SOLUTION: Solve the system by graphing. 4x+2y=10 y=4

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Question 73874: Solve the system by graphing.
4x+2y=10
y=4

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
4x+2y=10
y=4
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To solve this system by graphing, you draw the graphs of the two equations and look for
the point where the two graphs cross. At that crossing point the value of x and the value
of y are the values that make both of these equations "work."
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Let's talk a little bit about the graphs of these two equations before we take a look at the
the graphs.
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We can use some math analysis on the first equation to quickly decide what its graph is.
like. Let's set x equal to zero in this equation. If we do, what is the value of y? With
x equal to zero we are left with 2y = 10 and after dividing both sides of this by 2 we
have y = 5. But where is the point (0,5)? It's on the y-axis 5 units up from the origin!
Easy to locate it.
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Still on the first equation. Let's now set the value of y equal to zero. This reduces the
equation to 4x = 10 and after dividing both sides by 4 we get x = 2.5. Where is the point
(2.5, 0) and it's on the x-axis 2.5 units to the right of the origin. We now have two
points on the graph of the first equation. We can plot them both and draw a line through them
to obtain the graph.
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Now what about the second graph, the graph of y = 4? This equation says to us that no
matter what value we assign to x, the value of y is always +4. So the points (-5,4), (0,4),
10,4) are all points on this graph because they all have y values equal to 4. If you can
picture this, you can see that the graph of y = 4 is a horizontal line through the point +4
on the y-axis.
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Wherever the graphs cross, the intersection point will have to have a value of y = 4 because
of the second graph. So look for the crossing point and get the corresponding value of
x at that point.
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The graphs should look like this:
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graph+%28500%2C500%2C-8%2C8%2C-8%2C8%2C-2x%2B5%2C4%29
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Hope this helps you to understand the use of graphing to solve linear equation sets.
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