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charlesmb7985
22.09.2020 •
Mathematics
Please help, Solve Systems of Linear Equations
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Ответ:
x = 1 , y = 1 , z = -3
Step-by-step explanation using elimination:
Solve the following system:
{3 x + 2 y - z = 8 | (equation 1)
2 z + 2 x = -4 | (equation 2)
3 y + x = 4 | (equation 3)
Subtract 2/3 × (equation 1) from equation 2:
{3 x + 2 y - z = 8 | (equation 1)
0 x - (4 y)/3 + (8 z)/3 = -28/3 | (equation 2)
x + 3 y+0 z = 4 | (equation 3)
Multiply equation 2 by 3/4:
{3 x + 2 y - z = 8 | (equation 1)
0 x - y + 2 z = -7 | (equation 2)
x + 3 y+0 z = 4 | (equation 3)
Subtract 1/3 × (equation 1) from equation 3:
{3 x + 2 y - z = 8 | (equation 1)
0 x - y + 2 z = -7 | (equation 2)
0 x+(7 y)/3 + z/3 = 4/3 | (equation 3)
Multiply equation 3 by 3:
{3 x + 2 y - z = 8 | (equation 1)
0 x - y + 2 z = -7 | (equation 2)
0 x+7 y + z = 4 | (equation 3)
Swap equation 2 with equation 3:
{3 x + 2 y - z = 8 | (equation 1)
0 x+7 y + z = 4 | (equation 2)
0 x - y + 2 z = -7 | (equation 3)
Add 1/7 × (equation 2) to equation 3:
{3 x + 2 y - z = 8 | (equation 1)
0 x+7 y + z = 4 | (equation 2)
0 x+0 y+(15 z)/7 = -45/7 | (equation 3)
Multiply equation 3 by 7/15:
{3 x + 2 y - z = 8 | (equation 1)
0 x+7 y + z = 4 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Subtract equation 3 from equation 2:
{3 x + 2 y - z = 8 | (equation 1)
0 x+7 y+0 z = 7 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Divide equation 2 by 7:
{3 x + 2 y - z = 8 | (equation 1)
0 x+y+0 z = 1 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Subtract 2 × (equation 2) from equation 1:
{3 x + 0 y - z = 6 | (equation 1)
0 x+y+0 z = 1 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Add equation 3 to equation 1:
{3 x+0 y+0 z = 3 | (equation 1)
0 x+y+0 z = 1 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Divide equation 1 by 3:
{x+0 y+0 z = 1 | (equation 1)
0 x+y+0 z = 1 | (equation 2)
0 x+0 y+z = -3 | (equation 3)
Collect results:
{x = 1 , y = 1 , z = -3
Ответ:
3g^2 = 30+9g