To solve this system, we can use the method of substitution or elimination.
Let's solve it using the method of substitution.
From the first equation:
3x + y - 4z = -30 ---> equation (1)
From the second equation:
3x + 2y + 2z = -8 ---> equation (2)
We can solve equation (1) for y:
y = -30 - 3x + 4z
Now substitute y in equation (2) with its value:
3x + 2(-30 - 3x + 4z) + 2z = -8
Simplify the equation:
3x - 60 - 6x + 8z + 2z = -8
-3x + 10z = 52 ---> equation (3)
From the third equation:
5x + 5y + z = -26 ---> equation (4)
We can solve equation (4) for x:
x = (-26 - 5y - z)/5
Now substitute x in equation (3) with its value:
-3((-26 - 5y - z)/5) + 10z = 52
Simplify the equation:
62 + 15y + 3z + 10z = 260
15y + 13z = 198 ---> equation (5)
Now we have two equations with two variables:
-3x + 10z = 52 ---> equation (3)
15y + 13z = 198 ---> equation (5)
We can solve this system using the method of elimination by multiplying equation (5) by 3 and equation (3) by 15 to make the coefficients of z the same:
-9x + 30z = 156 ---> equation (6)
45y + 39z = 594 ---> equation (7)
Now we can subtract equation (6) from equation (7) to eliminate z:
45y + 39z - (-9x + 30z) = 594 - 156
45y - 39z + 9x - 30z = 438
45y + 9x - 69z = 438 ---> equation (8)
Now we can solve equations (3) and (8) as a system of two equations with two variables:
-3x + 10z = 52 ---> equation (3)
45y + 9x - 69z = 438 ---> equation (8)
Let's multiply equation (3) by 9 and equation (8) by 3 to make the coefficients of x the same:
-27x + 90z = 468 ---> equation (9)
27x - 207z = 1314 ---> equation (10)
Adding equation (9) to equation (10), we eliminate x:
(-27x + 90z) + (27x - 207z) = 468 + 1314
-117z = 1782
z = -15
Now substitute z in equation (3) to find x:
-3x + 10(-15) = 52
-3x - 150 = 52
-3x = 202
x = -67.33
Finally, substitute x and z in equation (5) to find y:
15y + 13(-15) = 198
15y - 195 = 198
15y = 393
y = 26.2
Therefore, the solution to the system of equations is:
x = -67.33
y = 26.2
z = -15
Solve the system
{3x+y-4z=-30
{3x+2y+2z=-8
{5x+5y+z=-26
1 answer