Question

if 3x/16+2y/7=420 and x+y=1800 find the value of x and y​

Answer 1

Answer:

We will find the value of x and y by substitution method.

Given,

$\frac{3x}{16} + \frac{2y}{7} = 420 \\\frac{(21x + 32y)}{112} = 420 \\ 21x + 32y = 47040 \\ 21x = 47040 - 32y \\ x = \frac{(47040 - 32y)}{21}$

Putting the value of ‘x’ in equation (ii),

$x + y = 1800 \\ y + \frac{(47040 - 32y)}{21} = 1800 \\ \frac{21y + 47040 - 32y}{21} = 1800\\47040 - 11y = 37800 \\ 11y = 47040 - 37800 \\ y = \frac{9240}{11} = 840$

Talking this value of ‘y’ obtained in equation (i), we get,

$x = \frac{(47040 - 32y)}{21} \\ \: \: \: \: \: \: \: x = \frac{(47040 - 32(840))}{21} \\ \: \: \: x = \frac{(47040 - 26880)}{21} \\x = \frac{20160}{21} \\ x = 960$

∴ x = 960

y = 840

Hope It Helps

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