Question 93716
A fairly straightforward and easy method of doing these two problems is to find three coordinate
pairs that solve each equation. After doing that, plot the points and then draw a straight
line through all three points.
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Take the first equation:
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{{{2x + 5y = 48}}}
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Suppose we set x equal to zero. Then the x term disappears and this equation reduces to:
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{{{5y = 48}}}
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Solve for y by dividing both sides by 5 to get that:
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{{{y = 48/5 = 9.6}}}
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So this means that the {x, y) point of (0, 9.6) is on the graph.
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Now lets return to the original equation and set y equal to zero. That makes the y term disappear
and the equation is reduced to:
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{{{2x = 48}}}
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Solve for x by dividing both sides of this equation by 2 to find that
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{{{x = 48/2 = 24}}}
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Therefore when x is 24, then y is 0 and this gives us that the (x, y) point (24, 0) is on the
graph.
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We can do this one more time to get a third point. This will serve as a good check since
all three points must be in a straight line or else one or more of our points is wrong.
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Return to the original equation and let's let x = 10. [We could choose any other number for
x, but I guess 10 is an easy number to work with.] When x is 10 the original equation becomes:
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{{{2*10 + 5y = 48}}}
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The 2 times 10 is 20 so the equation can be written as:
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{{{20 + 5y = 48}}}
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Subtract 20 from both sides to get rid of the 20 on the left side and the equation is
reduced to:
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{{{ 5y = 28}}}
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Solve for y by dividing both sides by 5 to get:
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{{{y = 28/5 = 5.6}}}
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This means that the (x, y) point (10, 5.6) is on the graph. 
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Plot the three points (0, 9.6), (24, 0) and (10, 5.6) and extend a straight line through them
to get the graph of the equation {{{2x+5y=48}}}. It should look like:
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{{{graph(300,300,-5,35,-5,15,(-2x+48)/5)}}}
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The second of your problems involves the equation {{{4x - 2y = 0}}}
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This equation can be graphed by selecting any three values of x and determining the corresponding
three values of y that will give three (x, y) points for the graph.  For example, if 
you set x = 0 then the term containing x goes to zero, and the equation is reduced to:
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{{{-2y = 0}}}
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If you divide both sides by -2 the equation becomes:
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{{{y = 0}}}
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Therefore, the (x, y) point of (0, 0) is on the graph meaning that the graph goes through 
the origin.
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Select another value for x ... say x = 2. If you substitute that value into the equation, it
becomes:
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{{{4*2 -2y = 0}}}
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Do the multiplication to get:
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{{{8 - 2y = 0}}}
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Subtract 8 from both sides to get rid of the 8 on the left side. This subtraction 
results in:
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{{{-2y = -8}}}
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and dividing both sides by -2 to solve for y leads to:
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{{{y = -8/-2 = +4}}}
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So the (x, y) point (2, 4) is on the graph.
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Finally let's set x = -2 and the equation becomes:
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{{{4*(-2) - 2y = 0}}}
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Do the multiplication and the equation simplifies to:
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{{{-8 - 2y = 0}}}
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Get rid of the -8 on the left side by adding 8 to both sides to get:
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{{{-2y = 8}}}
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Then solve for y by dividing both sides by -2 to get:
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{{{y = 8/(-2) = -4}}}
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So the (x, y) point (-2, -4) is on the graph.  
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Plot the three points (0, 0), (2, 4), and (-2, -4) and extend a straight line through them
to get the graph.  It should look like this:
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{{{graph(300,300,-8,8,-8,8,2x)}}}
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Hope these two exercises will show you one way of getting graphs from equations. That way
is assigning values to x (or y) and calculating the value of the corresponding coordinates so
that you can plot points on the graph and connect them to display an extended graph.