Solutions to Equations
Date: 01/16/97 at 19:38:59 From: Judy Holbrook Subject: Pre-algebra My problem is in pre-algebra. It is on equalities. I can do most of the ones that were assigned except for this one: (3/4)y = -x Here is one like it I have already done and know it is right: 3y = -7x. You make a table similar to this showing the results: x y 0 0 7 -3 -7 3 Thanks!
Date: 01/20/97 at 04:05:02 From: Doctor Wallace Subject: Re: Pre-algebra Hi Judy! You have been given an equation in two variables (x and y) and need to come up with some solutions for x and y. Since your example problem has three listed solutions, I'm assuming you need only three. (Do you know how many total solutions there are to a problem like this one?) While we're on the subject of your sample problem, I think you got the x and y values reversed. You mean that for 3y = -7x the values should be as follows: x y 0 0 -3 7 3 -7 Okay, now for (3/4)y = -x. This problem will be solved in exactly the same way as you solved 3y = -7x. Choose any x value that you like. Any one at all. Plug it into the equation where you see the x. Then, solve for y. For example, plug in 0. That's always pretty easy to work out. If we plug 0 in for x, what do we get for y? We would have (3/4)y = -0 which is (3/4)y = 0. This means that y must also be 0, since 0 is the only number we could multiply (3/4) by to get 0. So one solution is x=0, y=0. Let's try another one. Choose a y value that you like. Any one at all. Plug it into the equation where you see the y. Then, solve for x. (This will also find a solution. You may choose a value for either one, and then solve the equation to find the other.) Let's choose 4 for y. Plugging it in, we get: (3/4) times 4 = -x, which gives 3 = -x Now we multiply by -1, to get our answer: -3 = x. So our solution is x=-3, y=4. Notice that choosing 4 for y made simplifying the left side of the equation easier, since 4 is the denominator of 3/4. We could have chosen any number for y, say 13, but then we'd have had 39/4 on that side. You can save yourself a lot of extra work by a wise choice of x or y. I hope this helps. Since you weren't specific on exactly where you were having trouble solving this problem, I'm not sure I've answered you very well. If not, write back! -Doctor Wallace, The Math Forum Check out our web site! http://mathforum.org/dr.math
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum