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I like to subtract multiples of the equations to solve for one of the variables, then back substitute.
Okay I have an example can you show me how to do it? y=2x+6 and 4x-2y=13
it's in my math book. umm,
ok umm, i'll do the second one for you. 4x-2y=13 4x=15y x=15y/4
i think that's right
Well here let me type it the right way I gues y=2x+6 4x-2y=13 now is that easier??
umm that's the same as before.
I know but it's grouped together how do they work together or correlate?
Certainly. \[y=2x+6\implies 2y = 4x + 12 \implies -4x + 2y = 12\] Since we have another equation with a -2y we can just add the left and right sides of both equations together \(-4x+2y=12\) +\(4x-2y=13\) \[\implies 0 = 25\] So there is no solution.
Thankyou sooo much!! I'm gonna work on another problem like that one and I need you to see if I'm right okay?
You can actually note that those two equations have no solution because the lines are parallel so they'll never cross eachother. That's what you're finding when you solve these systems is where the graphs intersect.
Ohh! Okay I gotcha now that makes more sence.
Now I'm officially lost on this equation: 4x+2y=-10 -6x-2y=12 anad I got -2x+y=2 ;what do I do from there??
So close! But 2y -2y = 0 not y
Well I know but I didnt know whether to put y or 0y...so do I keep going from there or??
0y is correct. but that's the same as 0 which is the same as not putting it at all.
So you have -2x = 2
What is x?
So substitute that in for x in either one of your original equations to solve for y.
3, so the answer would be (-1,3)
Or would it be -3???
(-1,-3) is correct.
Alright. Well, thankyou so much for helping me!(: that makes way more sence now