How do you solve the following system: $- 2 x + 5 y = 20, x + y = 4$ $-2x + 5y = 20, x + y = 4$ ?

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Answer 1 · 2018-03-30T09:44:10

Just arrange the equation. Get $x = 0$ $x=0$ and $y = 4$ $y=4$

$- 2 x + 5 y = 20$ $-2x + 5y = 20$
$2 x + 2 y = 8$ $2x +2y = 8$

(after expanding the second equation by 2).

Now combine these equations:

$5 y + 2 y = 20 + 8$ $5y + 2 y = 20+8$

$7 y = 28$ $7y = 28$

$y = \frac{28}{7}$ $y=28/7$

$y = 4$ $y=4$

Now put this value into the second equation:

$x + y = 4$ $x+y=4$

$x = 4 - y$ $x=4-y$

$x = 4 - 4 = 0$ $x=4-4=0$

How do you solve the following system: -2x + 5y = 20, x + y = 4−2x+5y=20,x+y=4-2x + 5y = 20, x + y = 4?