Equation of the line passing through the point of intersection of  x+y−4=0and 2x+3y−10=0 , parallel to the line  5x−7y+5=0

The equation of any line passing through the point of intersection of two lines x+y−4=0 and 2x+3y−10=0 can be taken as x+y−4+λ2x+3y−10=0x1+2λ+y1+3λ−4−10λ=0 Since the line x(1+2λ)+y(1+3λ)−4−10λ=0 is parallel to the line 5x−7y+5=0, 1+2λ5=1+3λ−7Cross multiply and then simplify −71+2λ=51+3λ−7−14λ=5+15λ29λ=−12λ=−1229 Substitute λ=−1229 in the equation x+y−4+λ(2x+3y−10)=0It implies that x+y−4−1229(2x+3y−10)=029x+29y−116−24x−36y+120=05x−7y+4=0 Therefore, the equation of the required line is 5x−7y+4=0