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4 years ago

14

Solve for p: √p − 1 − √p = 1

1 answer:

astraxan [27]4 years ago

4 0

<span>√p − 1 − √p = 1
- 1 = 1 (</span>√p is canceled out)<span>
since -1 cant equal 1 so there's no solution</span>

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Solve the following system of equation

krok68 [10]

Answer:

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:

Answer: {x = 6 , y = -4 , Z = -1

6 0

4 years ago

Hey guy, can yall please help me?

Arisa [49]

Answer:

x is greater than or equal to 3

Step-by-step explanation:

x ≥ 3

x is greater than or equal to 3

so you got the plot right

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