Question

The point of intersection of lines 3x + 6y = 9 and 7x – 15y = 2 is

Answer 1

Answer:

Step-by-step explanation:

3x+6y=9-------eq 1

7x-15y=2--------eq 2

x=(2+15y)/7------eq 3

Putting value of X in eq 1,

3{(2+15y)/7}+6y=9

6+45y+42y=63

y= 57/87

y= 19/29

Putting value of Y in eq 3

x=2+15*19/29/7

x=2+285/29/7

x=58+285/29*7

x=343/203

x= 49/29

Answer 2

Answer:

The point of intersection of lines   $3x+6y=9$ and  $7x-15y=2$ is ( $\frac{49}{29}$,$\frac{57}{87}$)

Step-by-step explanation:

• A point where 2 lines meet is called a point of intersection.
• A straight line has only one point of the intersection while curved lines can have many points of intersection.

Step 1 of 1:

• The given equations are  $3x+6y=9$ and  $7x-15y=2$
• Put the equation number on the equations

        $3x+6y=9\ \ \ \ \ \ \ \ \ \ ----\ \ eq 1$

        $7x-15y=2\ \ \ \ \ \ \ \ \ \ ----\ \ eq 2$

• multiply eq 1 by 7 and eq 2 by 3. Thus equation becomes

        $21x+42y=63\ \ \ \ \ \ \ \ \ \ ----\ \ eq 3$

        $21x-45y=6\ \ \ \ \ \ \ \ \ \ ----\ \ eq 4$

• Subtract eq 4 from 3

      $21x+42y-21x-(-45y)=63-6\\87y=57\\y=\frac{57}{87}$

• Put the value of y in any equation, the value of x can be calculated as $\frac{49}{29}$

Conclusion:

The point of intersection of lines is ( $\frac{49}{29}$,$\frac{57}{87}$)