would someone confirm if I am right?

Question

4x-12y=5
-x+3y= -1

-x=3y+-1
4(3y+-1)-12y=5
12y+-4-12=5
0y=1
y=0
-x=12y+5
-x=12(0)+5
-x=0+5
-x=-5

Answer 1

No, I don't think you are right. I assume this is a system of simultaneous equations and you're trying to find a solution set.

4x - 12y = 5
-x + 3y = -1

First, multiply the second equation by 4 on both sides, giving

4x - 12y = 5
-4x + 12y = -4

Next, add the two equations, giving

4x-4x +12y -12y = 5 -4
0x + 0y = 1

There are no solutions (x,y) for which this is true, so the solution set is the empty set.

Taking the solution set you've determined, (5,0), putting that into the first equation gives
4*5 -12 * 0 = 5
20 -0 = 5 or 20 = 5, which is not true, so (5,0) cannot possibly be a solution.

You've made several errors, as follows:

4x-12y=5 OK
-x+3y= -1 OK

-x=3y+-1 NO!, should be -x = -3y + -1 or x = 3y +1
4(3y+-1)-12y=5 OK
12y+-4-12=5 NO!, should be 12y -4 -12y = 5 or -4 = 5
0y=1 I have no idea how you got this
y=0 NO!, if 0y =1, y is not 0, there is no y such that 0y =1
-x=12y+5 I have no idea how you got this
-x=12(0)+5 I understand how you got this, but y does not equal 0
-x=0+5 OK
-x=-5 previously you had -x = 0 + 5 , which was wrong but if it was right, you should get x = -5, not -x = -5

OMG

Answer 2

No, that can't be right. Check you work. If x = -5 and y = 0, then -x + 3y would equal 5, and it doesn't, It equals -1.

Let me analyze your solution.

-x + 3y = -1

-x = -1 -3y. NOT -1 + 3y. You have to SUBTRACT 3y from each side.

4(3y + 1) -12y = 5. You had substituted -x into this equation, when you needed to substitute x.

12y + 4 -12y = 5. You wrote this step down incorrectly (leaving the y off the second 12y), but it looks as though you actually did it right.

0y + 4 = 5--->0y = 1. Again, you seem to have some trouble with cancelling terms. I got positive 4 on the lefthand side of the equation, so I SUBTRACTED 4 from each side. You had negative 4. You shoud have ADDED 4 to each side.

Finally, 0y = 1 does not mean that y = 0. In fact, it's impossible. Nothing multiplied by 0 can equal 1. So either I've made a mistake too, or the system is unsolvable. I'm going to check my math.

[added] I didn't find a mistake. I think there are no solutions.

Answer 3

Be sure to check for dropped signs and dropped variables. Also be careful in the equation 0y=1 -- this equation isn't true regardless of value of y (you can't divide by zero), and it should signal to you that either there may be a mistake somewhere or that the system of equations doesn't have a solution.

**Edit** The two people immediately above have the right solution, i.e. there are no solutions.

Another way to see this is that if we take the equation -x + 3y = -1 and multiply both sides by -4, we get

4x - 12y = 4,

and there are no values of x and y which satisy this equation and

4x - 12y = 5

simultaneously.

Answer 4

these are not real pair of equations .
and you did many mistakes in solution , u make 9 wronge steps :
and the correct are :
step 1: -x=-3y-1
step2: 4(3y+1)-12y=5
step3:-12y-4-12y=5
step6:4x=12y+5
u must do that even its not real equations and havent answer

Answer 5

no, its incorrect you would end up with 0=1 which is no solution, and in ur first step when u substituted u would end up with (+3y+1) not (+3y-1)

Answer 6

i think you are wrong 0 *y = 1 now that's not TRUE. i think there is a mistake from the start. you can't solve the exercise