How do you solve the following system: $- 8 x + 2 y = 4, 2 x + 5 y = - 1$ $-8x+2y=4 , 2x+5y=-1$ ?

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Answer 1 · 2018-05-16T19:15:37

$x = - \frac{1}{2}, y = 0$ $x=-1/2,y=0$

$:.-8x+2y=4-------(1)$

$:.2x+5y=-1--------(2)$

$:.(2)xx4$

$:.8x+20y=-4-------(3)$

$:.(1)+(3)$

$:.22y=0$

$:.y=0$

substitute $y=0 in (2)$

$:.2x+5(0=-1)$

$:.2x=-1$

$:.x=-1/2$

~~~~~~~~~

check:-

substitute $x=-1/2$ and $y=0$ in $(1)$

$:.-8(-1/2)+2(0=4)$

$:.4+0=4$

$:.4=4$

How do you solve the following system: -8x+2y=4 , 2x+5y=-1 −8x+2y=4,2x+5y=−1 -8x+2y=4 , 2x+5y=-1 ?