If x and y are real numbers and (2x - 1)^2 + (3y - 1)^2 = 0, then [(1/x^2) + (1/y^2)], then 1. 25 2. 13 3. 1/13 4. -(1/13)

Given, ⇒ (2x - 1)2 + (3y - 1)2 = 0 ⇒ (2x - 1)2 = 0 and (3y - 1)2 = 0 ⇒ (2x - 1) = 0 and 3y - 1 = 0 ⇒ 2x = 1 and 3y = 1 ⇒ x = and y = . Substituting value of x and y in , we get : Hence, option 2 is the correct option.

Mathematics

If x and y are real numbers and (2x - 1) ² + (3y - 1) ² = 0, then $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) =$
25
13
$\dfrac{1}{13}$
$-\dfrac{1}{13}$

Linear Equations

1 Like

Answer

Given,

⇒ (2x - 1)² + (3y - 1)² = 0

⇒ (2x - 1)² = 0 and (3y - 1)² = 0

⇒ (2x - 1) = 0 and 3y - 1 = 0

⇒ 2x = 1 and 3y = 1

⇒ x = $\dfrac{1}{2}$ and y = $\dfrac{1}{3}$ .

Substituting value of x and y in $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big)$ , we get :

$\Rightarrow \Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) \\[1em] \Rightarrow \dfrac{1}{\Big(\dfrac{1}{2}\Big)^2} + \dfrac{1}{\Big(\dfrac{1}{3}\Big)^2} \\[1em] \Rightarrow \dfrac{1}{\dfrac{1}{4}} + \dfrac{1}{\dfrac{1}{9}} \\[1em] \Rightarrow 4 + 9 \\[1em] \Rightarrow 13.$

Hence, option 2 is the correct option.

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Mathematics

If x and y are real numbers and (2x - 1) ² + (3y - 1) ² = 0, then $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) =$
25
13
$\dfrac{1}{13}$
$-\dfrac{1}{13}$

Linear Equations

Answer

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Mathematics

If x and y are real numbers and (2x - 1) 2 + (3y - 1) 2 = 0, then (1x2+1y2)=\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) =(x21​+y21​)=2513113\dfrac{1}{13}131​−113-\dfrac{1}{13}−131​

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If x and y are real numbers and (2x - 1) ² + (3y - 1) ² = 0, then $\Big(\dfrac{1}{x^2} + \dfrac{1}{y^2}\Big) =$
25
13
$\dfrac{1}{13}$
$-\dfrac{1}{13}$

The solution of the simultaneous equations $2x + \dfrac{1}{y} = 0$ and $3x + \dfrac{1}{2y} = -2$ is :
x = 1, y = $\dfrac{1}{2}$
x = 1, y = $-\dfrac{1}{2}$
x = -1, y = $\dfrac{1}{2}$
x = -1, y = $-\dfrac12$

If $\dfrac{x}{6} + 6 = y, \dfrac{3x}{4} = 1 + y$ , then :
x = 12, y = 8
x = 10, y = 8
x = 8, y = 12
x = 12, y = 6

The solution of $\sqrt{5}x - \sqrt{7}y = 0$ and $\sqrt{3}y + \sqrt{13}x = 0$ is :
x = 0, y = 0
x = 0, y = 1
x = 1, y = 0
x = 1, y = -1

The solution of 0.4x + 3y = 1.2 and 7x - 2y = $\dfrac{17}{6}$ is :
$x = \dfrac{1}{3}, y = \dfrac{1}{2}$
$x = \dfrac{1}{2}, y = \dfrac{1}{3}$
$x = \dfrac{1}{3}, y = 1$
$x = 0, y = \dfrac{1}{2}$