Question

23. Solve by substitution method 1. 3x+2y=12,5x-2y=4, 2.x/7+y/3=5, x/2-y/9=6​

Answer 1

Answer:

$\huge{\boxed{\boxed{\sf{(1)x=2\:and\:y=3}}}}$

$\huge{\boxed{\boxed{\sf{(2)x=14\:and\:y=9}}}}$

Step-by-step explanation:

$\huge{\boxed{\boxed{\underline{\sf{SOLUTION-}}}}}$

First question:

$3x + 2y = 12 - - - - - (1) \\ 5x - 2y = 4 \\ 5x = 4 + 2y \\ x = \frac{4 + 2y}{5} - - - - - (2) \\ putting \: value \: of \: x \: in \: (1) \\\\ 3 \times \frac{(4 + 2y)}{5} + 2y = 12 \\ \frac{12 + 6y}{5} + 2y = 12 \\ \frac{12 + 6y + 10y}{5} = 12 \\ 16y + 12 = 60 \\ 16y = 60 - 12 \\ 16y = 48 \\ y = \frac{48}{16} \\ y = 3 \\ putting \: value \: of \: y \: in \: (2) \\ \\x = \frac{4 + 2y}{5} \\ x = \frac{4 + 2 \times 3}{5} \\ x = \frac{10}{5} \\ x = 2$

$\huge{\boxed{\boxed{\sf{(1)x=2\:and\:y=3}}}}$

Second question:

$\frac{x}{7} + \frac{y}{3} = 5 \\ \frac{3x + 7y}{21} = 5 \\ 3x + 7y = 105 - - - - - (1) \\ \frac{x}{2} - \frac{y}{9} = 6 \\ \frac{9x -2y }{18} = 6 \\ 9x - 2y = 108 \\ 9x = 108 + 2y \\ x = \frac{108 + 2y}{9} - - - - - (2) \\ \\ putting \: value \: of \: x \: in \:( 1) \\ 3 \times \frac{(108 + 2y)}{9} + 7y = 105 \\ \frac{324 + 6y}{9} + 7y = 105 \\ \frac{324 + 6y + 63y}{9} = 105 \\ 324 + 69y =94 5 \\ 69y = 945 - 324 \\ 69y = 621 \\ y = \frac{621}{69} \\ y = 9 \\ \\ putting \: value \: of \: y \: in \: (2) \\ x = \frac{108+ 2 \times 9}{9} \\ x = \frac{108+ 18}{9} \\ x = \frac{126}{9} \\ x = 14$

$\huge{\boxed{\boxed{\sf{(2)x=14\:and\:y=9}}}}$

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Answer 2

Answer:

Step-by-step explanation: