How do you solve $3 x + y = 9$ $3x+y=9$ and $2 x + y = 4$ $2x+y=4$ ?

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Answer 1 · 2015-08-02T19:45:54

Subtract the second equation from the first to get $x = 5$ $x=5$ , then substitute that into either equation and rearrange to find $y = - 6$ $y=-6$

Subtracting the second equation from the first we find:

$x = (3 x + y) - (2 x + y) = 9 - 4 = 5$ $x = (3x+y)-(2x+y) = 9 - 4 = 5$

Then substitute $x = 5$ $x=5$ in the second equation to get:

$4 = 2 \cdot 5 + y = 10 + y$ $4 = 2*5+y = 10 + y$

Subtract $10$ $10$ from both ends to get:

$y = 4 - 10 = - 6$ $y = 4 - 10 = -6$

How do you solve 3x+y=93x+y=9 and 2x+y=42x+y=4?