In each of these questions, two equations are given. You have to solve these equations and nd out the values of x and y and :
Give answer
a:if x < y
b: if x > y
c: if x ≤ y
d: if x ≥ y
e: if x = y
I. $$8x^{2} + 6x = 5$$
II. $$12y^{2} - 22y + 8 = 0$$
I : $$8x^{2} + 6x - 5 = 0$$
=> $$8x^2 - 4x + 10x - 5 = 0$$
=> $$4x (2x - 1) + 5 (2x - 1) = 0$$
=> $$(4x + 5) (2x - 1) = 0$$
=> $$x = \frac{-5}{4} , \frac{1}{2}$$
II : $$12y^{2} - 22y + 8 = 0$$
=> $$12y^2 - 6y - 16y + 8 = 0$$
=> $$6y (2y - 1) - 8 (2y - 1) = 0$$
=> $$(6y - 8) (2y - 1) = 0$$
=> $$y = \frac{1}{2} , \frac{4}{3}$$
$$\therefore x \leq y$$
Create a FREE account and get:
