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Question 400011: Could someone please explain how to do these problems, we are solving problems with two variables and I do not understand how to even start working on them, the directions state...Solve by the addition-or-subtraction method, if someone could explain how they are solved I think I could get it....
3x-y=-5
2x+2y=-6
I would greatly appreciate it...thanks!!!!
Found 3 solutions by ewatrrr, stanbon, solver91311:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!
Hi
Solving by the addition-or-subtraction method is the result of:
eliminating one variable when adding or subtrating the two equations
3x - y = -5 |Sets Up for eliminating y by adding the two equation
2x + 2y = -6
3x - y = -5 |dividing 2nd equation by 2 prior to adding the equations
x + y = -3
Adding the two equations together
4x = -8
x = -2
Substituting x-value in one the equations to find y-value
x + y = -3
-2 + y = -3
y = -1
Answer by stanbon(75887)  (Show Source):
You can put this solution on YOUR website! 3x-y=-5
2x+2y=-6
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It's arbitrary whether you eliminate the x-terms
or the y-terms.
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If you want to eliminate the x terms:
multiply thru the 1st equation by 2
multiply thru the 2nd equation by 3
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If you want to eliminate the y-terms;
multiply thru the 1st equation by 2
leave the 2nd equation alone.
==========================
Looks like eliminating the y-terms would be easier so do it:
6x-2y = 10
2x+2y=-6
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Add and solve for "x":
8x = 4
x = 1/2
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Solve for "y":
2x+2y = -6
x+y = -3
(1/2) + y = -3
y = -7/2
----------------
Hope that helps.
Check out the web using Google to search for
"solving linear systems" to see more on this
subject.
==============
Cheers,
Stan H.
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Step 1: Determine a factor that will transform one of the coefficients on one of the variables into the additive inverse of the coefficient on that same variable in the other equation.
For your example, note that multiplying the first equation by 2 would make the coefficient on  in the first equation be -2 which is the additive inverse of 2, the coefficient on in the second equation.
is the first equation after multiplying it by 2.
is the original second equation.
Step 2: Add the two equations, term-by-term to get:
\ +\ 2y\ =\ -10\ +\ (-6))
Did you notice how the contrivance to have the coefficients on be additive inverses caused the variable to disappear altogether?
Step 3: Solve for the remaining variable:
Step 4: Substitute the value of the variable you just determined back into either of the original equations:
\ +\ 2y\ =\ -6)
Step 5: Solve:
Step 6: Specify the solution set:
The solution set consists of a single element, namely the ordered pair )
\right\})
Now that you are done with your example problem, a few general notes.
1. Sometimes you will have to find two multipliers, one for each equation in order to achieve the additive inverse relationship between coefficients in the two equations. For example:
would require multiplication of the first equation by 3 and the second by 5 in order to eliminate the variable.
2. Cross out any reference to "addition-or-subtraction" method. Completely erase such a notion from your head. It is the Elimination method because the purpose of the method is to eliminate one of the variables.
3. While you are at it, eliminate the notions of "subtraction" and "division." You never have to do either. Always add. That is to say perform ) instead of  . You will make orders of magnitude fewer sign errors if you do that. As for division, you multiply by the reciprocal instead. For example, the process  instead of  will save you lots of grief when you start working with rational functions later on.
4. You will encounter situations where the result of the elimination process will be a triviality, generally  , or an absurdity, that is something that looks like  .
Every two-variable linear equation has a graph that is a straight line. For any given pair of equations there are three possibilities. One, they can be two lines that intersect in a single point. Such is the case with your example problem -- the solution set consists of a single ordered pair representing the single point of intersection of the graphs of the two lines. Two, they could be exactly the same line, such that any ordered pair that satisfies one of the equations also satisfies the other equation. Graphically, this is represented by the two equations having exactly the same graph, and this situation is indicated by arriving at the trivial result mentioned above. Three, they could be two different but parallel lines. They never intersect, and hence the solution set is the empty set. Such a situation is indicated by achiving an absurd result to the Elimination method.
John
My calculator said it, I believe it, that settles it
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