How do you solve the following system: $- 3 x + y = - 4, 4 x + 2 y = 0$ $-3x + y = -4, 4x + 2y = 0$ ?

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How do you solve the following system: $- 3 x + y = - 4, 4 x + 2 y = 0$ $-3x + y = -4, 4x + 2y = 0$ ?

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Answer 1 · 2016-03-16T22:21:00

$x = \frac{4}{5}$ $x=4/5$ and $y = - \frac{8}{5}$ $y=-8/5$

Explanation:

we begin with $- 3 x + y = - 4$ $-3x+y=-4$ and $4 x + 2 y = 0$ $4x+2y=0$

Now, I'm going to solve for $y$ $y$ in $- 3 x + y = - 4$ $-3x+y=-4$ which looks like $y = - 4 + 3 x$ $y=-4+3x$ .

So no we can replace the $y$ $y$ in the second equation with $- 4 + 3 x$ $-4+3x$ .

That gives us $4 x + 2 (- 4 + 3 x) = 0$ $4x+2(-4+3x)=0$ which we can simplify to $4 x - 8 + 6 x = 0$ $4x-8+6x=0$ . This becomes $10 x = 8$ $10x=8$ or $x = \frac{4}{5}$ $x=4/5$ .

No, we can solve for $y$ $y$ .

$y = - 4 + 3 (\frac{4}{5})$ $y=-4+3(4/5)$ , which is $- 4 + \frac{12}{5}$ $-4+12/5$ or $- \frac{8}{5}$ $-8/5$ .

Now we should double check our math. If we did this right then $4 (\frac{4}{5}) + 2 (- \frac{8}{5})$ $4(4/5)+2(-8/5)$ should equal $0$ $0$ . Let's see if it is: $\frac{16}{5} - \frac{16}{5}$ $16/5-16/5$ does in fact equal $0$ $0$ , which means that we were right!

Nice Job!

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How do you solve the following system: $- 3 x + y = - 4, 4 x + 2 y = 0$ $-3x + y = -4, 4x + 2y = 0$ ?

1 Answer

Explanation:

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How do you solve the following system: −3x+y=−4,4x+2y=0-3x + y = -4, 4x + 2y = 0 ?

1 Answer

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How do you solve the following system: $- 3 x + y = - 4, 4 x + 2 y = 0$ $-3x + y = -4, 4x + 2y = 0$ ?