How do you solve the system of equations $20 x + 3 y = 20$ $20x + 3y = 20$ and $- 20 x + 5 y = 60$ $- 20x + 5y = 60$ ?

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Answer 1 · 2017-03-16T21:50:38

$x = - \frac{1}{2}$ $x=-1/2$
$y = 10$ $y=10$

$20 x + 3 y = 20$ $20x+3y=20$
$- 20 x + 5 y = 60$ $-20x+5y=60$

You can add all.

$8 y = 80$ $8y=80$
$y = 10$ $y=10$

We know that y=10, so we can come back to the one of the first equations. I'll choose $20 x + 3 y = 20$ $20x+3y=20$ .

$20 x + 3 \cdot 10 = 20$ $20x+3*10=20$
$20 x + 30 = 20$ $20x+30=20$
$20 x = - 10$ $20x=-10$
$x = - \frac{1}{2}$ $x=-1/2$

How do you solve the system of equations 20x + 3y = 2020x+3y=2020x + 3y = 20 and - 20x + 5y = 60−20x+5y=60- 20x + 5y = 60?