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Question 73631This question is from textbook
: Please show the steps needed to solve this system of linear equation
2(y-3)=x+3y
x+2=3-y
thanks
This question is from textbook
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! 2(y-3)=x+3y
x+2=3-y
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Let's first find out what we are dealing with by arranging each of the equations so that
it is in the standard form of ax + by = c
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Multiply out the left side of the top equation to get:
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2y - 6 = x + 3y
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Move the x and y terms to the left side using the method of subtraction to eliminate
them from the right side. First subtract the x from the right side, but to do so you must
also subtract x from the left side. This subtraction results in:
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-x + 2y - 6 = 3y
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Next subtract the 3y from both sides to accomplish the same thing. When you subtract
3y from the left side it combines with the 2y that is already there to just give -y. As a
result of subtracting 3y from both sides the equation becomes:
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-x - y - 6 = 0
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Then move the 6 to the other side of the equal sign by adding +6 to both sides:
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-x - y = +6
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I have a personal preference to working with positives as much as possible on the left
side. You don't have to do this next thing, but multiply all the terms on both sides of
this equation by -1 to get:
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+x + y = -6
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There you are. This equation is now in standard form. Now do the same thing to the second
(or bottom) equation that you were given.
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x + 2 = 3 - y
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Move the -y to the left side by adding +y to both sides. On the right side adding
+y to the -y just cancels it out of existence. After you add the +y to the left side the
equation has become:
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+x + y + 2 = 3
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Now subtract 2 from the left side to remove the +2. Of course, you must then subtract
2 from the right side also. When you do this subtraction you get:
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+x + y = 1
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At this point let's list our two equations and see what we are working with:
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+x + y = -6 and
+x + y = 1
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Wait a second. How can x+y = -6 and the very same x+y also = 1? The answer is that you
can't. There is no common solution for these two equations. What you have here is two
equations whose graphs are parallel lines. By setting x equal to zero in both of these
equations you get two different answers ... y = -6 from the top equation and y = 1 from
the bottom equation. But when x is zero, the corresponding value of y has to be on the y-axis.
Therefore, we can say that the graph for the first equation crosses the y-axis at -6
[the point (0,-6)] and the graph for the second equation the graph crosses the y-axis at +1
[the point (0, +1)].
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You can do a similar thing for the x-axis. Set y equal to 0 in the first equation and you
find that the corresponding value of x is x = -6 [the point being (-6,0)]. And in the second
equation when y = 0, the corresponding value of x is x = 1 [the point being (1,0)].
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Now you can figure out what the graphs look like. Two points on the graph for the top
equation are (0,-6) and (-6,0). This graph has a negative slope (goes down as you move
toward the right). If you care to do the work, the slope is -1.
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Similarly, two points on the graph for the second equation are (0,1) and (1,0). It has
a slope of -1.
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Since the slopes of the graphs are both -1, the lines are parallel. To have a common
solution, the slopes must be different so that the lines cross at some point. That point
would consist of the x and y values that cause both equations to be satisfied.
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More than you might have wanted to know about this problem, but all the ideas discussed
here are important to your understanding of linear equations. Hope it helped.
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