Question

the point of intersection of line 7 x minus 15 Y minus 2 is equals to zero and 6 X + 12 Y - 18 is equals to zero is​

Answer 1

Answer:

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

Step-by-step explanation:

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

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

Substitute the value of x from A in B

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

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

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

Substitute the value of y in B

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

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

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

Answer 2

The point of intersection is $\frac{49}{29},\frac{19}{29}$

Step-by-step explanation:

The equation given are

$7x-15y-2=0$

$7x-15y=2$ ......(1)

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

$6x + 12y= 18$ ......(2)

Solving (1) and (2) by substitution method,

Substitute x from (1) in (2),

From (1), $x=\frac{2+15y}{7}$ .....(3)

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

$\frac{12+90y}{7}+ 12y= 18$

$\frac{12+90y+84y}{7}= 18$

$174y+12=126$

$174y=114$

$y=\frac{114}{174}$

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

Substitute the value of y in (3),

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

$x=\frac{2+\frac{285}{29}}{7}$

$x=\frac{\frac{58+285}{29}}{7}$

$x=\frac{\frac{343}{29}}{7}$

$x=\frac{343}{29\times 7}$

$x=\frac{343}{203}$

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

The point of intersection is $\frac{49}{29},\frac{19}{29}$

#Learn More

System of equations

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