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

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Answer 1 · 2016-03-07T02:56:44

$x = - 3 and y = 5$ $x = -3 and y = 5$

Since $x + y = 2$ $x+y = 2$ , we can say that $x = 2 - y$ $x = 2 - y$ .

We then substitute this into the equation $2 x + y = - 1$ $2x + y = -1$ .

So it becomes $2 (2 - y) + y = - 1$ $2(2 - y) + y = -1$ .

Open the bracket and simplify.

$4 - 2 y + y = - 1$ $4 - 2y + y = -1$

$4 - y = - 1$ $4 - y = -1$

Add $(1 + y)$ $(1+y)$ to both sides of the equation and simplify.

$4 - y + (1 + y) = - 1 + (1 + y)$ $4 - y + (1+y) = -1 + (1+y)$

$4 cancel(-y) + (1+cancel(y)) = cancel(-1) + (cancel(1)+y)$

$5 = y$

Substitute the value of $y$ back in the first equation.

$x = 2 - y$

$= 2 - 5$

$= -3$

So we have $x = -3$ and $y = 5$ . To check your answer. Substitute both values into both equations.

$x+y = -3 + 5 = 2$

$2x+y = 2(-3) + 5 = -1$

They match.

How do you solve x + y = 2x+y=2x + y = 2 and 2x + y = -12x+y=−12x + y = -1 using substitution?