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marika35
07.12.2019 •
Mathematics
Solve the following system of equation
2x + 3y - z = 1
3x + y + 2z = 12
x + 2y - 3 = -5
a) (3 , 1 , 2)
b) (-3 , 1 , 2)
c) (3 , -1 , 2)
d) (3 , 1 , -2)
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Ответ:
x = 6 , y = -4 , Z = -1
Step-by-step explanation:
Solve the following system:
{2 x + 3 y - Z = 1 | (equation 1)
3 x + y + 2 Z = 12 | (equation 2)
-3 + x + 2 y = -5 | (equation 3)
Express the system in standard form:
{2 x + 3 y - Z = 1 | (equation 1)
3 x + y + 2 Z = 12 | (equation 2)
x + 2 y+0 Z = -2 | (equation 3)
Swap equation 1 with equation 2:
{3 x + y + 2 Z = 12 | (equation 1)
2 x + 3 y - Z = 1 | (equation 2)
x + 2 y+0 Z = -2 | (equation 3)
Subtract 2/3 × (equation 1) from equation 2:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+(7 y)/3 - (7 Z)/3 = -7 | (equation 2)
x + 2 y+0 Z = -2 | (equation 3)
Multiply equation 2 by 3/7:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+y - Z = -3 | (equation 2)
x + 2 y+0 Z = -2 | (equation 3)
Subtract 1/3 × (equation 1) from equation 3:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+y - Z = -3 | (equation 2)
0 x+(5 y)/3 - (2 Z)/3 = -6 | (equation 3)
Multiply equation 3 by 3:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+y - Z = -3 | (equation 2)
0 x+5 y - 2 Z = -18 | (equation 3)
Swap equation 2 with equation 3:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+5 y - 2 Z = -18 | (equation 2)
0 x+y - Z = -3 | (equation 3)
Subtract 1/5 × (equation 2) from equation 3:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+5 y - 2 Z = -18 | (equation 2)
0 x+0 y - (3 Z)/5 = 3/5 | (equation 3)
Multiply equation 3 by 5/3:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+5 y - 2 Z = -18 | (equation 2)
0 x+0 y - Z = 1 | (equation 3)
Multiply equation 3 by -1:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+5 y - 2 Z = -18 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Add 2 × (equation 3) to equation 2:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+5 y+0 Z = -20 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Divide equation 2 by 5:
{3 x + y + 2 Z = 12 | (equation 1)
0 x+y+0 Z = -4 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Subtract equation 2 from equation 1:
{3 x + 0 y+2 Z = 16 | (equation 1)
0 x+y+0 Z = -4 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Subtract 2 × (equation 3) from equation 1:
{3 x+0 y+0 Z = 18 | (equation 1)
0 x+y+0 Z = -4 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Divide equation 1 by 3:
{x+0 y+0 Z = 6 | (equation 1)
0 x+y+0 Z = -4 | (equation 2)
0 x+0 y+Z = -1 | (equation 3)
Collect results:
{x = 6 , y = -4 , Z = -1
Ответ:
1. nucleus
2. ribosomes