How do you evaluate this system of equations: $y - 2 x = - 5; y - x = - 3$ $y-2x=-5; y-x=-3$ ?

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Answer 1 · 2017-06-14T04:34:49

$x = 2$ $x=2$ , $y = - 1$ $y=-1$ ; Ordered Pair: $(2, - 1)$ $(2,-1)$

To solve this specific system, the easiest method is substitution.
$y - 2 x = - 5$ $y-2x=-5$
$y - x = - 3$ $y-x=-3$
Choose one of the equations .. I will pick $y - x = - 3$ $y-x=-3$ .
The objective is to get one of the variables ( $x$ $x$ or $y$ $y$ ) by itself on one side of the equation .. Pick a variable to isolate .. I will pick $y$ $y$ .
To isolate $y$ $y$ , ADD $x$ $x$ to both sides of the equation:
$y - x = - 3$ $y-x=-3$
$y = - 3 + x$ $y=-3+x$
Look at the OTHER equation (the one you did not choose) .. That would be $y - 2 x = - 5$ $y-2x=-5$ for me.
See how $y$ $y$ is EQUAL to $(- 3 + x)$ $(-3+x)$ ?
You can substitute the $y$ $y$ in $y - 2 x = - 5$ $y-2x=-5$ with $(- 3 + x)$ $(-3+x)$
It should look like this: $(- 3 + x) - 2 x = - 5$ $(-3+x)-2x=-5$
SIMPLIFY! Now we only have ONE variable in the equation.
$(- 3 + x) - 2 x = - 5$ $(-3+x)-2x=-5$
$- 3 - x = - 5$ $-3-x=-5$
$- x = - 2$ $-x=-2$
$x = 2$ $x=2$ [THIS IS THE ANSWER FOR X]
Take one of the equations and substitute $x$ $x$ for $2$ $2$ to solve for $y$ $y$ .
$y - 2 x = - 5$ $y-2x=-5$
$y - 2 (2) = - 5$ $y-2(2)=-5$
$y - 4 = - 5$ $y-4=-5$
$y = - 1$ $y=-1$ [THIS IS THE ANSWER FOR Y]

If your teacher wants it as $(x, y)$ $(x,y)$ , it is $(2, - 1)$ $(2,-1)$ .

How do you evaluate this system of equations: y−2x=−5;y−x=−3y-2x=-5; y-x=-3?