Question

solve the following system of equations by substitution method
(1) 3x + 2y = 11 , 2x + 3y = 4
(2) 7x - 2y = 1 , 3x + 4y = 15​

Answer 1

Answer:

1) 3x + 2y = 11...…(I)

2x + 3y = 4...….(II)

from equation 1

y = (11 - 3x)/2

substituting in 2

2x + 3(11 - 3x)/2 = 4

4x + 33 - 9x = 8

-5x = -25

x = 5

and y = (-2)

2) 7x - 2y = 1...…(I)

3x + 4y = 15...…(II)

from equation 1

x = (1 + 2y)/7

substituting in 2

3(1+2y)/7 +4y = 15

3 + 6y + 28y = 105

34y = 102

y = 3

x = 1

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Answer 2

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$3x + 2y = 11..................(1) \\ 2x + 3y = 4.......................(2)$

from equation 1st,

$3x = 11 - 2y \\ x = \frac{11 - 2y}{3} ........................(3)$

substituting equation (3)in( 2),

=>$\frac{2(11 - 2y)}{3} + 3y = 4 \\ =>\frac{22 - 4y + 9y}{3} = 4 \\ \\=> 22 -5y = 12 \\=> \large \boxed{ \green{y = - 2}}$

$substituting \: value \: of \: y \: in \: (3)$

=>$x = \frac{11 - 2( - 2)}{3} \\=> x = \frac{15}{3} \\=> \boxed{ \green{x = 5}}$

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$7x - 2y = 1................(1)$

$3x + 4y = 15...............(2)$

from equation 1,

=>$7x = 1 + 2y \\ =>x = \frac{1 + 2y}{7} ..............(3)$

substituting equation 3 in 2,

=>$3( \frac{1 + 2y}{7} ) + 4y = 15 \\ =>\frac{3 + 6y}{7} + 4y = 15 \\ \\ =>\frac{3 + 6y + 28y}{7} = 15 \\ =>3 + 34y = 105 \\=> 34y = 102 \\ \boxed{ \green{ y = 3}}$

putting value of y in 3rd equation

$x = \frac{1 + 2(3)}{7} \\ \\ x = \frac{7}{7} \\ \\ \boxed{ \green{x = 1}}$

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