Question

Solve by elimination method 3x+4y=12,2x + 6y=42

Answer 1

$3x + 4y = 12.........(1) \\ 2x + 6y = 42 \\ both \: sides \: divided \: by \: 2\\ x + 3y = 21...........(2) \\ multiply \: (2) \: by \: - 3 \\ (2) \times - 3 = > \\ - 3x - 9y = - 63...............(3)\\ \\ + 3x + 4y = + 12.............(1) \\ - 3x - 9y = - 63.............(3) \\ ............................... \\ \: \: \: \: \: \: \: \: \: \: - 5y = - 51 \\ y = \frac{51}{5} \\ subtitute \: y \: in \: (2) \\ x + 3 \frac{51}{5} = 21 \\ x = 21 - \frac{153}{5} \\ x = \frac{105 - 153}{5} \\ x = \frac{ - 48}{5} \\ \\so, x = \frac{ - 48}{5} \: \:and \: \:y = \frac{51}{5}$

Answer 2

Answer:

Step-by-step explanation:

3x+4y=12......[1]

2x+6y=42......[2]

on multiplying eqn [1] buy 2 and eqn[2] by 3

thus, [3x+4y=12] 2

=6x+8y=24...............[3]

and,[2x+6y=42]

=6x+18y=126............[4]

on subtracting eqn[3]-[4]

6x+8y=24

6x+18y=126

    {-}     {-}

___________________

   -10y=-102

y=-102/-10

y=51/5

on substituting in eqn [2]

2x+6y=42

2x+306/5=42

2x=42-306/5

2x=96/5

x=96/10

x=48/5

thus x=51/5 and y=-48/5