If $3x -7y = 10$ and $xy = -1$, find the value of $9x^2 + 49y^2$.

Given:

$3x -7y = 10$ and $xy = -1$

To do:

We have to find the value of $9x^2 + 49y^2$.

Solution:

We know that,

$(a+b)^2=a^2+b^2+2ab$

$(a-b)^2=a^2+b^2-2ab$

$(a+b)(a-b)=a^2-b^2$

Therefore,

$3x - 7y = 10$

Squaring both sides, we get,

$(3x - 7y)^2 = (10)^2$

$(3x)^2 + (7y)^2 - 2 \times 3x \times 7y = 100$

$9x^2 + 49y^2 - 42xy=100$

$9x^2 + 49y^2 =100+42(-1)$

$9x^2 + 49y^2 = 100 - 42$

$9x^2 + 49y^2 = 58$

The value of $9x^2+49y^2$ is $58$. 

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Updated on: 10-Oct-2022

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