One more question. Can you please show the steps of how to solve this. I feel like the way I'm solving this is wrong.
Solve for x y and z
X+5y+3z=7
2x+y+4z=1
2x-y+2z=1
Not knowing how your are solving it, I would go to matrix methods on your calculator.
1,5,3,7
2,1,4,1
2,-1,2,1
http://www.mathportal.org/calculators/system-of-equations-solver/system-3x3.php
hard to say, since you don't show your work. One way is substitution.
x = 7-5y-3z, so using that in the last two equations, you get
2(7-5y-3z)+y+4z = 1
2(7-5y-3z)-y+2z = 1
shuffling things about, that gives
9y+2z = 13
11y+4z = 13
Now, since 2z = 13-9y, use that to get
11y+2(13-9y) = 13
7y = 13
y = 13/7
2z = 13-9y, so z = -13/7
x = 7-5y-3z = 23/7
Or, using elimination,
subtract 2*#1 from #2 and #3:
1x+5y+3z = 7
0x-9y-2z = -13
0x-11y-4z = = -13
add 5*#2 to 9*#1
subtract 11*#2 from 9*#3
1x+0y+17z = -2
0x-9y-2z = -13
0x+0y-14z = 26
Now you can see that z = -13/7
and use that to get y, then x.
Sure! I can help you with that. To solve these equations for x, y, and z, we can use a method called "elimination" or "substitution." Let's go through the steps:
Step 1: Consider the first equation:
X + 5y + 3z = 7
Step 2: Look for the variable with the simplest coefficient to eliminate. In this case, we can eliminate the x variable because it has a coefficient of 1 in the first equation.
Step 3: Multiply the second equation by the coefficient of x (2) and subtract it from the first equation. This will eliminate the x variable, allowing us to solve for y and z.
So, let's multiply the second equation by 2:
2(2x + y + 4z) = 2(1)
4x + 2y + 8z = 2
Now, subtract this equation from the first equation:
(X + 5y + 3z) - (4x + 2y + 8z) = 7 - 2
This simplifies to:
X - 4x + 5y - 2y + 3z - 8z = 5
Simplifying further:
-3x + 3y - 5z = 5 --> Equation 1
Step 4: Repeat the process to eliminate x between the first and third equations. Multiply the third equation by 2 and subtract it from the first equation:
(X + 5y + 3z) - (2x - y + 2z) = 7 - 1
This simplifies to:
X - 2x + 5y + y + 3z - 2z = 6
Simplifying further:
-X + 6y + z = 6 --> Equation 2
Step 5: Now, we have two simplified equations:
-3x + 3y - 5z = 5 --> Equation 1
-X + 6y + z = 6 --> Equation 2
Step 6: Next, we need to eliminate a variable between these two equations. In this case, we can eliminate the z variable by multiplying equation 1 by 5 and equation 2 by -3 to make the coefficients of z the same.
Multiply equation 1 by 5:
-15x + 15y - 25z = 25 --> Equation 3
Multiply equation 2 by -3:
3x - 18y - 3z = -18 --> Equation 4
Step 7: Add equation 3 and equation 4 together to eliminate the z variable:
(-15x + 15y - 25z) + (3x - 18y - 3z) = 25 + (-18)
Simplifying:
-12x - 3z = 7 --> Equation 5
Step 8: Now we have the following two equations:
-12x - 3z = 7 --> Equation 5
-X + 6y + z = 6 --> Equation 2
Step 9: We can now solve this system of equations by elimination or substitution, but let's use the elimination method again to eliminate the z variable.
Multiply equation 2 by 3:
-3x + 18y + 3z = 18 --> Equation 6
Now, add equation 5 and equation 6 together:
(-12x - 3z) + (-3x + 18y + 3z) = 7 + 18
Simplifying:
-15x + 18y = 25 --> Equation 7
Step 10: We now have two simplified equations:
-15x + 18y = 25 --> Equation 7
-X + 6y + z = 6 --> Equation 2
Step 11: To solve this system, we can either use substitution or elimination. I will use elimination one more time to eliminate the y variable.
Multiply equation 7 by 6:
-90x + 108y = 150 --> Equation 8
Multiply equation 2 by 15:
-15x + 90y + 15z = 90 --> Equation 9
Now, add equation 8 and equation 9 together:
(-90x + 108y) + (-15x + 90y + 15z) = 150 + 90
Simplifying:
-105x + 198y + 15z = 240 --> Equation 10
Step 12: Finally, we have the following equation:
-105x + 198y + 15z = 240 --> Equation 10
From here, you have a system of three linear equations with three variables. You can use various methods, such as substitution or elimination, to solve this system and find the values of x, y, and z.