Question

3x_4y=1 4x =3y+6 please solved this problem please

Answer 1

Now we have two equations :
1) 3x-4y=1 can also be written as
$x = \frac{4y + 1}{3} .......(1)$

2)4x=3y+6............(2)
Putting (1) in (2)
$4( \frac{4y + 1}{3} ) = 3y + 6 \\ \frac{16y + 4}{3} - 3y - 6 = 0 \\ \frac{16y + 4 - 9y - 18}{3} = 0 \\ 16y - 9y - 18 + 4 = 0 \\ 7y - 14 = 0 \\ y = \frac{14}{7} = 2$
Putting value of y in (1)
$x = \frac{4(2) + 1}{3} = \frac{8 + 1}{3} = \frac{9}{3} \\ x = 3$

Therefore the value of x=3 and y=2.
Hence solved.

Answer 2

Hola there..

Given equation are:

3x - 4y = 1. ..... equation 1
and
4x - 3y = 6. ..... equation 2

Adding both equation 1 and equation 2, we get

=> 7x - 7y = 7

=> x - y = 1. .... Equation 3

Subtracting both equation 1 and equation 2, we get

=> y - x = - 5. .... Equation 4

Adding equation 3 and equation 4, we get

=> 2y = -4

=> y = -2

Putting the value of y in equation 3, we get

=> x + 2 = 1

=> x = -1

Hope this helps....:)