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Mathematics 21 Online
OpenStudy (anonymous):

x+2y=4 How would i solve that and put it on a graph?

OpenStudy (anonymous):

Isolate y first.

OpenStudy (anonymous):

So it would be in the form y=mx + b, where m is the slope and b is the y-intercept, aka where the line hits the y-axis.

OpenStudy (anonymous):

so i would x+2y=4 -x -x 2y=4-x?

OpenStudy (anonymous):

No, you can't have any number in front of the y.

OpenStudy (anonymous):

Divide everything by 2.

OpenStudy (anonymous):

so /2 from y?

OpenStudy (anonymous):

so x+y=4

OpenStudy (anonymous):

no...

OpenStudy (anonymous):

You can't just divide one side by 2, because it's no longer equal.

OpenStudy (anonymous):

Divide both sides by 2, and leave y isolated.

OpenStudy (anonymous):

so x+y=2?

OpenStudy (anonymous):

What's 1/2(2y) = 1/2(4-x)?

OpenStudy (anonymous):

2=4-x?

OpenStudy (anonymous):

What is 1/2 times 2y?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

It's y.

OpenStudy (anonymous):

So then, what is 1/2*(4-x)?

OpenStudy (anonymous):

8-x?

OpenStudy (anonymous):

Look at it this way: what is (1/2)*4 -(1/2)*x?

OpenStudy (anonymous):

8-2x?

OpenStudy (anonymous):

what is 4 divided by 2.

OpenStudy (anonymous):

2

OpenStudy (anonymous):

what is x divided by 2.

OpenStudy (anonymous):

2x

OpenStudy (anonymous):

no, that's 2 times x.

OpenStudy (anonymous):

what is x divided by 2.

OpenStudy (anonymous):

x over 2?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

therefore, 1/2(4-x) is...

OpenStudy (anonymous):

2- x over 2

OpenStudy (anonymous):

yes. So the complete equation would be y=-(1/2)x + 2. From y = mx + b, we know that the line passes through the point (0, 2) with a slope of -1/2.

OpenStudy (anonymous):

You should be able to graph it now.

OpenStudy (anonymous):

ok the equation is y=2-x 2

OpenStudy (anonymous):

Yes, the equation is y = 2 - x/2, or y = -(1/2)x + 2

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

Remember to close the question and give medals if you're done.

OpenStudy (anonymous):

ok

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