Question

solve the following pair of linear equationby substitution and elimination method. 2x+3y=5 and 3x-2y=1​

Answer 1

* Elimination Method *

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2x+3y=5..............(1)

3x-2y=1................(2)

Multiplying eq. (1) by 2 and (2) by 3 we get,

4x+6y=10.........(3)

9x-6y=3...........(4)

Adding (3) and (4) we get,

13x= 13

x=13/13

x=1

Substituting the value of x in (2) we get,

3(1) -2y=1

3-2y=1

-2y=1-3

-2y=-2

y=-2/-2

y=2/2

y=1

* Substitution Method *

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2x+3y=5.........(1)

3x-2y=1............(2)

From (2) ,

3x-2y=1

3x=1+2y

x=1+2y/3

Substituting the value of x in (1)

2(1+2y/3) +3y=5

2+4y/3+3y=5.

2+4y+9y=15..

2+13y=15

13y=15-2

13y=13

y=13/13.

y=1

Substituting the value of y in (1) we get,

2x+3(1) =5.

2x+3=5

2x=5-3

2x=2..

x=2/2

x=1

(1, 1) is the solution of the given equations

Answer 2

Answer:

Elimination method

$2x + 3y = 5 - - - eqn(1) \times 3$

$3x - 2y = 1 - - - eqn(2) \times 2$

$6x + 9y = 15 \\ 6x - 4y= 2 \\ - \: \: \: + \: \: \: \: \: \: - \\ - - - - - - - \\ \: \: \: \: \: \: \: \: \: \: 13y = 13$

$y = \frac{13}{13}$

•°•y=1

put y In eqn 1

2x+3y=5

2x+3 (1)=5

2x=5-3=2

•°• x=1

Substitution method

2x+3y=5

$x = \frac{5 - 3y}{2}$

put in 3x-2y=1

$3( \frac{5 - 3y}{2} ) - 2y = 1$

$\frac{15 - 9y }{2} - 2y = 1$

$\frac{15 - 9y - 4y}{2} = 1$

$15 - 13y = 2$

$15 - 2 = 13y$

$13y = 13$

•°•y=1

put y in eqn 1

2x+3y=5

2x+3 (1)=5

2x=5-3=2

•°•x=1