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2086171

USA
Joined: Jan 25, 2007
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solve by subsititution

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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
on this if I subsitutite 3(2/3)+6=0; then I get lost Please help

[Feb 10, 2007 8:19:48 AM]

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0641650001

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Joined: Apr 25, 2007
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Re: solve by subsititution

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x=4(i think) , i dont know bout y

[Apr 29, 2007 3:01:26 PM]

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pskinner

Joined: Apr 2, 2005
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Re: solve by subsititution

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

If you start with the two equations:

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)
----------------------------------------
Principal Skinner

[Apr 29, 2007 6:22:41 PM]

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