if 1 is added to each …
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Sia ? 6 years, 4 months ago
Let the required numbers be x and y respectively.
Then,
{tex}\frac { x + 2 } { y + 2 } = \frac { 1 } { 2 } \Rightarrow 2 x + 4 = y + 2{/tex}
{tex}\Rightarrow 2 x - y = - 2{/tex}
and {tex}\frac { x - 4 } { y - 4 } = \frac { 5 } { 11 } \Rightarrow 11 x - 44 = 5 y - 20{/tex}
{tex}\Rightarrow 11 x - 5 y = 24{/tex}
{tex}\therefore 2x - y = -2{/tex} .......(i)
{tex}11x - 5y = 24{/tex} .....(ii)
Multiplying (i) by 5 and (ii) by 1,
{tex}10x - 15y = -10{/tex} .......(iii)
{tex}11x - 5y = 24{/tex} .........(iv)
Subtracting (iii) and (iv), we get
{tex}x = 34{/tex}
Putting {tex}x = 34{/tex} in (i), we get
2 {tex}\times{/tex} {tex}34 - y = 2{/tex}
{tex}\Rightarrow 68 - y = - 2{/tex}
{tex}\Rightarrow y = - 2 - 68{/tex}
{tex}\Rightarrow y = 70{/tex}
Hence, the required numbers are 34 and 70.
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