How do you solve the following system of equations?: $x+y= -2 , 4x+3y=8$ ?

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Answer 1 · 2016-02-02T03:01:16

the answer: $x=14$ and $y = -16$

In the two equations, get $y$ alone. So you have:

$y= -x-2$ and

$y= -4/3 x +8/3$ .

Then you set the two equations equal to each other:

$-x-2= -4/3 x +8/3$ .

Then you solve for $x$ , and you get $x=14$ . You then substitute $14$ into both equations to get $y$ , and then you have your coordinates.

Answer 2 · 2016-02-02T12:17:16

Solve by substitution and elimination:

$x+y=-2$

$4x+3y=8$

We can eliminate $4x$ from the second equation by $x$ from the first equation if we multiply it by $-4$ to get $-4x$ :

$rarr-4(x+y=-2)$

$rarr-4x-4y=8$

Now add both of the equations:

$rarr(-4x-4y=8)+(4x+3y=8)$

$rarr-1y=16$

$rarr-y=16$

So,if $-y =16$ then , $y =-16$

Now substitute the value of $y$ to the first equation:

$x+(-16)=-2$

$x-16=-2$

$x=-2+16=14$

So, $(x,y)=(14,-16)$

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How do you solve the following system of equations?: x+y= -2 , 4x+3y=8 x+y= -2 , 4x+3y=8 ?