Question 1034664: I need to solve this equation by the elimination method and keep coming up with zero. I feel I'm missing something. Help would be much appreciated.
4x-3y=10
2x-5=3/2y
Found 3 solutions by nerdybill, fractalier, Edwin McCravy:
Answer by nerdybill(7384)   (Show Source):
You can put this solution on YOUR website! You are correct... (see below)
4x-3y=10
2x-5=3/2y
.
move y's to the left and constants to the right
4x-3y=10
2x-3/2y=5
.
4x-3y=10
4x-3y=10
.
4x-3y=10
-4x+3y=-10
----------
0 = 0
If the elimination produces the equation 0=0, then the two equations are for the same line. This means that there are an infinite number of solutions.
Answer by fractalier(6550)  (Show Source):
You can put this solution on YOUR website! Well, this is kind of a tricky elimination problem, but here goes...from
4x-3y=10
2x-5=3/2y
let us multiply the bottom equation by two and subtract it from the top one...
4x - 3y = 10
-(4x - 10 = 3/y)
-----------------
-3y + 10 = 10 - 3/y
Now subtract 10
-3y = -3/y
Now divide by -3
y = 1/y
y^2 = 1
y = 1 or y = -1
If y = 1, we have
4x - 3(1) = 10
4x = 13
x = 13/4
If y = -1, we have
4x - 3(-1) = 10
4x + 3 = 10
4x = 7
x = 7/4
so it looks like there are two solutions...
(13/4, 1) and
(7/4, -1)
You can check...
Answer by Edwin McCravy(20060)  (Show Source):
You can put this solution on YOUR website!
What you're missing is that this is a dependent system and
there are infinitely many solutions to the system.
When you try to solve them by the elimination method,
If you clear the fraction of the second
2x-5=3/2y
you get:
4x-10=3y
And put it in standard order
4x-3y=10
which is identical to the first equation.
So when you try to solve them by elimination,
you get:
4x-3y = 10
-4x+3y =-10
-----------
0y = 0
So you don't get 0, you get
0y = 0
And every value of y will satisfy that equation.
So there are infinitely many solutions.
Graphically they are two lines one coinciding with
the other. Therefore they "intersect" everywhere.
The "solution" is "It is a dependent system with infinitely
many solutions."
Edwin
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