Question

The point of intersection of lines 7x-15y-2=0
and 6x + 12y - 18 = 0 is
(A)( -49/29 , 19/ 29 )
(B) ( 49 /29 , 19 / 29 )
(C) ( 49/29 , -19 / 29 )
(D) None of these
​

Answer 1

Answer:

$\frac{49}{29},\frac{19}{29}$

Step-by-step explanation:

Given :$7x-15y-2=0$

            $6x + 12y - 18 = 0$

To Find : Point of intersection

Solution :

$7x-15y-2=0$ --A

$6x + 12y - 18 = 0$ --B

Substitute the value of x from B in A

$7(\frac{18-12y}{6})-15y-2=0$

$\frac{126-84y}{6})-15y-2=0$

$y=\frac{19}{29}$

Substitute the value of y in A

$7x-15(\frac{19}{29})-2=0$

$x=\frac{49}{29}$

Hence the points of intersection are $\frac{49}{29},\frac{19}{29}$