Question

1/ 2(x+2y) + 5/ 3(3x-2 y) = -3/ 2​

Answer 1

Answer:

y2 - qy)/(py - r) = (t - 1)/(t + 1)

⇒ (t + 1)(y2 - qy) = (py - r)(t - 1)

⇒ ty2 - qty + y2 - qy = tpy - tr - py + r

⇒ ty2 + y2 - qty - qy - tpy + py + tr - r = 0

⇒ (t + 1)y2 + (p - qt - q - tp)y + (t - 1)r = 0

As the roots are numerically equal but of opposite sign

∴ p - qt - q - tp = 0

⇒ p - q = tp + qt

⇒ (p + q)t = p - q

⇒ t = (p - q)/( p + q)

Key Points

Let ax2 + bx + c = 0 be the quadratic equation in x where a, b and c are constants and a ≠ 0, If α and β are the roots of the equation, then

Sum of the roots: α + β = - b/a

If the roots are numerically equal but of opposite signs, then

α + β = 0

⇒ - b/a = 0

⇒ b = 0

Answer 2

Answer:

33x -14y + 9 = 0

Step-by-step explanation:

Given :

$\mathrm{\dfrac{1}{2}(x+2y) + \dfrac{5}{3}(3x-2y)=-\dfrac{3}{2}}$

To find :

Simplify the given expression

Solution :

$\longmapsto \mathrm{\dfrac{1}{2}(x+2y) + \dfrac{5}{3}(3x-2y)=-\dfrac{3}{2}}$

Equalizing the denominator on LHS,

$\implies \mathrm{\dfrac{3}{\not6^3}(x+2y) + \dfrac{10}{\not6^3}(3x-2y)=-\dfrac{3}{\not2}}$

$\implies \mathrm{3(x+2y) + 10(3x - 2y) = -9}$

$\implies \mathrm{3x + 6y + 30x - 20y = - 9}$

$\implies \mathrm{33x - 14y + 9 = 0}$

This is the finally obtained equation

Hope it helps!!