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Studentka2010 [4]
3 years ago
15

Consider the following system of equations. StartLayout Enlarged left-brace 1st row negative 10 x squared minus 10 y squared = n

egative 300 2nd row 5 x squared + 5 y squared = 150 EndLayout
Mathematics
2 answers:
lubasha [3.4K]3 years ago
5 0

Answer:

answer is D :)

Step-by-step explanation:

ziro4ka [17]3 years ago
4 0

These 2 equations has no solution and the equations are independent of each other.

<u>Step-by-step explanation:</u>

-10x² -10y² = -300 ----a

5x² + 5y² = 150 ---- b

While trying to solve this,

We can multiply the eq. b by 2 so we will get eq. c and then add to eq. a we will get 0 as the solution.

10x² + 10y² = 300 ----c

-10x² -10y² = -300 ---a

Everything cutoff, we will get 0, and there is no solution to these equations.

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