Unformatted text preview: of the remaining variable.
In words this method is not always very clear. Let’s work a couple of examples to see how this
method works. Example 1 Solve each of the following systems.
3x  y = 7
(a)
[Solution]
2x + 3 y = 1
(b) 5x + 4 y = 1
[Solution]
3x  6 y = 2 Solution
(a) 3x  y = 7
2x + 3 y = 1 So, this was the first system that we looked at above. We already know the solution, but this will
give us a chance to verify the values that we wrote down for the solution.
Now, the method says that we need to solve one of the equations for one of the variables. Which
equation we choose and which variable that we choose is up to you, but it’s usually best to pick
an equation and variable that will be easy to deal with. This means we should try to avoid
fractions if at all possible. © 2007 Paul Dawkins 317 http://tutorial.math.lamar.edu/terms.aspx College Algebra In this case it looks like it will be really easy to solve the first equation for y so let’s do that. 3x  7 = y Now, substi...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.
 Spring '12
 MrVinh

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