So lets first multiply the second equation by two x 2

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Unformatted text preview: Finally, plug this into either of the equations and solve for x. We will use the first equation this © 2007 Paul Dawkins 320 http://tutorial.math.lamar.edu/terms.aspx College Algebra time. 2 x + 4 ( -4 ) = -10 2 x - 16 = -10 2x = 6 x=3 So, the solution to this system is x = 3 and y = -4 . [Return to Problems] There is a third method that we’ll be looking at to solve systems of two equations, but it’s a little more complicated and is probably more useful for systems with at least three equations so we’ll look at it in a later section. Before leaving this section we should address a couple of special case in solving systems. Example 3 Solve the following systems of equations. x- y =6 -2 x + 2 y = 1 Solution We can use either method here, but it looks like substitution would probably be slightly easier. We’ll solve the first equation for x and substitute that into the second equation. x = 6+ y -2 ( 6 + y ) + 2 y = 1 -12 - 2 y + 2 y = 1 - 12 = 1 ?? So, this is clearly not true and there doesn’t appear to be a mistake anywhere in our work. So, what’s the problem? To see let’s graph these two lin...
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