How do you solve x+y=1 and 2x-y=2?

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How do you solve x+y=1 and 2x-y=2?

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Answer 1 · 2015-10-04T09:31:05

Add the two equations together to get $3 x = 3$ $3x=3$ , hence $x = 1$ $x = 1$ , then substitute this value of $x$ $x$ into the first equation to find $y = 0$ $y = 0$ .

Explanation:

If we add the two equations together, the left hand side becomes:

$x + y + 2 x - y = 3 x$ $x+y+2x-y = 3x$

and the right hand side becomes:

$1 + 2 = 3$ $1 + 2 = 3$

So $3 x = 3$ $3x = 3$ .

Divide both sides by $3$ $3$ to get $x = 1$ $x = 1$ .

Then substituting this value for $x$ $x$ in the first equation, we get:

$1 + y = 1$ $1 + y = 1$

Subtracting $1$ $1$ from both sides we get:

$y = 1 - 1 = 0$ $y = 1 - 1 = 0$

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How do you solve x+y=1 and 2x-y=2?

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