As we will see these can be fairly involved problems

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Unformatted text preview: will be a little more work that the method of substitution. To find y we need to substitute the value of x into either of the original equations and solve for y. Since x is a fraction let’s notice that, in this case, if we plug this value into the second equation we will lose the fractions at least temporarily. Note that often this won’t happen and we’ll be forced to deal with fractions whether we want to or not. æ1ö 3ç ÷ - 6 y = 2 è3ø 1- 6 y = 2 -6 y = 1 y=- 1 6 Again, this is the same value we found in previous example. [Return to Problems] (b) 2 x + 4 y = -10 6x + 3y = 6 In this part all the variables are positive so we’re going to have to force an opposite sign by multiplying by a negative number somewhere. Let’s also notice that in this case if we just multiply the first equation by -3 then the coefficients of the x will be -6 and 6. Sometimes we only need to multiply one of the equations and can leave the other one alone. Here is this work for this part. 2 x + 4 y = -10 6x + 3y = 6 ´ -3 uuuuu r same uuuuu r -6 x - 12 y = 30 6x + 3 y = 6 - 9 y = 36 y = -4...
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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.

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