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latresa31s

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Sep 4, 2005
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Solve 6x + (2/3)y = 8; first for x = – 2 and then for y = – 3



I am not even sure of how to do this one?

I am not sure if I am suppose to work the problem by adding the -2 first and -3 as a seperate problem . could someone help.
 
The first one you subsitute x=-2 into your equation:

6(-2) + (2/3)y = 8

And solve for y.

The second one you do the same except plug y=-3 in and solve for x.

mmkay?
 
Solve 6x + (2/3)y = 8; first for x = – 2 and then for y = – 3

ok

is this problem to be step up like this to solve for y

6(-2)+(2/3)y = 8

(-12/1)+(2/3)y = 8

then what
 
How about you have a stab in the dark... I think you can do it. (I'll be here if you're still stuck)
 
Latresa, you're going to have to show your work.

I'm unsure as to what you mean by y = (39,-3). You should have just one answer for each case.

What did you do from here:

-12 + (2/3)y = 8

?

Did you move the -12 to the right-hand side by adding 12 to both sides?

Did you then divide both sides by (2/3)?
 
latresa31s said:
Solve 6x + (2/3)y = 8; first for x = – 2 and then for y = – 3
I am not even sure of how to do this one?
I am not sure if I am suppose to work the problem by adding the -2 first and -3 as a seperate problem . could someone help.
Click to expand...
SIMPLIFY the equation first:
6x + (2/3)y = 8
multiply by 3 to get rid of fraction:
18x + 2y = 24
now divide by 2:
9x + y = 12 : isn't that now MUCH simpler?

x = -2:
9(-2) + y = 12
-18 + y = 12
y = 30

y = -3:
9x + (-3) = 12
9x = 15
3x = 5
x = 5/3

CLEAR?!
 
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