y^(2)=7y

Equation (3y+1)/(2)=(7y-5)/(4) is simplified to

Find the product of (y+(2)/(7)y^(2)) and (7y-y^(2)) and verify the result for y=3

Simplify: (1/3y^2-4/7y+11)-(1/7y-3+2y^2)-(2/7y-2/3y^2+2)

((1) / (3) y ^ (2) - (4) / (7) y + 5) - ((2) / (7) y- (2) / (3) y ^ (2) +2 ) - ((1) / (7) y-3 + 2y ^ (2))

Simplify: ((1)/(3)y^(2)-(4)/(7)y+11)-((1)/(7)y-3+2y^(2))-((2)/(7)y-(2)/(3)y^(2)+2)

Solve (7y^(2)+1)/(y^(2)-1)-4 ((y^(2)-1)/(7y^(2)+1))=-3

If y^2 = y + 7 , then what is the value of y^3 ? यदि y^2 = y + 7 है, तो y^3 का मान क्या है?

Simplify : (i) (5x - 9y) - (-7x + y) (ii) (x^(2) -x) -(1)/(2)(x - 3 + 3x^(2)) (iii) [7 - 2x + 5y - (x -y)]-(5x + 3y -7) (iv) ((1)/(3)y^(2) - (4)/(7)y + 5) - ((2)/(7)y - (2)/(3)y^(2) + 2) - ((1)/(7)y - 3 + 2y^(2))

3x^(2)y+7x^(2)y = _____