How do solve the following linear system?: $- 9 x + 5 y = 1, - 3 x + 7 y = - 1$ $-9x+5y=1 , -3x+7y=-1$ ?

Question

1331 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-12T10:59:38

$x = y = - \frac{1}{4}$ $x=y=-1/4$

$(- 9 x + 5 y) - 3 \cdot (- 3 x + 7 y) = 1 - 3 \cdot (- 1)$ $(-9x+5y)-3*(-3x+7y)=1-3*(-1)$

$- 9 x + 5 y + 9 x - 21 y = 1 + 3$ $-9x+5y+9x-21y=1+3$

$- 16 y = 4$ $-16y=4$ , so $y = \frac{4}{- 16} = - \frac{1}{4}$ $y=4/(-16)=-1/4$

Hence,

$- 9 x + 5 \cdot (- \frac{1}{4}) = 1$ $-9x+5*(-1/4)=1$

$- 9 x - \frac{5}{4} = 1$ $-9x-5/4=1$

$- 9 x = \frac{9}{4}$ $-9x=9/4$ , thus $y = \frac{\frac{9}{4}}{- 9} = - \frac{1}{4}$ $y=(9/4)/(-9)=-1/4$

How do solve the following linear system?: -9x+5y=1 , -3x+7y=-1 −9x+5y=1,−3x+7y=−1 -9x+5y=1 , -3x+7y=-1 ?