You want to take your 3 equations in 3 variables to 2 equations with the same 2 variables. Since the first equation has x and z, lets see if we can eliminate the y by combining the 2nd and 3rd equations.
To eliminate y you need the coefficients to be equal and opposite. So you would want a -10 and a + 10 since you are dealing with a -2 and 5.
Multiply the second equation by 5 to get
-5x-10y + 5z = 50
Multiply the 3rd equation by 2 to bet
6x + 10y + 4z = -14
Now, combine those two equations and you will bet x + 9z =36
Now, combine this equation with the first equation. If you want to eliminate x, then multiply the first equation by negative 1.
-x -3z = -12 combine this with x+9z = 36
The x will be eliminated.
6z =48 Solve for z.
Once you have z, put it into the first equation and solve for x.
Then put your value for x and z into the second equation to find y.
Once you have (x,y,z) you should check those value in all three equations.
can someone help me with this math question?Solve the system of equations given below.
x+3z=12
-x-2y+z=10
3x+5y+2z=-7
I am really stumped. Please help! Thanks in advance!
2 answers
from the first: x = 12-3z
plug that into the 2nd:
-12+3z-2y+z = 10
4z - 2y = 22
2z - y = 11 , #4
plut x = 12-3z into the 3rd
36-9z +5y + 2z = -7
-7z + 5y = -43 , #5
#4 times 5 ---> 10z - 5y = 55
add this to #5
3z = 12
z = 4
in #4: 8-y = 11
y = -3
back in x = 12-3z
x = 12-12 = 0
x=0 , y = -3, z = 4
As you can see, there are many ways to do this.
Just wiggle the variables back and forth a bit, look for simple combinations
plug that into the 2nd:
-12+3z-2y+z = 10
4z - 2y = 22
2z - y = 11 , #4
plut x = 12-3z into the 3rd
36-9z +5y + 2z = -7
-7z + 5y = -43 , #5
#4 times 5 ---> 10z - 5y = 55
add this to #5
3z = 12
z = 4
in #4: 8-y = 11
y = -3
back in x = 12-3z
x = 12-12 = 0
x=0 , y = -3, z = 4
As you can see, there are many ways to do this.
Just wiggle the variables back and forth a bit, look for simple combinations
2 answers