In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x > y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship between x and y cannot be established.
I. $$3x^{2}-10x+8=0$$
II. $$2y^{2}-19y+35=0$$
I. $$3x^{2}-10x+8=0$$
=> $$3x^2 - 6x - 4x + 8 = 0$$
=> $$3x (x - 2) - 4 (x - 2) = 0$$
=> $$(3x - 4) (x - 2) = 0$$
=> $$x = 2 , \frac{4}{3}$$
II. $$2y^{2}-19y+35=0$$
=> $$2y^2 - 14y - 5y + 35 = 0$$
=> $$2y (y - 7) - 5 (y - 7) = 0$$
=> $$(2y - 5) (y - 7) = 0$$
=> $$y = 7 , \frac{5}{2}$$
Therefore $$x < y$$
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