Which ordered pair is a solution to the system of equations? 2x - y = 5 3x + 2y = 4 1. (0,2) 2. (5,5) 3.(-1, -7) 4.(2, -1)
hey :D welcome to OpenStudy!!
(2, -1)
@314dokg please do not just post answers
x - y = 5 -------->equation 1 3x + 2y = 4 -------->equation 2 In equation 1-> x-y=5 x=y+5 substitute this value of x in equation 2 3x+2y=4 now we substitute x=y+5 in this equation 3(y+5)+2y=4 now open the brackets and simplify 3y+15+2y=4 5y+15=4 from here you wil get y once you get y you just put that value of y in any of equation 1 or equation 2 to get the value of x
You just have to substitute.. (x, y) to the corresponding x and y's in the equation... it's just basic trial and error. for example 1.) (0,2) subsititute 2x - y = 5 (2)(0) - (2) = 5 0 - 2 = 5 if one side is the same with the other side. therefore you got the answer.. but since (0.2) is not then try the other one... just continue until it satisfy both solution
@bubblegum. you copied the first equation incorrectly, which will confuse the OP mightily when they find answers that do not appear in their choices... should be \[2x-y=5\]
aww D:
Solve that for \(y\):\[2x-y=5\]\[y = 2x-5\]Substitute in other equation \[3x+2y=4\]\[3x+2(2x-5)=4\]\[3x+4x-10=4\]\[7x=14\]etc.
@s00sm can you solve that for the value of \(x\)? Once you have \(x\), plug it into one of the equations and find the corresponding value of \(y\). What do you get?
x=2.
Good. How about the value of \(y\)?
no, no 5/2?
Mmmm...no. show me your work..
2x−y+y=5+y 2x=y+5 2x/2 = y+5/2 x= 5/2
I need to go to bed, so I'm just going to show you my work: We have 3 different equations to choose from: \[2x-y=5\]\[3x+2y=4\]\[y=2x-5\] we know that \(x=2\), so just plug it into one of the equations and solve for \(y\) \[2x-y=5\]\[2(2) - y = 5\]\[4 - y = 5\]\[4 = 5+y\]\[y=4-5 = -1\] or \[3x+2y=4\]\[3(2) + 2y = 4\]\[6+2y=4\]\[2y=4-6\]\[2y=-2\]\[y=-1\] or \[y=2x-5\](our substitution equation, and likely to be the easiest) \[y = 2(2)-5\]\[y=4-5\]\[y=-1\] As you can see, the same answer any way that you do it. Now that we think we have a solution, we must test it in both equations. Testing in just one is NOT sufficient to ensure a correct answer. \[2x-y=5\]\[2(2) - (-1) = 5\]\[4+1=5\checkmark\] \[3x+2y=4\]\[3(2) + 2(-1) = 4\]\[6-2=4\checkmark\] Our solution made all of the equations work, so it is valid.
oh pellet, im sorry! i appreciate it and i'm really sorry! I don't have a scanner and im pretty new ion this site. your explanation helped me understand that i'm a complete dumbass and this is isn't an insult, it's a complement bahhh simsorry! a medal and fan for you definitely! thank you
I'll do some more of these with you, but it will have to wait until after I've slept. Just takes a bit of practice and you'll be doing them without much trouble at all.
Mostly it is a matter of being methodical and careful. In my opinion, that's one of the enduring values of math classes, even if you don't end up doing anything in your career that involves solving math problems — you learn how to pay attention to detail.
it's alright! no, no, you don't have to do these with me, i appreciate it though, it's really my fault in not trying to practice more of it, and kind of my duty to practice on my own, but I do get stuck and well here I am. I don't want to burden you or anything! but i really appreciate it, you staying long enough to read and reply trying to educate a bum on the internet.
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