Question

17. Solve by using elimination method :
3x - 2y = 5
2x + 3y = 12​

Answer 1

$\large\bf\underline\red{Answer \: :- }$

Let,

$\sf{3x - 2y = 5 </p><p> \: \: \: \: \: ...(1)} \\ \sf{2x + 3y = 12 \: \: \: ...(2)}$

Multiplying equation (1) by 3

We get,

$\sf{9x - 6y = 15 \: \: \: ...(3) }$

Multiplying equation (2) by 2

We get,

$\sf{4x + 6y = 24 \: \: \: ...(4) }$

Add both equation (3) and (4)

$\sf : \implies{9x - 6y + (4x + 6y) = 15 + 24} \\ \sf : \implies{9x + 4x - 6y + 6y = 39} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{13x = 39} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{x = \frac{39}{13} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{ \boxed{ \sf \: x = 3}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Substitute x = 3 in equation (2)

$\sf : \implies{2(3) + 3y = 12} \: \: \: \: \\ \sf : \implies{6 + 3y = 12} \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{6 + y = \frac{12}{3} } \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{ 6 + y = 4} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies{y = 4 - 6 = \boxed{ - 2}}$

Answer :

x = 3

y = -2

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Answer 2

Answer:

3x-2y=5---------i

2x+3y=12--------ii

i=> 3x = 5 + 2y

=> x = 5 + 2y / 3( 5 pluse 2y divided by 3 )

The value of x put in ii

ii=> 2x + 3y = 12

=> 2 * 5 + 2y / 3 + 3y = 12 ( 2 into 5 pluse 2y divided by 3 )

=> 10 + 4y / 3 + 3y = 12

=> 10 + 4y + 9y / 3 ( LCM) = 12

=>10 + 13y = 12 * 3

=> 10 + 13y = 36

=> 13y = 36 - 10

=> y = 26 / 13

=> y = 2

There for , y's value put in i :-

i=> 3x - 2y = 5

=> 3x - 2 * 2 = 5

=> 3x - 4 = 5

=> 3x = 5 + 4

=> x = 9 / 3

=> x = 3

There for

x = 3

y = 2