# How do you determine whether a linear system has one solution, many solutions, or no solution when given 3x - y = -6 and x + y = 2?

##### 1 Answer
Apr 17, 2018

One solution, (-1,3)

#### Explanation:

Although there are multiple ways one could solve this, I will begin with the algerbraic path to finding this solution. (Skip to bottom for shortcut)

We begin with two equations...
$3 x - y = - 6$ and
$x + y = 2$
Lets first put them into $y = m x + b$ format
1)$3 x - y = - 6$ Given
2)$- y = - 6 - 3 x$ Subtraction property of Equality
3) $y = 6 + 3 x$ Multiply both side by -1 to turn y positive
4)$y = 3 x + 6$ Commutative Property
Now the second equation
1)$x + y = 2$ Given
2)$y = - x + 2$ Subtract x from both sides.
Now that both are in y=mx+b format, set the equations equal to eachother.
$y = - x + 2 = 3 x + 6$
$- x + 2 = 3 x + 6$
Then solve for x.....
However there is a faster way to do this... using the process of elimination...
$3 x - y = - 6$
$x + y = 2$
_______ Add these equations together
4x=-4
x=-1
Then subsitute the x=-1 into any of the other equations to find the y value....