How do you find the solution of the system of equations $2x – y = 4$ and $3x + y = 1$ ?

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Answer 1 · 2015-05-02T07:20:56

We can add the two equations to eliminate $y$

Adding the 2 equations will give us:

$(2x-y) +(3x + y) = 4+ 1$

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

$5x = 5$

Dividing both sides by 5 gives us

$(cancel(5)x)/cancel(5) = cancel(5)/cancel(5)$

$color(green)(x = 1$

Now we can substitute $x =1$ in the second equation to get $y$

$3*1 + y = 1$

$3 + y = 1$

Subtracting 3 from both sides will give us:

$cancel(3) + y - cancel(3) = 1 - 3$

$color(green)(y = -2$

The solution to these system of equations is:

$color(green)(x = 1 ; y = -2$

How do you find the solution of the system of equations 2x – y = 4 2x – y = 4 and 3x + y = 1 3x + y = 1 ?