Question

Question No. 25
If 5x - 7y= 19 and xy = -2, then the value of 25x^2 + 49y^2 is equal to
• 210
•215
•218
• 221​

Answer 1

Answer:

221

$(5x - 7y)^{2} = {19}^{2} \\ {25x}^{2} + {49x}^{2} - 70xy = 361 \\ {25x}^{2} + {49x }^{2} = 361 - 140 \: (xy = - 2) \\ {25x}^{2} + {49x}^{2} = 221$

Answer 2

The value of 25x² + 49y² = 221.

Given,

5x - 7y = 19

xy = -2

To Find,

the value of 25x² + 49y²

Solution,

This problem could be solved using the following method.

We have been given two equations.

Equation (1): 5x - 7y = 19

This is a linear equation in two variables.

Equation (2): xy = -2

This is also an equation in two variables.

We can rearrange equation (2):

x = -2/y

Let this be equation (3)

Substitute equation (3) in (1):

⇒ 5 (-2/y) - 7 y = 19

⇒ -10/y - 7 y = 19

⇒ -10/y - 7 y = 19

$\implies 5 \times \frac{-2}{y} - 7y = 19\\\\\implies \frac{-10}{y} - 7y = 19\\\\\implies \frac{-10-7y^2}{y} = 19\\\\\implies -10-7y^2 = 19y\\\\\implies 7 y^2 + 19y + 10 = 0$

This is a quadratic equation in one variable y

The solution for this equation is given by:

$y = \frac{-b \pm\ \sqrt{b^2-4ac}}{2a}$

where a = 7, b = 19 and c = 10

$y = \frac{-19 \pm \sqrt{19^2-4 \times 7 \times 10}}{2 \times 7}\\\\y = \frac{-19 \pm \sqrt{361 - 280}}{14}\\\\y = \frac{-19 \pm \sqrt{81}}{14}\\\\y = \frac{-19 \pm 9}{14}\\\\y = \frac{-19 + 9}{14}, \hspace{0.1 cm} y = \frac{-19 - 9}{14}\\\\y = \frac{-10}{14}, \hspace{0.1 cm} y = \frac{-28}{14}\\\\y = -0.714, \hspace{0.1 cm} y = -2$

Hence, the integral value of y is y = -2.

Substitute this value in equation (3):

x = -2/-2

⇒ x = 1.

Hence we have x=1 and y=-2.

The value of 25x² + 49y² becomes

⇒ 25(1)² + 49(-2)²

⇒ 25(1) + 49(4)

⇒ 25 + 196

⇒ 221

Therefore, 25x² + 49y² = 221.

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