Find the equation of the line passing through the intersection of the lines `2x+y=5\ a n d\ x+3y+8=0` and parallel to the line `3x+4y=7.`

Answer

N/a

Explanation

Given equation of lines `2x+y=5`
and `x+3y=-8`
From Eq.(i), `y=5-2x`
Now, put the value of y in Eq, (ii),we get
`x+3(5-2x)=-8`
`rArr x+15-6x=-8`
`rArr -5x=-23rArrx=(23)/(5)`
Now, `x=(23)/(5)` put in Eq. (i) we get
`y=5-(46)/(5)=(25-46)/(5)=(-21)/(5)`
Since, the required line is parallel to the line `3x+4y=7`. So,slope of the line is `m=(-3)/(4)`
So, the equation of the line passing through the point `((23)/(5),(-21)/(5))`having slope `(-3)/(4)` is
`y+(21)/(5)=(-3)/(4)(x-(23)/(5))`
`rArr 4y+(84)/(5)=-3x+(69)/(5)`
`rArr3x+4y=(84-69)/(5)rArr3x+4y+(15)/(5)=0`
`rArr 3x+4y+3=0`