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Mathematics 18 Online
OpenStudy (anonymous):

Can someone please help me with this?

OpenStudy (anonymous):

-\[-\frac{ 1 }{ 4 }x + \frac{ 1 }{ 10 }y = \frac{ 2 }{ 5 }\]

OpenStudy (anonymous):

ok

OpenStudy (unklerhaukus):

If you multiply both sides of the equation by the lowest common multiple of those denominators {4, 5, 10}, all those fractions will go away

OpenStudy (anonymous):

The question wants me to match the equation to it's graph. Here are the graphs

OpenStudy (anonymous):

Isn't the LCM 20?

OpenStudy (unklerhaukus):

if you rearrange the equation to the form \[y = mx+b\] you can compare the slope \(m\), and y-intercept \(b\) of the equation with those plots

OpenStudy (unklerhaukus):

yeah, multiply both sides of the equation by 20

OpenStudy (anonymous):

So (20)-1/4x + (20)1/10y = (20)2/5 hold on im working it on paper

OpenStudy (danjs):

Let x = 0 , solve for y to get the point (0,y) Let y = 0, Solve for x to get the point (x,0) Those are the x and y axis intercepts

OpenStudy (danjs):

Or unkerhaukus way is fine too

OpenStudy (anonymous):

Okay I got -5x + 2 = 8 by multiplying it all by 20

OpenStudy (unklerhaukus):

i think you mean\[-5x + 2y = 8\] now can you arrange this into \(y=mx+b\) form?

OpenStudy (danjs):

If x=0, y = 20/5 or y=4 so The point x = 0 y=4 is on the line, (0,4) Now let y = 0 and solve for x to get the point (x,0)

OpenStudy (anonymous):

ok hold on

OpenStudy (danjs):

If y = 0, you get \[\frac{ -1 }{ 4 }x + 0 = \frac{ 2 }{ 5 }\] solve for x

OpenStudy (danjs):

That is your second point(x-intercept) y=0, x=? (x,0)

OpenStudy (anonymous):

The only graphs with a point at (0,4) have 1.5 or -1.5 as the x

OpenStudy (danjs):

yes, which one of those works with the equation when y = 0? (-1/4)x + 0 = 2/5

OpenStudy (danjs):

\[x = \frac{ 2 }{ 5 }*\frac{ -4 }{ 1 }\]

OpenStudy (danjs):

Graph 2 looks like it intersects the x-axis at -8/5 or about - 1.6

OpenStudy (danjs):

Just remember, if you need to choose a graph, the quickest way to do it, is just let x=0 and solve for y, then let y=0 and solve for x. Those 2 points (0 , y) and (x , 0) will define the line for you.

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