How do you solve $2 x - y = 7$ $2x-y=7$ and $3 x + y = 8$ $3x+y=8$ ?

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Answer 1 · 2015-12-01T10:59:23

$x = 3$ $x=3$ , $y = - 1$ $y=-1$

From the first equation, you know that $y = 2 x - 7$ $y=2x-7$ . Substitute this expression for $y$ $y$ in the second equation and get

$3 x + (2 x - 7) = 8$ $3x + (2x-7) = 8$

$5 x - 7 = 8$ $5x-7=8$

$5 x = 15$ $5x = 15$

$x = 3$ $x=3$

Once $x$ $x$ is known, we can derive $y$ $y$ from

$y = 2 x - 7 = 2 \cdot 3 - 7 = 6 - 7 = - 1$ $y=2x-7=2*3-7=6-7=-1$

How do you solve 2x−y=72x-y=7 and 3x+y=83x+y=8?