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Mathematics 15 Online
OpenStudy (lovelycharm):

What is one of the solutions to the following system of equations? y2 + x2 = 53 y − x = 5

OpenStudy (tkhunny):

Have you considered substituting? y = x + 5 Go!

OpenStudy (lovelycharm):

i dont get ti @tkhunny

OpenStudy (loser66):

he said you plug y =x +5 to the first equation to solve for x

OpenStudy (lovelycharm):

is it -7

OpenStudy (loser66):

show your work, please

OpenStudy (tkhunny):

No, "it" is NOT "-7". 1) It doesn't mean anything. 2) -7 what? 3) Show your work. Did you do the substitution? Let's see it.

OpenStudy (lovelycharm):

y^2 + x^2 = 53 (x+5)^2 + x^2 = 53 x^2 + 5x + 5x + 25 + x^2 = 53 2x^2 + 10x + 25 = 53 2x^2 + 10x - 28 = 0 x^2 + 5x - 14 = 0 (x+7)(x-2) = 0

OpenStudy (lovelycharm):

and it iz -7 that wat i gettin

OpenStudy (tkhunny):

No, "it" is still not "-7". You must use language that can be understood. You have two solutions for x. x = -7 or x = 2 <== See how that makes sense and says something? However, you need a "solution to the system". A value for x doesn't cut it. What else do you need?

OpenStudy (lovelycharm):

i dont get it math iz confusing

OpenStudy (mathmale):

Sorry. I could accept your saying "I don't get it" once. But if you truly want to learn this material, you'll need to become involved: ask questions, try working the problem yourself (showing your work)... do something. You are given a set of equations: y2 + x2 =53\[y^2+x^2=53~and~y-x=5,\] y − x = 5 ...and you need to find a point (or perhaps two points) whose x- and y-coordinates satisfy BOTH equations. Note that y-x=5 is easily solved for y; all we have to do is to add x to both sides of this equation. Why would we do this? To be able to eliminate y in the other equation. 1) Please solve y-x=5 for y, now. Type your answer here. 2) Once you have that, please substitute (x+5) for y in the other equation. type your result here.

OpenStudy (tkhunny):

I think if you would try to be a little more formal, use actual words, formulate actual sentences, you would find mathematics substantially less confusing.

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