SOLUTION: Solve the system by graphing. 3x-2y=4 -6x+4y=7

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Question 158543: Solve the system by graphing.
3x-2y=4
-6x+4y=7
Found 2 solutions by gonzo, Electrified_Levi:
Answer by gonzo(654) About Me  (Show Source):
You can put this solution on YOUR website!
solving the equation by algebraic means, we get
%283%2Ax-2%2Ay%29=4
%28-6%2Ax%2B4%2Ay%29=7
solving for y in the first equation, we get
y=%28%283%2Ax%29%2F2%29-2%29
substituting this for y in the second equation, we get
y=%28-6%2Ax%2B4%29%2B4%2A%283%2Ax%2F2+-+2%29+=+7
this becomes
y=-6%2Ax+%2B+6%2Ax+-+8+=+7
since the x cancels out we are left with
-8=7
which means there is no solution.
looking at both equations in y = slope intercept form, we see
y=%283%2Ax%2F2%29-2
and
y=%286%2Ax%2F4%29%2B%287%2F4%29
general form of the slope intercept form of the equation is y=m*x+b
where m is the slope and b is the y intercept when x = 0.
slope of the first equation is 3/2
slope of the second equation is 6/4 which simplifies to 3/2.
slopes are equal so lines are parallel.
solving by equation below shows that right away.
graph%28600%2C600%2C-5%2C5%2C-15%2C15%2C%283x-4%29%2F2%2C%286x%2B7%29%2F4%29

Answer by Electrified_Levi(103) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, Hope I can help
.
Solve the system by graphing.
+3x-2y=4+
+-6x%2B4y=7+
.
Points are given as (x,y), to find points on both of these lines, you just replace "x" with any number and solve for "y"
.
Lets find some points on the first line
.
+3x-2y=4+, (we can use any number for "x", then we solve for "y"
.
Lets replace "x" with "2", +3x-2y=4+ = +3%282%29-2y=4+ = +6+-2y=4+
.
We will move (-2y) over to the right side
.
+6+-2y=4+ = +6+-2y+%2B+2y=4+%2B+2y+ = +6+=+4+%2B+2y+, we will move "4" to the left side
.
+6+=+4+%2B+2y+ = +6+-+4+=+4+-+4+%2B+2y+ = +2+=++2y+ = +2y+=++2+
.
To get "y", we will divide each side by "2"
.
+2y+=++2+ = +2y%2F2+=++2%2F2+ = +y+=++1+,
.
"x" = 2
"y" = 1
.
Since points are given as (x,y), this point would be (2,1)
.
(2,1) is a point of the line(We can check by replacing "x" with "2", "y" with "1", +3x-2y=4+ = +3%282%29-2%281%29=4+ = +6-2=4+ = +4=4+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2Cblue+%28circle%282%2C1%2C0.1%29%29%29+
.
Lets find another point,
.
Lets replace "x" with "0"
.
+3x-2y=4+ = +3%280%29-2y=4+ = +0-2y=4+ = +%28-2y%29=4+
.
We can multiply each side by (-1) to get (-2y) positive
.
+%28-2y%29=4+ = +%28-2y%29%28-1%29=4%28-1%29+ = +2y+=+%28-4%29+, to find "y" we will divide each side by "2"
.
+2y+=+%28-4%29+ = +2y%2F2+=+%28-4%29%2F2+ = +y+=+%28-2%29+
.
"x" = 0
"y" = (-2)
.
Another point on the line is (0,-2)(x,y)(we can check by replacing "x" with "0", "y" with (-2), +3x-2y=4+ = +3%280%29-2%28-2%29=4+ = +0%2B4=4+ = +4+=+4+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2Cblue+%28circle%282%2C1%2C0.1%29%29%2Cblue+%28circle%280%2C-2%2C0.1%29%29%29+
.
Another point on the line is (-4,-8), We can check by replacing "x" with (-4), "y" with (-8), +3x-2y=4+ = +3%28-4%29-2%28-8%29=4+ = +%28-12%29%2B16=4+ = +4+=+4+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2Cblue+%28circle%282%2C1%2C0.1%29%29%2Cblue+%28circle%280%2C-2%2C0.1%29%29%2Cblue+%28circle%28-4%2C-8%2C0.1%29%29%29+
.
Lets draw a line through the points,
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2C+graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28%283%2F2%29x-2%29%29%2C+blue%28circle%282%2C1%2C0.1%29%29%2Cblue%28circle%280%2C-2%2C0.1%29%29%2C+blue%28circle%28-4%2C-8%2C0.1%29%29%29+
.
Now lets find some points for the second equation, +-6x%2B4y=7+
.
Lets replace "x" with +-1%2F2+, +-6x%2B4y=7+ = +-6%28-1%2F2%29%2B4y=7+ = +3%2B4y=7+
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"4y" has to equal "4", so +4y+=+4+, divide each side by "4", +4y+=+4+ = +4y%2F4+=+4%2F4+ = +y+=+1+
.
x = +-1%2F2+
y = 1
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One point on the line is (+-1%2F2,1 )(We can check by replacing "x" with +-1%2F2,"y" with "1", +-6x%2B4y=7+ = +-6%28-1%2F2%29%2B4%281%29=7+ = +3+%2B+4+=+7+ = +7+=+7+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2C+graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28%283%2F2%29x-2%29%29%2C+blue%28circle%282%2C1%2C0.1%29%29%2Cblue%28circle%280%2C-2%2C0.1%29%29%2C+blue%28circle%28-4%2C-8%2C0.1%29%29%2Cgreen%28circle%28-1%2F2%2C1%2C0.1%29%29%29+
.
Replace "x" with "4", +-6x%2B4y=7+ = +-6%284%29%2B4y=7+ = +%28-24%29%2B4y=7+
.
"4y" will have to equal "31", +4y+=+31+, dividing each side by "4" will get "y"
.
+4y+=+31+ = +4y%2F4+=+31%2F4+ = +y+=+31%2F4+
.
x = 4
y = +31%2F4+, or 7 +3%2F4+
.
The point is (4,31%2F4), (we can check by replacing "x" with "4", "y" with +31%2F4+, +-6x%2B4y=7+ = +-6%284%29%2B4%2831%2F4%29=7+ = +%28-24%29%2B31+=+7+ = +7+=7+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2C+graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28%283%2F2%29x-2%29%29%2C+blue%28circle%282%2C1%2C0.1%29%29%2Cblue%28circle%280%2C-2%2C0.1%29%29%2C+blue%28circle%28-4%2C-8%2C0.1%29%29%2Cgreen%28circle%28-1%2F2%2C1%2C0.1%29%29%2Cgreen%28circle%284%2C31%2F4%2C0.1%29%29%29+
.
Another point on the line is ( 0, +7%2F4+ ), ( we can check by replacing "x" with "0", "y" with "+7%2F4+", +-6x%2B4y=7+ = +-6%280%29%2B4%287%2F4%29=7+ = +0+%2B+7+=7+ = +7+=+7+ (True)
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2C+graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28%283%2F2%29x-2%29%29%2C+blue%28circle%282%2C1%2C0.1%29%29%2Cblue%28circle%280%2C-2%2C0.1%29%29%2C+blue%28circle%28-4%2C-8%2C0.1%29%29%2Cgreen%28circle%28-1%2F2%2C1%2C0.1%29%29%2Cgreen%28circle%284%2C31%2F4%2C0.1%29%29%2Cgreen%28circle%280%2C7%2F4%2C0.1%29%29%29+
.
Lets draw a line through the points
.
+drawing+%28600%2C600%2C-10%2C10%2C-10%2C10%2Cgrid%281%29%2C+graph%28600%2C600%2C-10%2C10%2C-10%2C10%2C%28%283%2F2%29x-2%29%2C%28%283%2F2%29x+%2B+%287%2F4%29%29%29%2C+blue%28circle%282%2C1%2C0.1%29%29%2Cblue%28circle%280%2C-2%2C0.1%29%29%2C+blue%28circle%28-4%2C-8%2C0.1%29%29%2Cgreen%28circle%28-1%2F2%2C1%2C0.1%29%29%2Cgreen%28circle%284%2C31%2F4%2C0.1%29%29%2Cgreen%28circle%280%2C7%2F4%2C0.1%29%29%29+
.
These lines are parallel, they don't have any points that intersect each other, there is no solution to the system of equations
.
Hope I helped, Levi