How do you solve the following system: $-2x + 5y = 20, x +4y = 16$ ?

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Answer 1 · 2018-03-21T22:43:50

Let's use substitution:

$-2x + 5y = 20$

$x + 4y = 16$

We need to solve for $x$ in the second equation

$x = 16 - 4y$

Now we substitute $(16 - 4y)$ for $x$ in the first equation

$-2(16 - 4y) + 5y = 20$

distribute the $-2$

$-32 + 8y + 5y = 20$

Solve for $y$ . Add $32$ to both sides

$13y = 52$

Divide by $13$ on both sides

$y = 4$

$* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *$

We have $y$ , let's find $x$ :

$x = 16 - 4y$

$x = 16 - 4(4)$

$x = 16 - 16$

$x = 0$

$* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *$

To check our work, let's plug our values for $x$ and $y$ into the first equation and see if it equals $20$ :

$-2(0) + 5(4)$

$0 + 20$

$20$ EQUALS $20$ ! We were right

How do you solve the following system: -2x + 5y = 20, x +4y = 16-2x + 5y = 20, x +4y = 16?