SOLUTION: I have a test tomorrow, I have no clue how to graph this: Solve the system graphically: -3x + y = 7 -4x +4y +12 I got this so far : y + 3x + 7 y = x + 3 Then they want me

Question 687311: I have a test tomorrow, I have no clue how to graph this:
Solve the system graphically:
-3x + y = 7
-4x +4y +12
I got this so far :
y + 3x + 7
y = x + 3
Then they want me to graph it. Everyone tries to tell help me, but only makes it more confusing.
Found 2 solutions by MathLover1, stanbon:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!

Solved by pluggable solver: Solve the System of Equations by Graphing

Start with the given system of equations:

In order to graph these equations, we need to solve for y for each equation.

So let's solve for y on the first equation

Start with the given equation

Add to both sides

Rearrange the equation

Divide both sides by

Break up the fraction

Reduce

Now lets graph (note: if you need help with graphing, check out this solver)

Graph of

So let's solve for y on the second equation

Start with the given equation

Add to both sides

Rearrange the equation

Divide both sides by

Break up the fraction

Reduce

Now lets add the graph of to our first plot to get:

Graph of (red) and (green)

From the graph, we can see that the two lines intersect at the point (,) (note: you might have to adjust the window to see the intersection)

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Solve the system graphically:
-3x + y = 7
-4x +4y = 12
-----
Graph the 1st equation
y = 3x + 7
If x = 0, y = 7
If x = 1, y = 10
----
Plot (0,7) and (1,10) and draw a line thru them.
====================================================
Graph the 2nd equation:
!st simplify by dividing thru by -4 to get:
x - y = -3
2nd: Solve for "y":
y = x+3
---
If x = 0, y = 3
If x = 1, y = 4
----
Plot (0,3) and (1,4) and draw a line thru them.
------

-----------------
Solution is the point of intersection of the 2 lines: (-2,1)
==================
Cheers,
Stan H.
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