Question

Solve for positive numbers x,y if 3x2+3y2=30 and 5x2−3y2=-22

Answer 1

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Answer 2

Answer:

$\large\boxed{\sf{x=\pm1\;\;and\;\;y=\pm3}}$

Step-by-step explanation:

Given a pair of quadratic equations,

$3 {x}^{2} + 3 {y}^{2} = 30 \: \: \: .......(1)$

$5 {x}^{2} - 3 {y}^{2} = - 22 \: \: \: \: ........(2)$

Adding eqn (1) and (2), we get,

$= > 3 {x}^{2} + 5 {x}^{2} = 30 - 22 \\ \\ = > 8 {x}^{2} = 8 \\ \\ = > {x}^{2} = \frac{8}{8} \\ \\ = > {x}^{2} = 1 \\ \\ = > {x}^{2} - 1 = 0 \\ \\ = > x = \pm1$

Now, when x = ±1

$= >( 5 \times { (\pm1)}^{2}) - 3 {y}^{2} = - 22 \\ \\ = > 3 {y}^{2} = 5 + 22 \\ \\ = > 3 {y}^{2} = 27 \\ \\ = > {y}^{2} = \frac{27}{ 3} \\ \\ = > {y}^{2} = 9 \\ \\ = > {y}^{2} - 9 = 0 \\ \\ = > y = \pm3$

Hence, the values of x = ±1 and y = ±3