systems of linear equations in 3 variables
1/3x-2/3y+z=0
1/2x-3/4y+z=1/4
-2x-y+z=1
Mult 1st eq by 3; x-2y+3z = 0
Mult 2nd eq by 4: 2x - 3y + 4z = 1
Add 2nd and 3rd: -4y+5z = 2
Mut 1st by 2: 2x-4y + 6z = 0
Add that answer and 3rd eq: -5y + 7z = 1
5(-4y + 5z = 2) + -4(-5y + 7z = 1)
-20y + 25z = 10 + 20y - 28z = -4
-3z =6
z = -2
Sub into the 1st eq: x - 2y -6 =0
Sub into the 3rd eq: -2x-y-2 = 1
2(x-2y = 6) + (-2x-y=3)
2x - 4y = 12 + -2x -y = 3
-5y =15
y = -3
sub into the 3rd eq
-2x -(-3) +(-2) = 1
-2x +1 = 1
x = 0
(0,-3,-2)
It might be easier if you
a) multiplied the first equation by 3
b) the second equation by 4
To solve the system of linear equations in three variables, you can use the method of substitution or elimination. Here, we will use the method of elimination to solve the given system of equations:
Step 1: Multiply each equation by the appropriate factor to eliminate the fractions.
Multiply Equation 1 by 3 to get:
1x - 2y + 3z = 0 (Equation 1)
Multiply Equation 2 by 4 to get:
2x - 3y + 4z = 1 (Equation 2)
Multiply Equation 3 by 12 to get:
-24x - 12y + 12z = 12 (Equation 3)
Step 2: Add or subtract the equations to eliminate a variable.
Add Equation 1 and Equation 2:
(1x - 2y + 3z) + (2x - 3y + 4z) = (0) + (1)
3x - 5y + 7z = 1 (Equation 4)
Subtract Equation 3 from Equation 4:
(3x - 5y + 7z) - (-24x - 12y + 12z) = 1 - 12
3x - 5y + 7z + 24x + 12y - 12z = 1 - 12
27x + 7y - 5z = -11 (Equation 5)
Step 3: Create a new system of equations with the eliminated variable.
We now have two equations:
Equation 4: 3x - 5y + 7z = 1 (Equation 4)
Equation 5: 27x + 7y - 5z = -11 (Equation 5)
Step 4: Solve the new system of equations.
At this point, we have reduced the system of equations to two equations with two variables. To solve this new system, you can use the method of substitution or elimination again. However, if you only need to find the values of x, y, and z, you can use an online system of equations solver or a calculator that can solve systems of equations to get the numerical values.
To solve a system of linear equations in three variables, you can use several methods such as substitution, elimination, or matrix methods like Gauss-Jordan elimination or the inverse matrix method. I will explain the process using the method of elimination.
Let's solve the given system of equations step by step:
1) Rewrite the equations in standard form:
First, let's clear the fractions by multiplying through by the least common denominator, which is 12. Multiply each equation by 12:
4x - 8y + 12z = 0 (equation 1)
6x - 9y + 12z = 3 (equation 2)
-24x - 12y + 12z = 12 (equation 3)
2) We can use elimination to eliminate one variable at a time. Our goal is to eliminate the x and y variables and find the value of z.
To eliminate x, we can multiply equation 1 by 6, and equation 2 by 4, so that when we subtract them, the x variable will be eliminated:
24x - 48y + 72z = 0 (equation 4)
24x - 36y + 48z = 12 (equation 5)
3) Subtract equation 5 from equation 4:
(24x - 48y + 72z) - (24x - 36y + 48z) = 0 - 12
-12y + 24z = -12 (equation 6)
4) Now, let's eliminate the y variable. Multiply equation 4 by 3, and equation 6 by 8, so that when we subtract them, the y variable will be eliminated:
72x - 144y + 216z = 0 (equation 7)
-12y + 24z = -12 (equation 8)
5) Subtract equation 8 from equation 7:
(72x - 144y + 216z) - (-12y + 24z) = 0 -(-12)
72x - 132y + 192z = 12 (equation 9)
6) At this point, we have eliminated both x and y, leaving only z. Now we solve equation 9 for z:
72x - 132y + 192z = 12
192z = 12
z = 12/192
z = 1/16
7) Substitute the value of z = 1/16 in equation 6 to find the value of y:
-12y + 24z = -12
-12y + 24(1/16) = -12
-12y + 3/2 = -12
-12y = -12 - 3/2
-12y = -24/2 - 3/2
-12y = -27/2
y = (-27/2) / (-12)
y = 27/24
y = 9/8
8) Substitute the values of y = 9/8 and z = 1/16 in equation 1 to find the value of x:
4x - 8y + 12z = 0
4x - 8(9/8) + 12(1/16) = 0
4x - 9 + 3/4 = 0
4x = 9 - 3/4
4x = (36/4) - 3/4
4x = 33/4
x = (33/4) / 4
x = 33/16
So, the solution to the system of equations is x = 33/16, y = 9/8, and z = 1/16.