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graph of two lines, one with a positive slope and one with a negative slope, that intersect at the point (negative 1, 1)
\(y-2x=-1\) \(x+3y=4\) Multiply the second one by \(2\) and we get \(y-2x=-1\) \(2x+6y=8\) now add them together \(7y=7\) so that \(y=1\) and now we find \(x\). \(x+3(1)=4\implies x=1\)
\((-1,1)\) is not their intersection. Make sense?
give me a moment please
no rush m8
so its either the second or third one
at the intersection the (x, y) given (-1, 1) satisfies both equations
\((-1,1)\) Does NOT satisfy either equation @triciaal
graph of two lines, one with a positive slope and one with a negative slope, that intersect at the point (negative 1, 1) I misunderstood then so is the question incomplete?
Choose the correct graph of the given system of equations. y − 2x = −1 x + 3y = 4
I think those are his answers :)
i chose http://learn.flvs.net/webdav/assessment_images/educator_algebraI_v20/06_01_09b.gif
is it correct
@GiulianaRestrepo we can't see your choices. the solution is the graph where the (x, y) satisfies both equation
sketch each graph on the same grid and find the point where they meet
the first equation y = 2x -1 hint slope 2 and point (0, -1) is on that line the 2nd eqn 3y = -x + 4 slope negative and when y = 0 x = 4 so (4, 0) is on that line can you sketch each line and find (x, y) where they meet?
i already solved it thanks !
ok close the question.