How do you solve by substitution $x \pm y = 1$ $x+-y=1$ and $- 2 x + y = 4$ $-2x+y=4$ ?

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Answer 1 · 2015-09-08T15:41:36

The problem has two solutions:
$x = - 1, y = 2$ $x=-1,y=2$ .
$x = - 5, y = - 6$ $x=-5,y=-6$ .

To solve by substitution, we isolate one of the variables in one equation, and use that to solve the other equation.

$x \pm y = 1$ $x+-y =1$ shows two cases: $x + y = 1$ $x+y=1$ or $x - y = 1$ $x-y=1$ .

Case 1:
Our system is:

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

Since the first equation is easier to work with, we can isolate a variable there.
$x + y = 1$ $x+y=1$
$x = 1 - y$ $x=1-y$

Now, we substitute $1 - y$ $1-y$ into $x$ $x$ in the other equation, and simplify.

$- 2 x + y = 4$ $-2x+y=4$
$- 2 (1 - y) + y = 4$ $-2(1-y)+y=4$
$- 2 + 2 y + y = 4$ $-2+2y+y=4$
$3 y = 6$ $3y=6$
$y = 2$ $y=2$

We now know the value of $y$ $y$ , which we can use to solve for $x$ $x$ using either equation:

$x + y = 1$ $x+y=1$
$x + 2 = 1$ $x+2=1$
$x = - 1$ $x=-1$

Therefore, the solution to Case 1 is $x = - 1, y = 2$ $x=-1,y=2$ .

Case 2:
Our system is:

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

Like before, we isolate one of the variables in one equation and solve using the other equation.

$x - y = 1$ $x-y=1$
$x = 1 + y$ $x=1+y$

$- 2 x + y = 4$ $-2x+y=4$
$- 2 (1 + y) + y = 4$ $-2(1+y)+y=4$
$- 2 - 2 y + y = 4$ $-2-2y+y=4$
$- y = 6$ $-y=6$
$y = - 6$ $y=-6$

At this point, we know the $y = - 6$ $y=-6$ , so we can solve for $x$ $x$ using either equation:

$x - y = 1$ $x-y=1$
$x + 6 = 1$ $x+6=1$
$x = - 5$ $x=-5$

Therefore the solution to case 2 is $x = - 5, y = - 6$ $x=-5,y=-6$ .

The problem has two solutions:
$x = - 1, y = 2$ $x=-1,y=2$ .
$x = - 5, y = - 6$ $x=-5,y=-6$ .

How do you solve by substitution x+-y=1x±y=1 x+-y=1 and -2x+y=4−2x+y=4-2x+y=4?