Question

plz answer it guys l need help​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\sf{ \: If \: x+ iy = a + ib \: then \: x = a \: \: and \: \: y = b\: }$

TO DETERMINE

The values of a and b when

$\sf { \: (a + b)(2 + i) = b + 1 + (10 + 2a)i \: }$

$\implies \: \sf { \: 2(a + b) + i(a + b)= b + 1 + (10 + 2a)i \: }$

Comparing real and imaginary parts we get

$\implies \: \sf { \: 2(a + b) = b + 1 \: }$

$\implies \: \sf { \: 2a + b = 1 \: } \: ......(1)$

Also

$\: \sf { \: (a + b)= (10 + 2a) \: }$

$\implies \: \: \sf { \: a - b= - 10 \: }.......(2)$

Equation (1) + Equation (2) gives

$\: \sf { \: 3a= - 9 \: }$

$\implies \: \: \sf { \: a= - 3 \: }$

From Equation (2)

$\: \sf { \: b= - 3 + 10 = 7\: }$

RESULT

$\boxed{ \: \: \sf { \: a= - 3 \: \: and \: \: b = 7 \: } \: }$

Answer 2

Answer:

a = -3 and b = 7

Step-by-step explanation:

$:\implies\sf (a + b) \: (2 + i) = b + 1 + (10 + 2a)i$

$:\implies\sf 2a + 2b + \: ia + ib= b + 1 + 10i + 2ai$

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Therefore,

$:\implies\sf 2a + 2b = b + 1$

$:\implies\sf 2a + 2b - b = 1$

$:\implies\sf 2a + b = 1 \: \: \: \: \: \: \: \bigg \lgroup \bf equation \: (1) \bigg \rgroup$

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$:\implies\sf a + b = 10 + 2a$

$:\implies\sf b = 10 + 2a - a$

$:\implies\sf b = 10 + a\: \: \: \: \: \: \: \bigg \lgroup \bf equation \: (2) \bigg \rgroup$

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Now, Putting equation (2) in equation (1) we get:

$:\implies\sf 2a + b = 1$

$:\implies\sf 2a + 10 + a = 1$

$:\implies\sf 3a+ 10 = 1$

$:\implies\sf 3a = 1 - 10$

$:\implies\sf 3a = - 9$

$:\implies\sf a = \frac{ - 9}{3}$

$:\implies \underline{ \boxed{\sf a = - 3 }}$

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Substituting the value of a = -3 in equation (1) we get :

$:\implies\sf 2a + b = 1$

$:\implies\sf 2 ( - 3)+ b = 1$

$:\implies\sf - 6 + b = 1$

$:\implies\sf b = 1 + 6$

$:\implies \underline{ \boxed{\sf b = 7 }}$