Equations with FractionsDate: 04/21/97 at 21:15:10 From: Alisia Fu'e Subject: Adding fractions How do find the common denominator for a fraction that has a variable in it? This is what I know so far: To solve the equation x/2 + x/3 = 1/6, first find the common denominator. After finding the commom denominator, distribute it to both sides of the equation: 6(x/2 + x/3) = 6(1/6) Right? If so, what next? Thanks, Alisia Fu'e Date: 05/03/97 at 16:50:25 From: Doctor Kathryn Subject: Re: adding fractions Hi Alisia, You have the right idea to start, but do you understand WHY you need to find a common denominator? Whether trying to solve an equation or simplify an expression, you need to turn "unlike fractions" into "like fractions." Fractions can be thought of as parts/whole. So 1/2 is one part of the two parts that make the whole. Similarly, 5/8 is five parts of the eight parts that make a whole. Improper fractions can also be thought of the same way. 9/4 is nine parts of the four parts that make a whole. So that must mean that there are two wholes and one part of four left over: 2 1/4! Now, your question: x/2 + x/3 = 1/6 You need a common denominator so that you can combine the two fractions. How can you know how many pieces (OF EQUAL SIZE) you have if some are halves and some are thirds? You need to find a common denominator, which is 6. Your shortcut works, but let's see why: First let's change each fraction so that all have the same denominator (I often think that the purpose of getting a common denominator in equations is so that you can get rid of it. Watch!): x/2 = 3x/6 x/3 = 2x/6 1/6 = 1/6 So, we have: 3x/6 + 2x/6 = 1/6 Now we want to get rid of the denominator so that the equation is easier to solve. This means that we multiply everything by the denominator (note that this does not change the equation): 6(3x/6 + 2x/6 = 1/6) Distribute the 6: 6(3x/6) + 6(2x/6) = 6(1/6) Simplify each fraction (note that the 6's cancel): 3x + 2x = 1 Combine like terms: 5x = 1 Isolate the variable by dividing by 5: x = 1/5 I hope this helps! -Doctor Kathryn, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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