Can anyone help me ??

Question

1.┌ 3x + y + 4z = 11
│ x - 2y + z = -2
└ x+ 3y + 3z = 14

2. (1+i)^ 4n-1 / (1-i)^ 4n-3

the 2nd one is just about "Imaginary Unit" =)

like, x^2 = -1
x = i

Answer 1

(1) 3x + y + 4z = 11
(2) x - 2y + z = -2
(3) x + 3y + 3z = 14
(4=3-2) 5y + 2z = 16
(5=2) 3x - 6y + 3z = -6
(6=1-5) 7y + z = 17
(7=6) 14y + 2z = 34
(8=7-4) 9y = 18 ==> y = 2 Answer
(4) 10 + 2z = 16 ==> z = 3 Answer
(2) x - 4 + 3 = -2 ===> x = -1 Answer

In Problem 2, note that Curious's third step above is a mistake. You can't add exponents with unlike bases.

[(1+i)^(4n-1)] / [(1-i)^(4n-3)]

First, we're going to get rid of the 4n exponents. Start with
(1+i)^(4n) / (1-i)^(4n) = [(1+i)/(1-i)]^(4n)

Now simplify (1+I)/(1-i) = [(1+i)^2 / (1^2 - i^2)]
= (1 + 2i + i^2) / [1 - (-1)] = 2i / 2 = i

So (1+i)^(4n) / (1-i)^(4n) = i^(4n) = 1

and a whole big part of this problem -- everything involving the 4n exponent -- just equals 1.

The rest of the problem has negative exponents, so let's flip it over so all the exponents are positive. When we do that, the problem becomes

(1-i)^3 / (1+i)

Multiply top & bottom by (1-i) to get

(1-i)^4 / (1^2 - i^2) = [(1-i)^4] / 2 because 1 - i^2 = 2

Let's expand that quartic binomial. It becomes

1^4 - 4*1^3*i + 6*1^2*i^2 - 4*1*i^3 + i^4

= 1 - 4i + 6i^2 - 4i^3 + i^4

= 1 - 4i - 6 + 4i + 1 = -4

So the problem reduces to

[(1-i)^4] / 2 = -4 / 2 = -2 That's your answer.

Answer 2

1.
3x+y+4z=11
x-2y+z=-2 so, x=-2+2y-z
x+3y+3z=14

3(-2+2y-z)+y+4z=11
-2+2y-z+3y+3z=14

-6+7y+z=11
-2+5y+2z=14

7y+z=11+6=17
5y+2z=14+2=16

-14y-2z=-34
5y+2z=16 so, -9y=-18 so y=2

so
5*2(substituted with y)+2z=16

16-10=2z
6=2z
z=3

so
x-2*2+3(substituted with first equation)=-2
x=-2+4-3=-1






2.(1+i)^4n-1/(1-i)^4n-3
=(1+i)^4n-1*(1-i)^-4n+3
=(1+i)^4n-1-4n+3
=(1+i)^2
=1+i^2+2i

Answer 3

1 meh... do it youself... not in the mood.

2

multiply the top and bottom by the conjugate

(1+i)^ 4n-1 / (1-i)^ 4n-3

(1+i)^ 4n-1 * (1+i)^ 4n-3 / [ (1-i)^ 4n-3 * (1+i)^ 4n-3]

Simplify

(1+i)^(8n-4) / 2^(4n-3)

Note (1+i)^(8n-4) = [(1+i)^4]^(2n-1) = (-4)^(n-2)

(-4)^(2n-1) / 2^(4n-3)

Answer 4

1. x = -1; y = 2; z = 3

2. not specific enough

Answer 5

this is a good way to solve your problem.but do your homework by own

Answer 6

Do your own homework.