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 This topic has been viewed 1222 times and has 2 replies
2086171

USA
Joined: Jan 25, 2007
Posts: 18
Status: Offline

first one can be solved either by addition or subsititution: is
3x-y=-7and x+y=-9
-x-y=-9
2x=-9; x=9/2
3(9/2)-y=-7; 27/2-y=-7 and now I am confused

the second on is 3x+6y=0 and x=2/3
 [Feb 10, 2007 8:19:48 AM] [Link]
0641650001

United States
Joined: Apr 25, 2007
Posts: 1
Status: Offline

x=4(i think) , i dont know bout y
 [Apr 29, 2007 3:01:26 PM] [Link]
pskinner

Joined: Apr 2, 2005
Posts: 797
Status: Offline

 first one can be solved either by addition or subsititution: is 3x-y=-7and x+y=-9-x-y=-92x=-9; x=9/23(9/2)-y=-7; 27/2-y=-7 and now I am confusedthe second on is 3x+6y=0 and x=2/3on this if I subsitutite 3(2/3)+6=0; then I get lost Please help

Let's just do one at a time, because things are all messed up.

3x-y=-7
x+y=9

It "appears" that you wanted to use the addition method, but things got messed up:

If you just add the left sides on the left and the right sides on the right, you get:

(3x+x)-y+y=-7+9

So, 4x=2, in which case x=1/2

Now, you can substitute that value in either equation and solve for y:

(1/2)+y=9, so y=9-(1/2)=17/2.

Now, for the second one:

3x+6y=0
x=2/3

When I substitute for x, I get:

3(2/3)+6y=0 so, 2+6y=0, in which case, 6y=-2, so y=-(1/3)
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Principal Skinner
 [Apr 29, 2007 6:22:41 PM] [Link]