SOLUTION: I need a little help with the following problems: 2x + 3y = 8 2x - y = -8 3x - 2y = -5 3y - 4x = 8 3x - y = 12 y - 3x = 15

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Question 121940: I need a little help with the following problems:
2x + 3y = 8
2x - y = -8

3x - 2y = -5
3y - 4x = 8

3x - y = 12
y - 3x = 15

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
All three of these problems require the same techniques to solve, so I'll do one of them and leave the other two for you to work.

2x+%2B+3y+=+8
2x+-+y+=+-8

You want to eliminate one of the variables by multiplying one (or both) of the equations by a constant so that the coefficient on one of the variables will be the additive inverse of the coefficient on that same variable in the other equation.

For this system, multiply either equation by -1. I'll use the second equation:

2x+%2B+3y+=+8
-2x+%2B+y+=+8

Now add the equations, term by term, to get a single equation with a 0 coefficient on the x term:

cartoon%28red%280x%29%2C0x%2Bred%284y%29%2C0x%2B4y=red%2816%29%29

Divide by the coefficient on y:

y=4

Now that you have a value for y, you can do either of two things. You can substitute the value of y into either of the original equations then solve for x, or you can repeat the above process using a different multiplier so that you can eliminate the y variable in the same manner we eliminated the x variable. Either process, correctly performed, will give the same result.

Substitution:
, giving us a solution set containing the ordered pair (-2,4)

Repeat elimination:
2x+%2B+3y+=+8
2x+-+y+=+-8

Multiply 2nd equation by 3:
2x+%2B+3y+=+8
6x+-+3y+=+-24

Add the equations, term by term:
cartoon%28red%288x%29%2C8x%2Bred%280y%29%2C8x%2B0y=red%28-16%29%29

Divide by the coefficient on x:
x=-2, and again, the solution set is (-2,4).

Follow this same procedure for the other two problems. Hint: For both of the other problems you will have to rearrange one of the equations so that the variables are in the same order. Standard form is x then y then the equal sign and then the constant, but any arrangement will work as long as both equations are arranged the same. On the second problem, you need to multiply both of the equations by constants. Try 4 for the first equation and 3 for the second, giving you 12 and -12 as the additive inverse coefficients for x.

When you do the third problem, you will come to the absurdity that 0=27. What this means is that there is no elements in the solution set for this system. This is called an inconsistent system. If you graphed these two equations, you will find that the lines are parallel. Had they been equations for the same line, you would have arrived at the result 0=0, meaning that the system is undetermined, and that the solution set has an infinite number of elements.