3x+y+z=14
-x+2y-3z=-9
5x-y+5z=30
linear equations solve by elimination
so in elimination you always add first, then subtract the for the next equation?
thank you, Aaron
To solve this system of linear equations by elimination, we will manipulate the equations to eliminate one variable at a time. Here are the steps:
Step 1: Choose two equations and eliminate the same variable.
Let's eliminate the variable x from the first and second equations. We can do this by multiplying the second equation by 3 and then adding it to the first equation:
3 * (-x + 2y - 3z) = 3 * (-9)
-3x + 6y - 9z = -27
Now, add the first and modified second equation:
(3x + y + z) + (-3x + 6y - 9z) = 14 + (-27)
7y - 8z = -13
Step 2: Choose another pair of equations and eliminate the same variable.
Next, let's eliminate the variable x from the first and third equations. This time, we will multiply the first equation by 5 and the third equation by -3 to get opposite coefficients for x:
5 * (3x + y + z) = 5 * 14
-3 * (5x - y + 5z) = -3 * 30
15x + 5y + 5z = 70
-15x + 3y - 15z = -90
Now, add the first and modified third equation:
(15x + 5y + 5z) + (-15x + 3y - 15z) = 70 + (-90)
8y - 10z = -20
Step 3: Solve the resulting system of equations.
We now have two equations with two variables:
7y - 8z = -13 ...(equation 1)
8y - 10z = -20 ...(equation 2)
To eliminate z, we can multiply equation 1 by 8 and equation 2 by -7:
8 * (7y - 8z) = 8 * (-13)
-7 * (8y - 10z) = -7 * (-20)
This gives us:
56y - 64z = -104
-56y + 70z = 140
Adding the two equations gives:
6z = 36
Divide both sides by 6:
z = 6
Now, substitute the value of z into one of the original equations to solve for y. Let's use equation 1:
7y - 8(6) = -13
7y - 48 = -13
7y = -13 + 48
7y = 35
y = 5
Finally, substitute the values of y and z into any of the original equations to solve for x. Let's use equation 1:
3x + 5 + 6 = 14
3x + 11 = 14
3x = 14 - 11
3x = 3
x = 1
So the solution to the system of linear equations is x = 1, y = 5, and z = 6.
looks like the y's would be the first line of attack
add 1st and last ----> 8x + 6z = 44
4x + 3z = 22
double the first ---> 6x + 2y + 2z = 28
keep second as is --> -x + 2y - 3z = -9
subtract them:
7x + 5z = 37
4x + 3z = 22 --times 5 --> 20x + 15z = 110
7x + 5z = 37 --times 3 --> 21x + 15z = 111
subtract the last two
x = 1
sub into 7x+5z=37
7+5z = 37
5z = 30
z = 6
back into the 1st:
3 + y + 6 = 14
y = 5
x=1 , y=5 , z=6