How do solve the following linear system?: $- 6 x + 5 y = 1, 7 x + 2 y = 2$ $-6x+5y=1 , 7x+2y=2$ ?

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Answer 1 · 2018-02-21T00:30:29

By substitution.
$x = \frac{8}{47}$ $x = 8/47$ , $y = \frac{19}{47}$ $y = 19/47$

$- 6 x + 5 y = 1$ $-6x+5y = 1$
$+ 7 x + 2 y = 2$ $+7x+2y = 2$

We can take out x of the first equation, for example ;

$x = \frac{5 y - 1}{6}$ $x = (5y-1)/6$

and we substitute it into the second equation and make all denominators equal :

$7 (\frac{5 y - 1}{6}) + 2 y = 2$ $7((5y-1)/6) + 2y =2$

$\frac{35 y - 7}{6} + \frac{12 y}{6} = \frac{12}{6}$ $(35y-7)/6 + (12y)/6 = 12/6$

Multiply both sides by $6$ $6$ .

$35 y - 7 + 12 y = 12$ $35y-7 + 12 y = 12$

$47 y = 19$ $47y = 19$

$y = \frac{19}{47}$ $y = 19/47$

Then we take this value and put it into our equation for $x$ $x$ .

$x = \frac{(\frac{95}{47}) - 1}{6}$ $x=((95/47) - 1)/6$

$x=(48/47)/6 = cancel(48)/47 * 1/cancel(6) = 8/47$

How do solve the following linear system?: -6x+5y=1 , 7x+2y=2 −6x+5y=1,7x+2y=2 -6x+5y=1 , 7x+2y=2 ?