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Unformatted text preview: 5x + 4 y = 1
3x  6 y = 2 This is the system in the previous set of examples that made us work with fractions. Working it
here will show the differences between the two methods and it will also show that either method
can be used to get the solution to a system.
So, we need to multiply one or both equations by constants so that one of the variables has the
same coefficient with opposite signs. So, since the y terms already have opposite signs let’s work
with these terms. It looks like if we multiply the first equation by 3 and the second equation by 2
© 2007 Paul Dawkins 319 http://tutorial.math.lamar.edu/terms.aspx College Algebra the y terms will have coefficients of 12 and 12 which is what we need for this method.
Here is the work for this step. 5x + 4 y = 1 3x  6 y = 2 ´3
uuu
r
´2
uuu
r 15 x + 12 y = 3
6 x  12 y = 4
21x =7 So, as the description of the method promised we have an equation that can be solved for x.
Doing this gives, x = 1
which is exactly what we found in the previous example. Notice
3 however, that the only fraction that we had to deal with to this point is the answer itself which is
different from the method of substitution.
Now, again don’t forget to find y. In this case it...
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 Spring '12
 MrVinh

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