SOLUTION: Hello, I am trying to help my daughter with her algebra homework, but Im a little rusty myself. I've been able to help her understand a few of the problems but these have stumped t
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Question 226538: Hello, I am trying to help my daughter with her algebra homework, but Im a little rusty myself. I've been able to help her understand a few of the problems but these have stumped the both of us. Help would be greatly appreciated! Thank you!
- Burg family
linear equation by any convenient method
2x + y - 3z = 5
x + y - 3z = 1
x - z = 4
linear equation by method of elimination
9x - 2y = 7
-18x + 4y = -7
linear equation by method of elimination
-3x - y = -3
x + 2y = -4
linear equation by any convenient method
x - 2y = -13
-2x - 2y = -4
Found 2 solutions by Alan3354, solver91311:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
2x + y - 3z = 5
x + y - 3z = 1
x - z = 4
---------------
2x + y - 3z = 5
x + y - 3z = 1
-------------- Subtract
x = 4
-----------
x - z = 4
z = 0
----------
x + y - 3z = 1
4 + y = 1
y = -3
-------------
--------------
linear equation by method of elimination
9x - 2y = 7 times -2 --> -18x + 4y = -14
-18x + 4y = -7
These are inconsistent, no solution. It can't equal both -7 and -14
------------------------
linear equation by method of elimination
-3x - y = -3 times 2 --> -6x - 2y = -6
x + 2y = -4
-6x - 2y = -6
----------------- Add
-5x = -10
x = 2
---------
2 + 2y = -4
y = -3
------------
linear equation by any convenient method
x - 2y = -13
-2x - 2y = -4
----------------- Subtract
3x = -9
x = -3
--------
y = 5
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Solve the third one for 
:
Use this expression for to substitute into the first two equations:
 + y - 3z = 5\cr\ \,(z+4) + y - 3z = 1\right)
Collect terms:
You can go either of two ways from here. Either solve either of the above equations for one of the variables in terms of the other and make another substitution that will result in a single variable equation, or use the elimination method.
Which can then be substituted into the original third equation to get 
, which can then be substituted, along with 
into either the first or second original equation to get 
To use the elimination method, you would multiply 
by -1 to get 
which would then be added to 
term by term, resulting in 
, which value could then be used to proceed as above.
=========================================================
Multiply the first equation by 2:
Add the equations, term-by-term:
Which leads to the absurd result:
What this means is that the solution set to the given system is the empty set, that is, there are no solutions. Graphically speaking, these two equations represent a pair of parallel lines. The given system is said to be inconsistent.
=============================================
For your third problem, multiply the first equation by 2 and then add the two equations term by term. The y variable will be eliminated leaving you with a single equation in x that can be solved by ordinary means. Once the value of x has been determined, substitute that value back into either of the original equations to create a single variable equation in y and solve. Express your solution in terms of an ordered pair (x,y), because this represents the point of intersection of the graphs of the two given equations.
=============================================
The fourth problem also lends itself to the elimination method. In this case, I would use a multiplier of -1 on either of the equations, setting up for elimination of the y variable. However, you will be every bit as successful if you multiply the first equation by 2, setting up for the elimination of the x variable. Works both ways just as well.
John
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