SOLUTION: Hi: I wondered if you could check this problem for me...
Solve each system by the addition method. 1/3x -1/6y = 1/3 1/6x+1/4y = 0
This is what I've got, but I'm not sure it
Algebra ->
Coordinate Systems and Linear Equations
-> SOLUTION: Hi: I wondered if you could check this problem for me...
Solve each system by the addition method. 1/3x -1/6y = 1/3 1/6x+1/4y = 0
This is what I've got, but I'm not sure it
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Question 321428: Hi: I wondered if you could check this problem for me...
Solve each system by the addition method. 1/3x -1/6y = 1/3 1/6x+1/4y = 0
This is what I've got, but I'm not sure it's correct...
I started by multiplying both equations by their LCD to get rid of all the fractions.
1/3x – 1/6y = 1/3
(6)1/3x – (6)1/6y = (6) 1/3
2x-1y=2
So our first equation is 2x-1y=2
I simplified the second equation by multiplying it by 12:
1/6x+1/4y=0
(12)1/6x+(12)1/4y=(12)0
2x+3y=0
The second equation is 2x+3y=0
Now our two equations are:
2x-1y=2
2x+3y=2
I multiplied the second equation by -1 to eliminate one of the variables:
(-1)2x+(-1)3y=(-1)0
-2x-3y=0
Added the two equations together and solved:
2x-1y=2
-2x-3y=0
-4y= 2
-4 -4 Reduce the fraction
y= -1/2
Let’s plug this value for y (-1/2) back into the second original equation:
1/6x+1/4(-1/2)=0
1/6x-1/8=0 Let’s multiply this equation by 24 (the LCD) to get rid of fractions
(24)1/6x-(24)1/8=(24)0
4x-3=0
+3 +3
4x= 3
4 4
x= ¾
So, the final solution is (3/4,-1/2)
Because this system only had one solution, this line is independent.
Thanks in advance for your help.
