How do solve the following linear system?: $6 x + 4 y = - 1, 5 x - 5 y = 10$ $6x+4y=-1 , 5x-5y=10$ ?

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How do solve the following linear system?: $6 x + 4 y = - 1, 5 x - 5 y = 10$ $6x+4y=-1 , 5x-5y=10$ ?

1 Answer

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Answer 1 · 2016-01-12T16:54:22

$x = \frac{7}{10}$ $x=7/10$ , y= $- \frac{13}{10}$ $-13/10$

Explanation:

This can be done either by elimination or substitution. Fro second equation it is x= 2+y. Substituting it for x in the first equation it would be 6(2+y) +4y=-1 which gives y= $- \frac{13}{10}$ $-13/10$ and hence x= $2 - \frac{13}{10} = \frac{7}{10}$ $2-13/10 = 7/10$

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How do solve the following linear system?: $6 x + 4 y = - 1, 5 x - 5 y = 10$ $6x+4y=-1 , 5x-5y=10$ ?

1 Answer

Explanation:

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How do solve the following linear system?: 6x+4y=-1 , 5x-5y=10 6x+4y=−1,5x−5y=10 6x+4y=-1 , 5x-5y=10 ?

1 Answer

Explanation:

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How do solve the following linear system?: $6 x + 4 y = - 1, 5 x - 5 y = 10$ $6x+4y=-1 , 5x-5y=10$ ?