Answer:
x = -1
, y = -4
, z = -4
Step-by-step explanation:
Solve the following system:
{-x - 5 y + z = 17 | (equation 1)
-5 x - 5 y + 5 z = 5 | (equation 2)
2 x + 5 y - 3 z = -10 | (equation 3)
Swap equation 1 with equation 2:
{-(5 x) - 5 y + 5 z = 5 | (equation 1)
-x - 5 y + z = 17 | (equation 2)
2 x + 5 y - 3 z = -10 | (equation 3)
Subtract 1/5 × (equation 1) from equation 2:
{-(5 x) - 5 y + 5 z = 5 | (equation 1)
0 x - 4 y+0 z = 16 | (equation 2)
2 x + 5 y - 3 z = -10 | (equation 3)
Divide equation 1 by 5:
{-x - y + z = 1 | (equation 1)
0 x - 4 y+0 z = 16 | (equation 2)
2 x + 5 y - 3 z = -10 | (equation 3)
Divide equation 2 by 4:
{-x - y + z = 1 | (equation 1)
0 x - y+0 z = 4 | (equation 2)
2 x + 5 y - 3 z = -10 | (equation 3)
Add 2 × (equation 1) to equation 3:
{-x - y + z = 1 | (equation 1)
0 x - y+0 z = 4 | (equation 2)
0 x+3 y - z = -8 | (equation 3)
Swap equation 2 with equation 3:
{-x - y + z = 1 | (equation 1)
0 x+3 y - z = -8 | (equation 2)
0 x - y+0 z = 4 | (equation 3)
Add 1/3 × (equation 2) to equation 3:
{-x - y + z = 1 | (equation 1)
0 x+3 y - z = -8 | (equation 2)
0 x+0 y - z/3 = 4/3 | (equation 3)
Multiply equation 3 by 3:
{-x - y + z = 1 | (equation 1)
0 x+3 y - z = -8 | (equation 2)
0 x+0 y - z = 4 | (equation 3)
Multiply equation 3 by -1:
{-x - y + z = 1 | (equation 1)
0 x+3 y - z = -8 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Add equation 3 to equation 2:
{-x - y + z = 1 | (equation 1)
0 x+3 y+0 z = -12 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Divide equation 2 by 3:
{-x - y + z = 1 | (equation 1)
0 x+y+0 z = -4 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Add equation 2 to equation 1:
{-x + 0 y+z = -3 | (equation 1)
0 x+y+0 z = -4 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Subtract equation 3 from equation 1:
{-x+0 y+0 z = 1 | (equation 1)
0 x+y+0 z = -4 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Multiply equation 1 by -1:
{x+0 y+0 z = -1 | (equation 1)
0 x+y+0 z = -4 | (equation 2)
0 x+0 y+z = -4 | (equation 3)
Collect results:
Answer: {x = -1
, y = -4
, z = -4
, which equals 
. Now, solve the integral of 
. You can check this by taking the integral of what you got. Now by the Fundamental Theorem![\int\limits^2_0 {4x} \, dx=[2x^2] ^{2}_{0}=2(2)^{2}-2(0)^2=8](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E2_0%20%7B4x%7D%20%5C%2C%20dx%3D%5B2x%5E2%5D%20%5E%7B2%7D_%7B0%7D%3D2%282%29%5E%7B2%7D-2%280%29%5E2%3D8)
.