Where do the graphs of the following equations intersect: x-y=3 x=-2y-6 please someone who is actually willing to help me i struggle on this

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Where do the graphs of the following equations intersect: x-y=3 x=-2y-6 please someone who is actually willing to help me i struggle on this

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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ok. first you need to have one of the equations in terms of x or y. Lets take x-y=3. solve for x ineedyouubiebs
so it doesnt matter which one i start with?
no, it does not matter. but x-y=3 is easier to work with.

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Other answers:

Solve for y, x-y=3 -x -y=-x+3 times -1 y=x-3 x=-2y-6 -6 x-6=-2y divde buy -2 -1/2x+3=y or y=-1/2x+3
would it be (-2y-6)-y=3 ?? please correct me if im wrong
@ineedyouubiebs , that would be correct... ^^^
omgggg yesss!!
yes that is correct
@dpaInc and how would i add like terms to that...i mean would i have to distribute?
technically you are distributing a "1" in 1(-2y-6)-y=3 but since it is only a positive 1, just get rid of the parenthesis to get: -2y - 6 - y = 3 now combine like terms and solve....
y=-3
correct... now solve for x...
and how would i do that?
you have x = -2y - 6 from the beginning....
just plug in the value that you found for y: x = -2(-3) - 6
x=6-6
simplify that....
x=0
you're basically done... and whenever you need to see if your answer is correct or not, plug in the values of both x and y into BOTH equations. if they both turn out to be true statements, your answer is correct.
okay thank you soo much=D can you help me in another problem?
sure.... close this question and post a new one.... :)

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