Question

Solve the simultaneous linear equation

Answer 1

x = 5 and y =-2

Explanation:

2x+3y = 4

3x+2y = 11

Multiply first equation by 2 throughout and second equation by 3 we get

4x + 6y = 8

9x + 6y = 33

subtracting the equations we get

9x - 4x = 33-8

5x = 25

x = 5

substituting x= 5 in 1st equation

10 + 3y = 4

3y = 4-10

3y = -6

y = -6/3

y= -2

Answer 2

Required Solution :

For first equation,

• 2x + 3y = 4

`: \longmapsto \: \sf{2x \: = \: 4 - 3y}`

`: \longmapsto \: \boxed{\sf{x \: = \: (4 \: - \: 3y)/(2)}}`

Now, substitute this value of x in second equation.

`: \longmapsto \: \sf{3 \bigg( (4 - 3y)/(2) \bigg) + 2y \: = \: 11}`

`: \longmapsto \: \sf{3 * \bigg( (4 - 3y)/(2) \bigg) + 2y \: = \: 11}`

`: \longmapsto \: \sf{ (12 \: - \: 9y )/(2) + 2y \: = \: 11}`

`: \longmapsto \: \sf{ (12 \: - \: 9y \: + 4y )/(2) \: = \: 11}`

`: \longmapsto \: \sf{ (12 \: - \: 5y )/(2) \: = \: 11}`

`: \longmapsto \: \sf{ 12 \: - \: 5y \: = \: 11 *2}`

`: \longmapsto \: \sf{ 12 \: - \: 5y \: = \: 22}`

`: \longmapsto \: \sf{ - \: 5y \: = \: 22 \: - \: 12}`

`: \longmapsto \: \sf{ - \: 5y \: = \:10}`

`: \longmapsto \: \sf{ y \: = \: - (10)/(5) }`

`: \longmapsto \: \red{ \boxed{\bf{ y \: = \: - 2 }}}`

Finding out value of x :

`: \longmapsto \: \sf{x \: = \: (4 - 3( - 2))/(2) }`

`: \longmapsto \: \sf{x \: = \: (4 - 3 * ( - 2))/(2) }`

`: \longmapsto \: \sf{x \: = \: (4 + 6)/(2) }`

`: \longmapsto \: \sf{x \: = \: (10)/(2) }`

`: \longmapsto \: \sf{x \: = \: \cancel(10)/(2) }`

`: \longmapsto \: \red{ \boxed{\bf{ x \: = \: 5 }}}`