SOLUTION: -x + y = 4 2x - 5y = -14 I am completely stuck on how to go about doing this problem. I know that the Solutions x and y are always intergers between 5 and -5. Each time i

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: -x + y = 4 2x - 5y = -14 I am completely stuck on how to go about doing this problem. I know that the Solutions x and y are always intergers between 5 and -5. Each time i      Log On


   



Question 149211: -x + y = 4
2x - 5y = -14
I am completely stuck on how to go about doing this problem.
I know that the Solutions x and y are always intergers between 5 and -5.
Each time i try I come up with No Solution.
But I am being told that is wrong.
x = ?
Y = ?
If you could, please show me a few steps on How to get the answer
to this problem.
I would greatly appreciate your help
Thank You,
Kerry G.

Found 2 solutions by Fombitz, josmiceli:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
There are several methods to solve this system of linear equations.
We could graph both lines and look for the intersection point.
Put both into slope-intercept form, y=mx%2Bb, and then graph.
-x%2By=4
y=x%2B4
+graph%28+300%2C+300%2C+-5%2C+5%2C+-10%2C+10%2C+x%2B4%29+
2x-5y=-14
-5y=-2x-14
y=%28-2x-14%29%2F%28-5%29
y=%282x%2B14%29%2F5
+graph%28+300%2C+300%2C+-5%2C+5%2C+-10%2C+10%2C+x%2B4%2C%282x%2B14%29%2F5%29+
Looks like a solution exists at (-2,2).
Although this method worked for this system, it's not always clear what the exact answer is.
The graphical method explains what you're really trying to do, find where lines intersect, if they do.
We'll work it out using the substitution method, to get an exact answer.
1.-x%2By=4
2.2x-5y=-14
Use eq. 1 to get the variable y as an expression with x.
1.-x%2By=4
y=x%2B4
Now substitute this expression into eq. 2 and solve for x,
2.2x-5y=-14
2x-5%28x%2B4%29=-14
2x-5x-20=-14
-3x=6
highlight%28x=-2%29
Now back substitute into eq. 1 or 2 to solve for y,
y=x%2B4
y=-2%2B4
highlight%28y=2%29

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
The basics:
You have 2 equations and 2 unknowns x, and y
In that situation, you can ALWAYS solve for the unknowns
Even if you had 30 equations and 30 unknowns, it would be
more difficult, but you could still solve for each of
your unknowns
----------------------------------------------
In this problem you can use the simple laws of algebra
to solve for x and y, mainly:
Whatever you do to one side of the equation, you MUST
do the same thing to the other side in order for the
equation to still be true
----------------------------------------------
-x+%2B+y+=+4
Add x to both sides
-x+%2B+x+%2B+y+=+4+%2B+x
On the left side, the x's cancel
y+=+4+%2B+x
That's what I want: one of the unknowns all by itself
Now I can replace y in the other equation with
4+%2B+x because it equals y
2x+-+5%2A%284+%2B+x%29+=+-14
Now I carry out the multiplication
2x+-+20+-+5x+=+-14
-3x+-+20+=+-14
Multiply both sides by -1
3x+%2B+20+=+14
Subtract 20 from both sides
3x+=+-6
x+=+-2
Now substitute this value for x back into either
of the equations
-x+%2B+y+=+4
-%28-2%29+%2B+y+=+4
2+%2B+y+=+4
Subtract 2 from both sides
y+=+2
The solutions are x = -2 and y = 2
Now check the answers:
-x+%2B+y+=+4
-%28-2%29+%2B+2+=+4
2+%2B+2+=+4
4+=+4
and
2x+-+5y+=+-14
2%2A%28-2%29+-+5%2A2+=+-14
-4+-+10+=+-14
-14+=+-14
OK