How do solve the following linear system?: $- 9 x + 5 y = 1, - x + 5 y = - 13$ $-9x+5y=1 , -x+5y=-13$ ?

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Answer 1 · 2015-11-26T12:15:44

$x = - \frac{7}{4}$ $x=-7/4$ , $y = - \frac{59}{20}$ $y=-59/20$

From the second equation, you have that $x = 5 y + 13$ $x=5y+13$ . Substitute this expression for $x$ $x$ in the first equation to have

$- 9 (5 y + 13) + 5 y = 1$ $-9(5y+13)+5y=1$ . Expand:

$- 45 y - 117 + 5 y = 1$ $-45y-117+5y=1$ . Isolate the $y$ $y$ terms:

$- 40 y = 118$ $-40y = 118$ . Solve for $y$ $y$ :

$y = - \frac{118}{40} = - \frac{59}{20}$ $y=-118/40 = -59/20$ .

Once $y$ $y$ is known, we can find back $x$ $x$ :

$x = 5 y + 13 = 5 \cdot (- \frac{59}{20}) + 13 = - \frac{7}{4}$ $x=5y+13 = 5*(-59/20)+13 = -7/4$ .

How do solve the following linear system?: −9x+5y=1,−x+5y=−13 -9x+5y=1 , -x+5y=-13 ?