What is the value of the x variable in the solution to the following system of equations? 2x + 3y = 4 x - 2y = -5 answers choices.. 2 -2 1 -1
Which system of equations am I supposed to use. Which lesson is this assignment from?
chapter 6, lesson 3.
Which system of equations do I use? Substitution or Elimination!
Ok. I'll use substitution and I'm working on the problem right now!
2x + 3y=4 1st sol x - 2y = -5 2nd sol multiply 2nd sol by 2 and use elimination method
So, Kantharaj! It's easier to use the elimination process not substitution?
It's personal preference. I always do substitution.
Yeah, me too. Thanks for that, ChmE!
@alienababy How are you doing? did you get an answer?
@ChmE, im working on it. Ive got y= 14, but im looking for x. I dont know what to do now?
@alienbaby, you don't need my help anymore? I was almost finished with the equation!
my bad, @alienababy
@Johnjakile1998, i do! im trying to learn.
hold on. Your y is wrong I got 2
You don't have to shout! Lol. Alright, I'll put down the steps and then tell if you understand, and you can tell me.
@Chme. I got the same y, @alienababy it's 2!
im so confused. how?
@ChmE, do you want to explain or me?
@Johnjakile1998 will show the steps for the substitution and I'll show the steps for elimination
sounds good to me :)
Cool! I do better with substitution anyway! Quick question thought @ChmE. You substitute the first equation into the second one right
I did 2nd into first cuz there is already an x alone so it made it quicker
Yeah. Thanks, don't know how I missed it.
\[2x+3y=4\]\[x-2y=-5\]We need to get one of the variables with the same coefficient. So lets multiply the second equation by 2.\[2x-4y=-10\]Now that we have a 2x in both equations we can ELIMINATE that term by subtracting the equations old school. One on top of the other and subtract like terms.\[7y=14\]2x-2x=0, 3x-(-4x)=7x, 4-(-10)=14 Now we solve for y and get 2. Plugging it into the 2nd equation because it is simpler we get -1 for x
(3y**-(-4y**)=7y**)
2x + 3y = 4 x - 2y = (-5) +2y +2y x = -5 + 2y 2(-5 + 2y) + 3y = 4 -10 + 4y + 3y = 4 +10 +10 4y + 3y = 14 7y = 14 -- --- 7 7 y = 2
2x + 3y = 4 2x + 3(2) = 4 2x + 6 = 4 - 6 -6 2x = -2 -- --- 2 2 x = (-1)
So, the solution is (-1,2) The value of x is (-1)
@ChmE, so is my answer correct? I think so!
How would you get x by itself if there is a number in front of it, like lets say.. 5x + 4y = 1 ?
You can't x by itself, so you would have to do this:
you can... x=(-4y+1)/5
5x + 4y = 1 -5x -5x 4y = 1 - 5x And solve your problem using that equation!
@ChmE, is that right?
ya\[x=-\frac{ 4y }{ 5 }+\frac{ 1 }{ 5 }\]You are just solving for x even though there is another unknown. It is possible to get it by itself but not get a value. You always need the same amount of equations as unknowns to solve for a value.
Thanks, @ChmE. @alienababy needs help on another problem. Are you going to help with it?
@Johnjakile1998 & @ChmE ; What is the value of the y variable in the solution to the following system of equations? 5x+4y=1 4x+3y=-1
Are you ready for this @ChmE?
I'm solving the problem @alienababy
thankyou @Johnjakile1998
Welcome! While, I'm solving this problem. I wanted to know if you needed help on another question?
Use the Substitution Method to solve the following system of equations. 2x + y = 3 -3x + 4y = 1
Which do I solve first?
the first one i gave you.
Cool. Come on @ChmE!
For the above question for which I was mentioned in. I would use the elimination method because both variables have a coefficient. It will begin to get messy by substitution illustrated by me solving for x just a couple lines earlier. I'm going to let you attempt it first. If you follow my instructions for this method in the beginning I'm confident you'll get the right answer.
@ChmE which one should i multiply first?
Thanks! That's why I'm so lost.
like in fractions find the lowest common multiple. In this case it is 12. The above equation needs to be multiplied by 3 and the bottom by 4. remember to multiple every term
@alienababy. I'm going to solve the second equation first! I already got the answer!
multiply**
okay :) @Johnjakile1998 & @ChmE i got 15x+12y=3 then 16x+12=-4. is that right?
For which problem?
mine
the first one, using the elimination method.
its correct now subtract old school
You're so weird @ChmE! Lol
i got -x+y=-1 ? that doesnt seem right to me.
@ChmE
you shouldn't get a y term try again
-x=-1 @ChmE
@alienababy. I got the answer for the second equation. Want me to post it?
yes please @Johnjakile1998
what is -4-3?
-7 @ChmE
-7.
x=-7
you can put whatever equation u want on top when subtracting. i put the 16x equation on top so that my x would be positive after subtraction.
eliminates possible stupid mistakes
oh.. @ChmE i didnt know that.
if you want to see it for yourself. do the subtraction both ways. you should get -x=7 and x=-7. those or equal, no?
yes, i did. @ChmE
are**. Nice
now just plug ur value of x into one of the equations, choosing the easier one (why make it even harder), and solve for y
the first equations, or the second equations after you solved it? @ChmE
whatever you want in this case. both equations look the same to me. It was more of a general statement
okay.
you want to use the original equations because it is simpler. smaller numbers
2x + y = 3 -2x -2x y = 3 - 2x -3x + 4(3 - 2x) = 1 -3x + 12 - 8x = 1 -11x + 12 = 1 - 12 -12 -11x = -11 x = 1 2x + y = 3 2(1) + y = 3 2 + y = 3 -2 -2 y = 1 So, the solution is (1,1)
gotcha. so it doesnt matter though? @ChmE
no. You should get the same answer in either case. I'd encourage you to try it to see for yourself.
thankyou @Johnjakile1998
So, @alienababy that's the answer for the second equation. Is the answer right @ChmE?
Any other questions @alienababy
looks correct to me
can i just be a math genious ? ):
@ChmE i got -9?
thanks @ChmE! You are @alienababy, just Algebra 1 isn't the math that is you're strongest. For all we know your strongest math is Calculus or Alebra 2 even Geometry!
thanks :) @Johnjakile1998
any others questions @alienababy
Just wait.. Here's a problem I had on my last test\[y-\ln y=x ^{2}+1\]is this a solution of \[\frac{ dy }{ dx }=\frac{ 2xy }{ y-1 }?\]
What kinda math is this?
I got positive 9
Differential Equations (Calc 4)
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