Solve the system of equations by substitution. What is the solution for x? 2x + y = 1 4x + 2y = −2 x = 0 x = 2 There is no x value as there is no solution. x can be any value as there is an infinite number of solutions.
Hello, welcome to open study! I can help you through the question. This is asking if there is a point in the (x,y) plane where these 2 lines cross. If there is, that point (x,y) is a solution to the system. If there is not a point where they cross, they are parallel lines, and there are no solutions (x,y). If they are both the same line, overlap each other, then there will be an infnite number of solutions where the lines intersect, being that they are the same line.
inconsistent, dependent, independent Those words sound familiar from class?
Yes, thanks so much for your help! @DanJS
@TheatreKid
if you want to go through the rest of it , let me know
there is a quick way to tell what the answer is, i can see they are parallel
Can you please help with one more if you don't mind?
Oh yes sure!
You want to know an easy way to tell the answer to these probs?
sure!
that would be awesome
Well it will tell you if they are parallel (no solutions), or the same line (infinite solutions) real fast...here are the equations again 2x + y = 1 4x + 2y = −2
reload page for thoseweird things to go away
Try this, multiply the first equation , everything, by 2. As long as you do that to both sides, the equation doesnt change.
2(2x+y) = 2*1 ---------> 4x + 2y = 2 right?
Yes thats right
k, now we have this for the system, 4x + 2y = 2 4x + 2y = -2 notice the left hand side...what happened?
everything doubled...
right, everything in the first equation we doubled, now look at both equations together 4x + 2y = 2 4x + 2y = -2 The left sides are exactly the same!
If you have this situation, the lines are parallel, and there are zero solutions
Ahhhh... that makes sense.
If you solved for y = mx+b form, they both will have the same slope, but different y-intercepts....parallel lines, zero solutions
so looking at the original problem 2x + y = 1 4x + 2y = -2 i think to myself, if i doubled 2x, i get the 4x in the second equation. Now if i double the y term also, does that equal the y term in the second equation. In this case it does, so the lines are parallel
In general, If you can multiply an equation by a constant number, to get the left hand side of the second equation, the lines are parallel, and have no solutions.
Make sense?
Wow, thank you! That helps.
One small step further
Yup!
If you do the exact same thing, and BOTH sides of the equations become the same, then they are the same line and have infinitely many solutions
I knew that one, thanks for the reminder.
In this case the right sides werent the same, BUT Lets pretend the first equation had a -1 instead of the +1 on the right 2x + y = 1 4x + 2y = -2
2x + y = -1 4x + 2y = -2 multiply the first equation by 2, and you get the second equation, they are the same line, infinitely many solutions
So that is an easy, off the cuff, sort of way to tell real fast if you have parallel lines, or the same line.
There is a way to tell if they are perpendicular too that is pretty quick too , if you want to know that
Thank you for your help with that! Would you mind helping me with one more question instead of telling me about the way to find if they are perpendicular or not, because I don't need to know that right now, plus I think it's in my lessons, but thank you :)
sure
Can you help me to solve this system of equations. x+3y = -1 2x + 2y = 6
For future reference when you come to it, a*x + b*y = c d*x + e*y = f a,b,c,d,e,f are all numbers in the 2 equations for a line. IF (a*d) + (b*e) = 0 - the lines are perpendicular
x+3y = -1 2x + 2y = 6 There are a few ways to solve this ,, i would multiply the first equation by -2 first, then add the two equations together
-2x - 6y = +2 Multiplied first equation by (-2) 2x + 2y = 6 ----------------- 0x - 4y = 8 Add them together, Notice -2x+2x = 0x so -4y = 8 y = -2
Now just use that y = -2 in one of the equations to get the x value
Thank you so so so much!!
This will help me not only with these questions but with others as well; so thanks.
awesome, you can fan me so you can tell when i am online if you want :)
feel free to ask questions whenever i am on here, you can type in @DanJS into the question and it will get my attention
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