Question

solve it plzzzz.....

Answer 1

$\underline{\bold{Answer:-}}$

Solving by Substitution method:-

$29x + 28y = 49150 - - (1) \\ \\ 28x + 29y = 50600 - - (2)$

From eq'n ( 1 )

$29x + 28y = 49150 \\ \\ = > 28y = 49150 - 29x \\ \\ = > y = (\frac{49150 - 29x}{28} ) - - (3)$

Substitute eq'n ( 3 ) in eq'n ( 2 )

$28x + 29y = 50600 \\ \\ = > 28x + 29(\frac{49150 - 29x}{28} ) = 50600 \\ \\ = > 28x + ( \frac{1425350 - 841x}{28} ) = 50600 \\ \\ = > 784x + 1425350 - 841x = 1416800 \\ \\ = > - 57x = 1416800 - 1425350 \\ \\ = > - 57x = - 8550 \\ \\ = > x = \frac{8550}{57} \\ \\ = > x = 150$

From eq'n ( 3 )

$y = (\frac{49150 - 29x}{28} ) \\ \\ = > y = ( \frac{49150 - 29(150)}{28} ) \\ \\ = > y = \frac{49150 - 4350}{28} \\ \\ = > y = \frac{44800}{28} \\ \\ = > y = 1600$

Therefore, the value x = 150 and y = 1600

$\textbf{Hope it helps!}$