Solving an Equation with FractionsDate: 1/30/96 at 21:4:44 From: Mongolian Gypsy Subject: pre algebra I have spent the last half hour on this problem: v = 6 - 5/4 v I don't know if the question is asking 6 minus 1 1/4 or 6 * 1 1/4. Please solve for v. I am in seventh grade and these problems are starting to haunt me. Amy Steiner email: sfduq@scfn.thpl.lib.fl.us Date: 1/31/96 at 12:7:40 From: Doctor Elise Subject: Re: pre algebra Hi! These problems will haunt you for many years yet to come, too! :-) What's called the "order of operations" takes a little getting used to. Basically, unless you have parentheses in your equation that tell you how to group things, you do all the multiplication and division in the problem before any addition and subtraction. Here are some examples: 6 - 5/4 v Here there are no parentheses, so you do multiplication and division first. I'm being told to take 5, divide by 4, multiply the result by v, and subtract the whole thing from 6. (6 - 5/4) v Here the parentheses tell me to take 5, divide it by 4, subtract the result from 6, and then multiply the whole thing by v. (6 - 5)/4 v Here I'm being told to subtract 6 from 5, then divide the result by 4, and multiply the whole thing by v. Your problem reads: v = 6 - 5/4 v If I add parentheses to show what this really means, I have: v = 6 - ((5/4) * v) In words, 'v' equals 6, minus 1 1/4 times 'v' To solve it you have to put all the 'v's on one side and all the numbers on the other. (Actually, that pretty much sums up Algebra. Any more questions? Just kidding!) To solve it, I can do anything I want to this equation, as long as I do the same thing on both sides of the equals sign. So I'll add (5/4) * v to both sides to put all the 'v's on the same side: v + (5/4)v = 6 - (5/4)v + (5/4)v v + (5/4)v = 6 (1 + 1 1/4)v = 6 (2 1/4) v = 6 (9/4)v = 6 Now I want to put all the numbers on the other side of the 'v', so I'll multiply both sides by 4/9: (4/9) * (9/4) v = 6 * (4/9) v = 6 * (4/9) v = (6 * 4) / 9 v = 24/9 v = 8/3 v = 2 2/3 Does this help? -Doctor Elise, The Math Forum |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/