Hello,
If you begin by creating a table of possible values for x and y, you can start to understand what is going on here.
For the first equation, x+y=6, we need to pick a number as a value for x. Any number will do. For this example, I'll pick 2. So, x=2
Now, for our y value, we've got to pick a number that you can add 2, our x value, to in order to make 6. In other words, 2+y=6 (I just put replaced x with 2). If you solve the equation by subtracting 2 from both sides, you'll see that y must be 4. Then, y=4.
But y changes when x changes, right? When x is 1, y has to be 5. When x is 0, y has to be 6. So we need to make a table to clarify this problem.
When you make the table, you're simply writing all the possible values of x down, along with the y values that you'll need to make the equation work.
Do the same for the second equation. Pick some x values, and solve for y. Write out a table for these values.
So, what you're trying to do here is find values for x and y that work for both equations. What you need to do, then, is look through your tables and see if you can find x and y values that match up from each table. Then you've got your answer.
At first, it's kind of challenging. After a little bit of effort, though, you'll eventually be able to solve this sort of problem in your head.
Good luck!
2007-04-12 21:29:39
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answer #1
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answered by Mystery Viscera 2
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I'm not exactly sure what you are asking, but...
x + y = 6
2x - y = 6
You try to cancel one of the terms out by combining the two equations, in this case adding both of them straight down
(x+2x) + (y-y) = (6+6)
3x = 12
x = 4
plug it in to one of the equations
4 + y = 6
y = 2
test the theory in the other equation
2(4) - 2 = 6
CHECK
Linear or non-linear, it should be linear because there are no x^2 terms
2007-04-13 04:10:37
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answer #2
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answered by rakham22 1
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I think this is what you're looking for...
x+ y = 6
+ 2x- y = 6
3x = 12
3x/3 = 12/3
x = 4
x + y = 6
(4) + y = 6
y = 6 - 4
y = 2
2007-04-13 04:10:24
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answer #3
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answered by markbesada 2
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using a mini table of values,
first solve each equation for y
y = 6 - x
y = 2x - 6
(to make it clearer, substitute z for y in the 2nd equation)
z = 2x - 6
then set up your table
x y z
o 6 -6
1 5 -4
2 4 -2
3 3 0
4 2 2
When the y value equals the "z" value you have your solution:
(x,y) = (4,2)
Solution by substitution or elimination is almost always easier.
2007-04-13 04:24:10
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answer #4
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answered by Helmut 7
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using 2nd eqn, y= 2x - 6
put this value of y in first eqn
x + 2x - 6 = 6
3x = 12
so x=4
y= 2(4) - 6
= 8 - 6 = 2
2007-04-13 04:10:57
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answer #5
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answered by PaRtY AnGeL 3
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