x = 4, y = 1 WARNING! i will write the first equation as $$4x^2 4y^2 8x 24y 140 = 0 \tag 1$$ form the second equation, we have $$2x = 5 3y, 4x^2 = (53y)^2 = 9y^2 30y 25 \tag 2$$ subbing $(2)$ in $(1)$ gives us $$(9y^230y25) 4y^24(53y)24y 140 = 0$$ this simplifies to $$0=13y^26y135= (y3)(13y45) $$ i hope you can take it from here Nature of Simultaneous Linear Equations The general form of a pair of linear equations in two variables is a1x1 b1y1 c1 = 0 (1) a2x2 b2y2 c2 = 0 (2) There are three conditions 1 If a1/ a2 ≠ b1/ b2, then both the equations have a
40 X Y 2 X Y 5 25 X Y 3 X Y 1
