SOLUTION: 5. Solve the following system of equations algebraically. A trial-and-error or calculator solution is not acceptable! 1. 4x + y + 2z = 24 2. 2x − y − 2z = &#8722

Algebra ->  Trigonometry-basics -> SOLUTION: 5. Solve the following system of equations algebraically. A trial-and-error or calculator solution is not acceptable! 1. 4x + y + 2z = 24 2. 2x − y − 2z = &#8722      Log On


   

Question 1118569: 5. Solve the following system of equations algebraically. A trial-and-error or calculator solution is not acceptable!
1. 4x + y + 2z = 24
2. 2x − y − 2z = −6
3. −x + 2y − z = −4










Found 2 solutions by rothauserc, greenestamps:
Answer by rothauserc(4718) About Me  (Show Source):
You can put this solution on YOUR website!
add equations 1 and 2 together
:
6x = 18
:
x = 3
:
substitute for x in equations 2 and 3
:
(2) 6 -y -2z = -6
:
(3) -3 +2y -z = -4
:
solve equation 2 for y
:
y = 12 -2z
:
substitute for y in equation 3
:
:
-3 +2(12-2z) -z = -4
:
24 -4z -z = -1
:
-5z = -25
:
z = 5
:
substitute for x and z in equation 1
:
4(3) +y + 2(5) = 24
:
12 +y +10 = 24
:
y = 2
:
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x = 3, y = 2, z = 5
:
check the answer
:
equation 1
:
4(3) +2 +2(5) = 24
:
12 +2 + 10 = 24
:
24 = 24
:
equation 2
:
2(3) -2 -2(5) = -6
:
6 -2 -10 = -6
:
-6 = -6
:
equation 3
:
-3 +2(2) -5 = -4
:
-3 +4 -5 = -4
:
-4 = -4
:
answer checks
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Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There are innumerable ways to solve a system like this algebraically. Different methods will be appropriate for different systems of equations.

You always want to look for ways to make the problem as easy as possible. In your example, a quick look shows that adding the first two equations together eliminates both y and z, allowing you to find the value of x and thus immediately reduce the system of three equations and three unknowns to a system of two equations and two unknowns.

4x+y+2z = 24; 2x-y-2z = -6 --> 6x = 18 --> x = 3

Substitute x=3 into either of the first two equations and the third. (Since you found x=3 using the first two equations, substituting x=3 in those same two equations won't get you anywhere.)

6-y-2z = -6 --> y+2z = 12;
-3+2y-z = -4 --> 2y-z = -1

The solution from the other tutor then uses substitution to finish the problem -- solving one equation for one of the variables and substituting the resulting expression for that variable in the other equation.

When the two equations are both in the form Ax+By=C, I think a solution using elimination is much easier. Multiply the second equation by 2 and add the two equations to eliminate y:

y+2z = 12; 4y-2z = -2 --> 5y = 10 --> y = 2

Then substitute y=2 in either of those last two equations to find z:

2+2z = 12 --> 2z = 10 --> z = 5

Answer: x=3; y=2; z=5