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9.6 Solving Rational Equations

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1 9.6 Solving Rational Equations
4/30/2014

2 Vocabulary 2 3 = 6 9 If π‘Ž 𝑏 = 𝑐 𝑑 Then aβˆ™π‘‘=π‘βˆ™π‘ Then 2βˆ™9=3βˆ™6 18 = 18
Rational Equation: Equation that shows two rational expressions or fractions are equal. 2 3 = 6 9 Then 2βˆ™9=3βˆ™6 18 = 18 Example: If π‘Ž 𝑏 = 𝑐 𝑑 Then aβˆ™π‘‘=π‘βˆ™π‘ Cross Multiply:

3 Example 1 Solve for x: = π‘₯ 7 3βˆ™7=4βˆ™π‘₯ 21=4π‘₯ 4 21 4 =π‘₯

4 Example 2 2(π‘₯+3)=3βˆ™π‘₯ 2π‘₯+6=3π‘₯ βˆ’2π‘₯ βˆ’2π‘₯ 6=π‘₯ 2 3 = 6 9 2βˆ™9=3βˆ™6 18 = 18
Solve for x: = π‘₯ π‘₯+3 To check: = 6 6+3 2 3 = 6 9 2βˆ™9=3βˆ™6 18 = 18 2(π‘₯+3)=3βˆ™π‘₯ 2π‘₯+6=3π‘₯ βˆ’2π‘₯ βˆ’2π‘₯ 6=π‘₯

5 Example 3 2 π‘₯βˆ’2 =π‘₯ π‘₯βˆ’2 2π‘₯βˆ’4= π‘₯ 2 βˆ’2π‘₯ βˆ’2π‘₯+4 βˆ’2π‘₯+4 0= π‘₯ 2 βˆ’4π‘₯+4
Solve for x: 2 π‘₯βˆ’2 = π‘₯ π‘₯βˆ’2 Or if you just look at the problem you can easily see that for the 2 fractions to be equal x must be 2! 2 π‘₯βˆ’2 =π‘₯ π‘₯βˆ’2 2π‘₯βˆ’4= π‘₯ 2 βˆ’2π‘₯ βˆ’2π‘₯ βˆ’2π‘₯+4 0= π‘₯ 2 βˆ’4π‘₯+4 0=(π‘₯βˆ’2)(π‘₯βˆ’2) 2=π‘₯ 2 is called an Extraneous solution because it leads to a division by 0 in the original equation. Always check for extraneous solutions! If x = 2, what happens to the denominator? Yes, it becomes 0 and you cannot divide by 0. So there is NO SOLUTION!

6 Example 4 π‘₯ π‘₯βˆ’1 =2 π‘₯+5 π‘₯ 2 βˆ’π‘₯ =2π‘₯+10 βˆ’2π‘₯βˆ’10 βˆ’2π‘₯βˆ’10 π‘₯ 2 βˆ’3π‘₯βˆ’10=0
Solve for x: π‘₯ π‘₯+5 = 2 π‘₯βˆ’1 To check: = 2 5βˆ’1 5 10 = 2 4 20=20 π‘₯ π‘₯βˆ’1 =2 π‘₯+5 π‘₯ 2 βˆ’π‘₯ =2π‘₯+10 βˆ’2π‘₯βˆ’10 βˆ’2π‘₯βˆ’10 π‘₯ 2 βˆ’3π‘₯βˆ’10=0 (π‘₯βˆ’5)(π‘₯+2) =0 π‘₯= 5,βˆ’2 To check: βˆ’2 βˆ’2+5 = 2 βˆ’2βˆ’1 βˆ’2 3 = 2 βˆ’3

7 Now we’re going to solve rational equations where one side contains addition or subtraction. Ex. 2 π‘₯ π‘₯ =2 How? By multiplying each term by the least common denominator (LCD). That will get rid of all fractions!!!

8 Example 5 2 π‘₯ 2 + 3 π‘₯ =2 2+3π‘₯= 2π‘₯ 2 βˆ’2βˆ’3π‘₯ βˆ’3π‘₯βˆ’2 0 = 2π‘₯ 2 βˆ’3π‘₯βˆ’2
Solve for x: π‘₯ π‘₯ =2 What’s the LCD? LCD: π‘₯ 2 2 π‘₯ π‘₯ =2 2+3π‘₯= 2π‘₯ 2 βˆ’2βˆ’3π‘₯ βˆ’3π‘₯βˆ’2 0 = 2π‘₯ 2 βˆ’3π‘₯βˆ’2 0=(π‘₯βˆ’2)(2π‘₯+1) βˆ’ 1 2 , 2=π‘₯ β€’π‘₯ 2 β€’π‘₯ 2 β€’π‘₯ 2 1 2(-2) = 2 2 -4 1 -2 -3

9 Example 6 Solve for x: 8+ 2 π‘₯βˆ’1 = 2π‘₯ π‘₯βˆ’1
8+ 2 π‘₯βˆ’1 = 2π‘₯ π‘₯βˆ’1 8(π‘₯βˆ’1)+ 2 π‘₯βˆ’1 (π‘₯βˆ’1)= 2π‘₯ π‘₯βˆ’1 (π‘₯βˆ’1) 8π‘₯βˆ’8+2=2π‘₯ 8π‘₯βˆ’6=2π‘₯ βˆ’2π‘₯+6 βˆ’2π‘₯+6 6π‘₯=6 π‘₯=1 LCD: π‘₯βˆ’1 Multiply each term by LCD Distributive Prop No Solution

10 Example 7 Solve for x: 2 π‘₯βˆ’1 + 3 π‘₯ =2
LCD: π‘₯(π‘₯βˆ’1) Multiply each term by LCD Distributive Prop Factor using Big X

11 Homework: 9.6 p.503 #21-41 odd only Since you can check your answers in the back of the book (and you should), you must show work!!! β€œI’m glad I know sign language, it’s pretty handy”


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