Question 256054
I need help with this one: 
Find the common solution to the following pair of lines by graphing them on the same axes. 
4x + y = 1
y = x + 6 
The lines intersect at (?,? ). 
<pre><font size = 4 color = "indigo"><b>
Get some points on the first line:

 x  |  y
-2  |  9
 0  |  1
 1  | -3

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10),

line(-2+.1,9,-2-.1,9), line(-2,9+.1,-2,9-.1), line(-2+.1,9+.1,-2-.1,9-.1), line(-2+.1,9-.1,-2-.1,9+.1), 

line(0+.1,1,0-.1,1), line(0,1+.1,0,1-.1), line(0+.1,1+.1,0-.1,1-.1), line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,-3,1-.1,-3), line(1,-3+.1,1,-3-.1), line(1+.1,-3+.1,1-.1,-3-.1),
line(1+.1,-3-.1,1-.1,-3+.1) )}}} 

Get a ruler and draw a straight line through those three points,
like the green line below:

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,-11, 1-4x),

line(-2+.1,9,-2-.1,9), line(-2,9+.1,-2,9-.1), line(-2+.1,9+.1,-2-.1,9-.1), line(-2+.1,9-.1,-2-.1,9+.1), 

line(0+.1,1,0-.1,1), line(0,1+.1,0,1-.1), line(0+.1,1+.1,0-.1,1-.1), line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,-3,1-.1,-3), line(1,-3+.1,1,-3-.1), line(1+.1,-3+.1,1-.1,-3-.1),
line(1+.1,-3-.1,1-.1,-3+.1) )}}}

---

Get some points on the second line:

 x  |  y
-3  |  3
 0  |  6
 1  |  7

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,-11, 1-4x),

line(-2+.1,9,-2-.1,9), line(-2,9+.1,-2,9-.1), line(-2+.1,9+.1,-2-.1,9-.1), line(-2+.1,9-.1,-2-.1,9+.1), 

line(0+.1,1,0-.1,1), line(0,1+.1,0,1-.1), line(0+.1,1+.1,0-.1,1-.1), line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,-3,1-.1,-3), line(1,-3+.1,1,-3-.1), line(1+.1,-3+.1,1-.1,-3-.1),
line(1+.1,-3-.1,1-.1,-3+.1),

line(-3+.1,3,-3-.1,3), line(-3,3+.1,-3,3-.1), line(-3+.1,3+.1,-3-.1,3-.1), line(-3+.1,3-.1,-3-.1,3+.1), 

line(0+.1,6,0-.1,6), line(0,6+.1,0,6-.1), line(0+.1,6+.1,0-.1,6-.1), line(0+.1,6-.1,0-.1,6+.1), 

line(1+.1,7,1-.1,7), line(1,7+.1,1,7-.1), line(1+.1,7+.1,1-.1,7-.1),
line(1+.1,7-.1,1-.1,7+.1) )}}}
Get a ruler and draw a straight line through those three points,
like the blue line below:

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,-11, 1-4x, x+6),

line(-2+.1,9,-2-.1,9), line(-2,9+.1,-2,9-.1), line(-2+.1,9+.1,-2-.1,9-.1), line(-2+.1,9-.1,-2-.1,9+.1), 

line(0+.1,1,0-.1,1), line(0,1+.1,0,1-.1), line(0+.1,1+.1,0-.1,1-.1), line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,-3,1-.1,-3), line(1,-3+.1,1,-3-.1), line(1+.1,-3+.1,1-.1,-3-.1),
line(1+.1,-3-.1,1-.1,-3+.1),

line(-3+.1,3,-3-.1,3), line(-3,3+.1,-3,3-.1), line(-3+.1,3+.1,-3-.1,3-.1), line(-3+.1,3-.1,-3-.1,3+.1), 

line(0+.1,6,0-.1,6), line(0,6+.1,0,6-.1), line(0+.1,6+.1,0-.1,6-.1), line(0+.1,6-.1,0-.1,6+.1), 

line(1+.1,7,1-.1,7), line(1,7+.1,1,7-.1), line(1+.1,7+.1,1-.1,7-.1),
line(1+.1,7-.1,1-.1,7+.1) )}}}

From the point where those lines cross, draw a line segment
perpendicular to the x-axis, and another perpendicular to the
y-axis (the two black line segments below:

{{{drawing(400,400,-10,10,-10,10,

graph(400,400,-10,10,-10,10,-11, 1-4x, x+6),

line(-2+.1,9,-2-.1,9), line(-2,9+.1,-2,9-.1), line(-2+.1,9+.1,-2-.1,9-.1), line(-2+.1,9-.1,-2-.1,9+.1), 

line(0+.1,1,0-.1,1), line(0,1+.1,0,1-.1), line(0+.1,1+.1,0-.1,1-.1), line(0+.1,1-.1,0-.1,1+.1), 

line(1+.1,-3,1-.1,-3), line(1,-3+.1,1,-3-.1), line(1+.1,-3+.1,1-.1,-3-.1),
line(1+.1,-3-.1,1-.1,-3+.1),

line(-3+.1,3,-3-.1,3), line(-3,3+.1,-3,3-.1), line(-3+.1,3+.1,-3-.1,3-.1), line(-3+.1,3-.1,-3-.1,3+.1), 

line(0+.1,6,0-.1,6), line(0,6+.1,0,6-.1), line(0+.1,6+.1,0-.1,6-.1), line(0+.1,6-.1,0-.1,6+.1), 
line(-1,5,-1,0), line(-1,5,0,5),
line(1+.1,7,1-.1,7), line(1,7+.1,1,7-.1), line(1+.1,7+.1,1-.1,7-.1),
line(1+.1,7-.1,1-.1,7+.1) )}}}

Notice that the vertical black line segment touches the x-axis at -1,
and the horizontal black line segment touches the y-axis at 5.

The lines intersect at (-1,5). 

Therefore the solution is x=-1 and y=5.  The solution is
often written (x,y) = (-1,5).

To check, we substitute -1 for x and 5 for y in both original equations:

{{{system(4x + y = 1, y = x + 6)}}}

Substituting in the first:

{{{4x + y = 1}}}
{{{4(-1) + (5)=1}}}
{{{-4+5=1}}}
{{{1=1}}}

That is an identity.  

Substituting in the second:

{{{y = x+6}}}
{{{(5) = (-1)+6}}}
{{{5 = -1+6}}}
{{{5=5}}}

That is also an identity, so we have found the common
solution to those two equations.

Edwin</pre>