βˆ’3,6

12 Solving a Rational Equation
Example 6 Solve the equation: βˆ’ 12 π‘₯ 2 +6π‘₯ = 2 π‘₯+6 + π‘₯βˆ’2 π‘₯ Step 1: Factorize the denominator Step 2: State the excluded values for x. Step 3: Find the LCM of all the denominators. Step 4: Multiply both sides by the LCM. Step 5: Simplify the denominators. Step 6: Solve for π‘₯.

13 Solving a Rational Equation
Example 6 Solve the equation: βˆ’ 12 π‘₯ 2 +6π‘₯ = 2 π‘₯+6 + π‘₯βˆ’2 π‘₯ βˆ’ 12 π‘₯(π‘₯+6) = 2 π‘₯+6 + π‘₯βˆ’2 π‘₯ 𝐿𝐢𝑀=π‘₯(π‘₯+6) Multiply π‘₯(π‘₯+6) to both sides π‘₯(π‘₯+6)βˆ™βˆ’ 12 π‘₯(π‘₯+6) = 2 π‘₯+6 βˆ™π‘₯(π‘₯+6)+ π‘₯βˆ’2 π‘₯ βˆ™π‘₯(π‘₯+6) βˆ’12=2π‘₯+ π‘₯βˆ’2 π‘₯+6 βˆ’12=2π‘₯+ π‘₯ 2 +4π‘₯βˆ’12 π‘₯ 2 +6π‘₯=0 π‘₯ π‘₯+6 =0 π‘₯=0 π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘ π‘œπ‘Ÿ π‘₯=βˆ’6(π‘Ÿπ‘’π‘—π‘’π‘π‘‘π‘’π‘‘) No solutions π‘₯β‰ 0,βˆ’6

14 Practice from Textbook
Textbook A58 Q.45, 46, 77, 78 Once you are done, you can work on A58 Q. 79, 80 and check the answer at AN120 (A.6) middle Work at your own pace: Review on Add/Subtract RE: A40 Example 8, 9 Textbook A44 Q.61-68 Review on Complex Fractions: A41 Example 10, 11 Textbook A45 Q.73-80


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