Question

ax+4y+3=0, bx+5y+3=0 , cx+6y+3=0 then a+c is what?

Answer 1

Hey there !

Solution:

Given Equations:

• ax + 4y + 3 = 0 .......( 1 )
• bx + 5y + 3 = 0 ......( 2 )
• cx + 6y + 3 = 0 .......( 3 )

Now let us subtract ( 1 ) from ( 3 )

=> cx + 6y + 3 - ( ax + 4y + 3 ) = 0

=> cx - ax + 6y - 4y + 3 - 3 = 0

=> ( c - a ) x + 2y = 0 .......( 4 )

Now let us subtract ( 2 ) from ( 3 ). We get,

=> cx + 6y + 3 - ( bx + 5y + 3 ) = 0

=> cx + 6y + 3 - bx - 5y - 3 = 0

=> cx - bx + 6y - 5y + 3 - 3 = 0

=> ( c - b ) x + y = 0 ........( 5 )

Now let us subtract ( 1 ) from ( 2 ), We get,

=> bx + 5y + 3 - ( ax + 4y + 3 ) = 0

=> bx + 5y + 3 - ax - 4y - 3 = 0

=> bx - ax + 5y - 4y + 3 - 3 = 0

=> ( b - a ) x + y = 0 .......( 6 )

Subtracting ( 5 ) from ( 6 )  we get,

=> ( b - a ) x + y - ( ( c - b ) x + y ) = 0

=> ( bx - ax ) + y - ( cx - bx + y ) = 0

=> bx - ax + y - cx + bx - y = 0

=> 2bx - ax - cx = 0

=> 2bx = ax + cx

=> ( 2b ) x = ( a + c ) x

x gets cancelled on both sides. Hence we get,

=> 2b = a + c

Which is the required answer.

Hope my answer helped !