Solve for x; Find 5 Ordered Pairs
Date: 11/02/2002 at 13:09:50 From: Rachel Brock Subject: Two questions I have two questions. The first one reads: Solve for x. 3x/4 - 4 = 7 + x/2 The second one reads: Determine 5 ordered pairs that will satisfy the equation 4y = 8 - 2x. I am really stuck and I appreciate the way you explain things instead of making me feel stupid. You are wonderful and I am so glad my sister recommended you. Rachel
Date: 11/02/2002 at 14:35:13 From: Doctor Ian Subject: Re: Two questions Hi Rachel, Let's look at your second question first. An ordered pair is a pair of values for x and y that satisfy some equation. Suppose we have an equation like 6y = 9 - 3x Now, we could _choose_ to give x a value like x=1. If we do that, what happens to the equation? 6y = 9 - 3*1 6y = 9 - 3 6y = 6 Now, for _that_ choice of x, the value of y is determined: it can only be y=1. So the values x=1 and y=1 are one possible solution to the equation, and we can write that as an ordered pair: (x=1,y=1), or just (1,1). But what about other choices? What if we give x the value x=2? Now we have 6y = 9 - 3*2 6y = 9 - 6 6y = 3 Now the corresponding value of y must be y=1/2, so (2,1/2) is another ordered pair that satisfies the equation. It's always a good idea to check your work by substituting _both_ values into the equation, to see if you end up with something true: 6(1/2) = 9 - 3(2) 3 = 9 - 6 3 = 3 (Okay) Note that if you substitute two incompatible values, you'll end up with something that doesn't make sense. For example, let's try the ordered pair (5,4): 6(4) = 9 - 3(5) 24 = 9 - 15 24 = -6 (Oops!) When you get something like this, you need to go back and find out what you did wrong. When you have two variables but only one equation, you can have an infinite number of ordered pairs like this. Note that you don't have to choose an x and figure out a y. You can choose a y and figure out an x if that seems easier. More importantly, you can 'solve for' either variable in terms of the other, which trades a little work up front for an easier time later. In this case, we can divide everything on both sides of the equation by 6 to get (6y)/6 = (9 - 3x)/6 y = (9 - 3x)/6 Now we can just plug in a value for x, and the corresponding value of y pops out: y = (9 - 3*3)/6 = 0/6 = 0 y = (9 - 3*4)/6 = -3/6 = -1/2 and so on. It's like turning a crank. Which sort of leads us back to your first question. To 'solve for' a variable, you look for things that you can do to both sides of an equation to make it look like some variable = some expression not involving the variable What kinds of things can you do? Basically, anything except divide by zero! In practice, the three main things you do are add the same thing to both sides of an equation; multiply both sides of an equation by the same thing; and use the distributive property to combine or split things up. For example, if you have something like 3x = 4y - 2x you can add 2x to both sides of the equation to get 3x + 2x = 4y - 2x + 2x (3+2)x = 4y 5x = 4y which is close to solving for x, but there is one more step: multiply both side of the equation by (1/5) to leave x by itself on one side: (1/5)5x = (1/5)4y x = 4y/5 Note that we could have solved for y instead, by making the last step (1/4)5x = (1/4)4y (5/4)x = y The earlier steps are still necessary: 1. Get all the terms involving the variable you want to solve for on one side of the equation. 2. Simplify that side of the equation to get something that looks like this*variable + that = .... 3. Add (-that) to both sides of the equation. 4. Multiply both sides of the equation by (1/this). If this seems new or strange to you, you might want to look at this: Basic Tips on Solving for X http://mathforum.org/library/drmath/view/57265.html Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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