Choose the system of equations that matches the following graph: https://lss.brainhoney.com/Resource/22181743,AD2,0,C,3,0/Assets/72257_53b5b170/02_13_mod_2_test_bankg_23.gif
A. 5x + 2y = –14 7x + y = 7 B. 5x + 2y = –14 7x – y = 7 C. 5x – 2y = –14 7x + y = 7 D. 5x – 2y = –14 7x – y = 7
is that the whole picture?
yeah
@amistre64 Can you check my answer? I think it's D
looks right
Plug in (0,-7) into the equations and see for which answer choice does it work in both equations.
Yeah ok that's what I did and I got D. Thanks! :)
keep in mind that there are many lines that can pass thru that point; so plugging in might not be a sufficient measure in this case.
@TheSmartOne wait so I can't plug in to find the answer? Is D right?
D is wrong.
@amistre64 There are two lines and they both intersect at (0,-7) so plugging in that point to test which 2 lines in the answer choices work it the logical way to solve it.
Or you can graph the two lines in each answer choice, but that would take longer.
how do we know they gave the equations of the 2 lines correctly? after all, each option has a different setup. |dw:1428862653439:dw|
suppose we are given y = L1 and y=L2 in one option then y = L'1 and y=L2 in another since they all have a common point, the point satisfies both options
They graphed the two lines and asked us what are the two lines :P
yes, but not all the options define the lines correctly; but some option MIGHT give us a setup where more than one option is true for a given point.
in this case: plugging in MIGHT not work out if we dont know the equation of the lines to start with
I think it would be good if you changed the equations to y=mx+b and check with this website: https://www.desmos.com/calculator to make sure. Or if you have a graphic calculator with you, you can graph them there.
You don't even need to change it in the format of y=mx+b It still graphs it as the equation is already :P
|dw:1428863008091:dw| spose L1 and L2 are given,we know they have a common point for a solution we also know that for some ungraphed line L'1, we have the same solution option A: y=L1 and y=L2 option B: y=L'1 and y=L2 we plug in the common point, and both options seem valid, but unless we know the properties of the lines we cant know for certain which is valid by just plugging in the common point.
now, if only one option gives a true solution, then yeah, its that one :) but im just demonstrating that plugging in is not always the best option here.
lets give option C: y=L1 and y=L'1 this also satisfies the common point .... now all 3 options seem equally valid.
oh, i mistype and said D worked didnt i ..... my brain said one thing my fingers typed another . hence the diatribe lol
@amistre64 So it is D?
You seem to only want the final answer.
And D is wrong.
Oh...but @amistre64 said it was right?
5x + 2y = –14 7x + y = 7 @amistre64 is this the answer?
"oh, i mistype and said D worked didnt i" ~Amistre64
5x + 2y = –14 7x + y = 7 so is this the answer?
Nope.
5x – 2y = –14 7x + y = 7 ?
So we are just going to guess?
How did you arrive to the answer being D?
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