SOLUTION: OK, so I know the following is incorrect, so could you help me to figure out what I am doing wrong. I am trying to solve the linear equations by using linear combinations. 9X-3Z

Question 271088: OK, so I know the following is incorrect, so could you help me to figure out what I am doing wrong. I am trying to solve the linear equations by using linear combinations.
9X-3Z=20
3X+6Z=2
My answers were X=2 and Z= -.66 or -2/3
I would appreciate your help!

Found 2 solutions by LarissaRichardson, unlockmath:
Answer by LarissaRichardson(34) (Show Source): You can put this solution on YOUR website!
9x-3z=20
3x+6z=2
A.Choose a variable to eliminate first. In this example, y will be eliminated.
B. Multiply the equations so that there is the same number of that variable in both equations. Find the least common multiple (LCM) of the coefficients of the variable you chose to eliminate. Multiply the equations accordingly so that both of the coefficients are of the same value, except one is negative.
The LCM would be six seen as 3x+6z=2 and six is the LCM
Multiply first equation by two.
2(9x-3z=20)
18x-6z=40
C. Now add the equations leaving out the variable with z in it.
18x-6z=40
+3x+6z=2
___________
21x=42
x=2
D. Solve for the remaining variable. This can be done using any methods to solve for a one-variable equation.

9(2)-3z=20
18-3z=20
-3z=2
z=-2/3
I believe you said you had the same answer so I don't know why you say it's wrong.
Answer by unlockmath(1688) (Show Source): You can put this solution on YOUR website!
Hello,
Here we go.
9X-3Z=20
3X+6Z=2
Multiply the first equation by 2 to get:
18X-6Z=40 Now, add this to the second:
3X+6Z=2
And we get:
21X=42 Divide each side by 21 to get:
X=2 Plug 2 into the first equation to get:
18-3Z=20 Subtract 18 from both sides and divide by -3 to get:
Z=-2/3
You'll see that plugging these answers into the equations works out right.
So don't be so hard on yourself. You are RIGHT!
RJ
Check out a book I wrote at:
www.math-unlock.com
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