12x  3y = 3 6x + y = 1Please explain how you got your answer,thanks! :)
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7 Answers
Given the system of equations:
12x  3y = 3.........(1)
6x + y = 1...........(2)
We will use the elimination method to solve.
First, we will multiply (2) by 2:
==> 2 [ 6x + y = 1 ]
==> 12x 2y = 2......(3)
Now we will add (1) to (3).
==> 5y = 5
Now we will divide by 5
==> y= 1
Now we will substitute into (2) to find x.
==> 6x + 1 = 1
==> 6x = 0
==> x = 0
Then the solution is the pair (x, y) = ( 0, 1)
12x3y=3 You subtract12 on both side to let the y on one side.
12x3y=3+12x
3y=3+12x
3y/3=3/3=12x/3
y=1+6x You plug this equation on the other equation where y is.
6x+1+6x=1
12x+1=1
12x=0 Divide both side by 12 to let x by itself
12x/12=0/12
x=0
You plug the 0 on where the x is to solve for y.
5(0)+y=1
0+y=1
0+y0=10
y=1
12x3y=3
12x3y=3+12x
3y=3+12x
3y/3=3/3=12x/3
y=1+6x
6x+1+6x=1
12x+1=1
12x=0
12x/12=0/12
x=0
5(0)+y=1
0+y=1
0+y0=10
y=1
12x  3y = 3
6x + y = 1
First multiply everything in the second equation by 3
By multiplying, you should get
12x  3y = 3
18x + 3y = 3 now add 3y with 3y ( which means also add 12x with 18x and 3 with 3 )
By adding, you should get
30x = 0 divide both sides by 30
By dividing you should get
x = 0/30 simplify
x = 0 which is your answer for " x "
Now plug 0 into one of the equation
6 ( 0 ) + y = 1 Multiply 6 with 0
By multiplying you should get
y = 1 which is your answer for " y "
So your answer is x = 0 ; y = 1
12x  3y = 3
6x + y = 1
you can work with any problem first by I am going to work with the first problem first:
Try to get 3y alone by subtracting 12x
3y = 3  12x
now divide by 3 to get y alone:
y = 1 + 4x
now plug this in as y in any of the problems:
6x + 1 + 4x =1
combine like terms
10x = 0
divide by 10
x = 0
now plug in 0 as x to get y
6(0) + y = 1
0 + y =1
y = 1
(1) 12x  3y = 3
(2) 6x + y = 1
Multiply the (2) by 2
(3) 12x + 2y = 2
subtract (3) from (1)
12x 3y = 3

12x + 2y = 2
5y = 5
Solve for y
y = 1
Plug in the y value for either (1) or (2) and solve for x!
given eq:
12x3y=3
6x+y=1
by solving these substituting method we get
12x=3y3
x=(3y3)/12
x=3(y1)/3*4
x=(y1)/4 substitute in6x+y=1
6*(y1)/2*2+y=1
(3y3)/2+y=1
(3y3+2y)/2=1
5y3=2
5y=5
y=1
(x=y1)/4
x=0/4=0
******x=0*************
******y=1*************
y=1