A motorboat can travel 30 km …

Let the speed of the boat in still water be 'x' km/hr and speed of the stream be 'y' km/

Speed = Distance / Time

{tex}\therefore{/tex} {tex}\frac { 30 } { x - y } + \frac { 28 } { x + y } = 7{/tex}

and {tex}\frac { 21 } { x - y } + \frac { 21 } { x + y } = 5{/tex} 

Let {tex}\frac { 1 } { x - y } \text { be } a \text { and } \frac { 1 } { x + y } \text { be } b{/tex}

30a + 28b = 7  ......(i)

21a + 21b = 5  ......(ii)

Multiplying (i) by 3 and (ii) by 4 and then subtracting.

{tex}90a+84b=21{/tex} ..............(iii)

{tex}84a+84b=20 {/tex} ..............(iv)

By solving (iii) and (iv)

{tex}90a-21=84a-20{/tex}

{tex}\Rightarrow{/tex}6a= 1

{tex}\Rightarrow{/tex} {tex}a = \frac { 1 } { 6 }{/tex}

Putting this value of ,a in eqn., (i),

{tex}30 \times \frac { 1 } { 6 } + 28 b = 7{/tex}

{tex}28 b = 7 - 30 \times \frac { 1 } { 6 } = 2{/tex}

{tex}\therefore{/tex}{tex}b = \frac { 1 } { 14 }{/tex}

x + y = 14 ...(iv)

Now, {tex}a = \frac { 1 } { x - y } = \frac { 1 } { 6 }{/tex}

{tex}\Rightarrow{/tex} x - y = 6 

{tex}\Rightarrow{/tex}x = y + 6 .....(v)

Putting (iv) in (v)

y + 6 + y = 14

{tex}\Rightarrow{/tex} y = 4

Hence, speed of the boat in still water = 10 km/hr and speed of the stream = 4 km/hr.