What are a and b if (a+2)/3+(b-1)/i-(a-1)*i^-1-(b+1)*2^-1 ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It is not possible to find the values of a and b from the expression you have provided. I have solved what I think would have been the equation: (a+2)/3 + (b-1)/i = (a-1)*i^-1 - (b+1)*2^-1

(a+2)/3+(b-1)/i = (a-1)*i^-1 - (b+1)*2^-1

=> (a+2)/3 + (b-1)/i = (a-1)/i - (b+1)/2

=> [i(a+2) + 3(b - 1)]/3i = [2(a - 1) - i(b + 1)]/2i

=> [2i(a+2) + 6(b - 1)] = [6(a - 1) - 3i(b + 1)]

=> [2ai + 4i + 6b - 6] = [6a - 6 - 3bi - 3i]

equate the real and complex coefficients

=> 2a + 4 = -3b - 3 and 6b - 6 = 6a - 6

6b - 6 = 6a - 6

=> a = b

substitute a for b in 2a + 4 = -3b - 3

=> 2b + 4 = -3b - 3

=> 5b = -7

=> b = -7/5

a = b = -7/5

The required values of a and b are a = -7/5 and b = -7/5

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll re-write the terms of the algebraic sum, using the property of negative power:

x^-y = 1/x^y

We'll multiply all over by 6i:

6i*(a+2)/3 + 6i*(b-1)/i - 6i*(a-1)/i - 6i*(b+1)/2 = 0

We'll simplify and we'll get:

2ia + 4i + 6b - 6 - 6a + 6 - 3ib - 3i = 0

We'll combine the real parts and the imaginary parts:

(6b - 6a) + i(2a + 4 - 3b - 3) = 0 + 0*i

We'll compare and we'll get:

6b - 6a = 0

b - a = 0

a - b = 0

a = b (1)

2a + 4 - 3b - 3 = 0

2a - 3b = -1 (2)

But a = b (1)

We'll substitute (1) in (2):

2a - 3a = -1

-a = -1

a = 1

Since a = b => b = 1

The values for a and b are equal and they are: a = 1 and b = 1.

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