```Question 49153
<pre><font size = 6><b>solve & graph the
equations

3x - 6y =  12
2x - 5y = -10

To eliminate the y's:
The coefficients of y in absolute
value are 6 and 5.  The smallest
positive integer which both 6 and 5
will divide evenly into is 30. So
we need to multiply each equation
thru by whatever will cause one
coefficient of y to be 30, and the
other coefficient of y to be -30.
There are two choices.  We can
either multiply the first equation
thru by -5 and the second thru by
+6 or the first thru by +5 and the
second by -6.  I'll arbitrarily
choose to multiply the first one
thru by +5 and the second thru by
-6:

5[3x - 6y =  12]
-6[2x - 5y = -10]

term by term:

15x - 30y =  60
-12x + 30y =  60
-----------------
3x       = 120
x = 40

------------------------------

3x - 6y =  12
2x - 5y = -10

To eliminate the x's:
The coefficients of x in absolute
value are 3 and 2.  The smallest
positive integer which both 3 and 2
will divide evenly into is 6. So
we need to multiply each equation
thru by whatever will cause one
coefficient of x to be 6, and the
other coefficient of x to be -6.
There are two choices.  We can
either multiply the first equation
thru by -2 and the second thru by
+3 or the first thru by +2 and the
second by -3.  I'll arbitrarily
choose to multiply the first one
thru by -2 and the second thru by
+3:

-2[3x - 6y =  12]
3[2x - 5y = -10]

term by term:

-6x + 12y = -24
6x - 15y = -30
-----------------
-3y = -54
y = 18

----------------------------

x = 40
y = 18

So the two lines will
cross at the point (40,18)
---------------------------

To find the x-coordinate
of the x-axis intercept,
substitute 0 for y and
solve for x.

To find the y-coordinate
of the y-axis intercept,
substitute 0 for x and
solve for y.

For equation 1,

3x - 6y =  12

substitute 0 for y

3x - 6(0) = 12
3x - 0 = 12
3x = 12
x = 4

So the x-coordinate of
the x-axis intercept is 4.
The x-axis intercept is
the point (4,0).  That's
the point at which the
graph of equation 1
crosses the x-axis

equation 1 x-axis intercept = (4,0)

For equation 1,

3x - 6y =  12

substitute 0 for x

3(0) - 6y = 12
0  - 6y = 12
-6y = 12
y = -2

So the y-coordinate of
the y-axis intercept is -2.
The y-axis intercept is
the point (0,-2).  That's
the point at which the
graph of equation 1
crosses the y-axis.

equation 1 y-axis intercept = (0,-2)

For equation 2,

2x - 5y = -10

substitute 0 for y

2x - 5(0) = -10
2x - 0 = -10
2x =  10
x = 5

So the x-coordinate of
the x-axis intercept is 5.
The x-axis intercept is
the point (5,0).  That's
the point at which the
graph of equation 2
crosses the x-axis.

equation 2 x-axis intercept = (5,0)

For equation 2,

2x - 5y = -10

substitute 0 for x

2(0) - 5y = -10
0  - 5y = -10
-5y = -10
y = 2

So the y-coordinate of
the y-axis intercept is 2.
The y-axis intercept is
the point (0,2).  That's
the point at which the
graph of equation 2
crosses the y-axis.

equation 2 y-axis intercept = (0,2)

To graph the equations, get some
points.  You already have two on
each line.  Trouble is, the point
where the lines cross has such
large coefficients (40,18).  The
graph will have to be so big that
we will need some other points
since those are so small relative
to the size we'll have to make the
graph.  So let's get some other
points with bigger numbers:

equation 1
x  |  y
-------
24 | 10
30 | 13

So we plot (24,10) and
(30,13)

and we get this line

{{{ graph( 300, 300, -10, 90, -10, 90, (12-3x)/(-6)) }}}

Then for equation 2, we find a
couple of points with large
coefficients:

equation 2
x  |  y
-------
20 | 10
35 | 16

So we plot (20,10) and
(35,16)

and we draw that line on the
same set of axes

{{{ graph( 300, 300, -10, 90, -10, 90, (12-3x)/(-6), (-10-2x)/(-5)) }}}

and we see that they cross at the
point (x,y) = (40,18).

Edwin</pre></font></b>
```