SOLUTION: how will you graph 2x-2y=6 and 3x+4y=12

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Question 85928: how will you graph 2x-2y=6 and 3x+4y=12
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Graphing Linear Equations


2%2Ax-2%2Ay=6Start with the given equation



-2%2Ay=6-2%2Ax Subtract 2%2Ax from both sides

y=%28-1%2F2%29%286-2%2Ax%29 Multiply both sides by -1%2F2

y=%28-1%2F2%29%286%29%2B%281%2F2%29%282%29x%29 Distribute -1%2F2

y=-6%2F2%2B%282%2F2%29x Multiply

y=%282%2F2%29%2Ax-6%2F2 Rearrange the terms

y=1%2Ax-3 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=1 (the slope) and b=-3 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-6

y=1%2A%28-6%29-3

y=-6-3 Multiply

y=-9 Add

So here's one point (-6,-9)



drawing%28+600%2C+600%2C+-16%2C5%2C+-19%2C+2%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=-5

y=1%2A%28-5%29-3

y=-5-3 Multiply

y=-8 Add

So here's another point (-5,-8). Add this to our graph


drawing%28+600%2C+600%2C+-16%2C5%2C+-19%2C+2%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-5%2C-8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-5%2C-8%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-16%2C5%2C+-19%2C+2%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-16-10%2C1%2A%28-16-10%29%2B-3%2C5%2B10%2C1%2A%285%2B10%29%2B-3%29%2C%0D%0A++++blue%28+circle%28+-5%2C-8%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-5%2C-8%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-6%2C-9%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=1%2Ax-3 through the points (-6,-9) and (-5,-8)


So from the graph we can see that the slope is 1%2F1 (which tells us that in order to go from point to point we have to start at one point and go up 1 units and to the right 1 units to get to the next point) the y-intercept is (0,-3)and the x-intercept is (3,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=-3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,-3).


So we have one point (0,-3)


drawing%28+600%2C+600%2C+-10%2C11%2C+-13%2C+8%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C-3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.1%2C+1.5+%29+%29%0D%0A++%29



Now since the slope is 1%2F1, this means that in order to go from point to point we can use the slope to do so. So starting at (0,-3), we can go up 1 units

drawing%28++600%2C+600%2C+-10%2C11%2C+-13%2C+8%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+-3%2B%281%2F2%29%2C+2%2C+1%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 1 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C11%2C+-13%2C+8%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+-3%2B%281%2F2%29%2C+2%2C+1%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-2%2C+1%2C+2%2C+180%2C+360+%29+%29%0D%0A++%0D%0A++%29

Now draw a line through those points to graph y=1%2Ax-3


drawing%28+600%2C+600%2C+-10%2C11%2C+-13%2C+8%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C1%2A%28-10-10%29%2B-3%2C11%2B10%2C1%2A%2811%2B10%29%2B-3%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+1%2C-2%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C-3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+-3%2B%281%2F2%29%2C+2%2C+1%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%281%2F2%29%2C+-2%2C+1%2C+2%2C+180%2C+360+%29+%29%0D%0A++%29 So this is the graph of y=1%2Ax-3 through the points (0,-3) and (1,-2)



Solved by pluggable solver: Graphing Linear Equations


3%2Ax%2B4%2Ay=12Start with the given equation



4%2Ay=12-3%2Ax Subtract 3%2Ax from both sides

y=%281%2F4%29%2812-3%2Ax%29 Multiply both sides by 1%2F4

y=%281%2F4%29%2812%29-%281%2F4%29%283%29x%29 Distribute 1%2F4

y=12%2F4-%283%2F4%29x Multiply

y=%28-3%2F4%29%2Ax%2B12%2F4 Rearrange the terms

y=%28-3%2F4%29%2Ax%2B3 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-3%2F4 (the slope) and b=3 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-8

y=%28-3%2F4%29%2A%28-8%29%2B3

y=24%2F4%2B3 Multiply

y=36%2F4 Add

y=9 Reduce

So here's one point (-8,9)



drawing%28+600%2C+600%2C+-18%2C6%2C+-4%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=-4

y=%28-3%2F4%29%2A%28-4%29%2B3

y=12%2F4%2B3 Multiply

y=24%2F4 Add

y=6 Reduce

So here's another point (-4,6). Add this to our graph


drawing%28+600%2C+600%2C+-18%2C6%2C+-4%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-4%2C6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-18%2C6%2C+-4%2C+19%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-18-10%2C-0.75%2A%28-18-10%29%2B3%2C6%2B10%2C-0.75%2A%286%2B10%29%2B3%29%2C%0D%0A++++blue%28+circle%28+-4%2C6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-8%2C9%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=%28-3%2F4%29%2Ax%2B3 through the points (-8,9) and (-4,6)


So from the graph we can see that the slope is -3%2F4 (which tells us that in order to go from point to point we have to start at one point and go down -3 units and to the right 4 units to get to the next point), the y-intercept is (0,3)and the x-intercept is (4,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=3 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,3).


So we have one point (0,3)


drawing%28+600%2C+600%2C+-10%2C14%2C+-10%2C+13%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.1%2C+1.5+%29+%29%0D%0A++%29



Now since the slope is -3%2F4, this means that in order to go from point to point we can use the slope to do so. So starting at (0,3), we can go down 3 units

drawing%28++600%2C+600%2C+-10%2C14%2C+-10%2C+13%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+3%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 4 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C14%2C+-10%2C+13%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+4%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+4%2C0%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+3%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%284%2F2%29%2C+0%2C+4%2C+2%2C+0%2C+180+%29+%29%0D%0A++%0D%0A++%29

Now draw a line through those points to graph y=%28-3%2F4%29%2Ax%2B3


drawing%28+600%2C+600%2C+-10%2C14%2C+-10%2C+13%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C-0.75%2A%28-10-10%29%2B3%2C14%2B10%2C-0.75%2A%2814%2B10%29%2B3%29%2C%0D%0A++++blue%28+circle%28+4%2C0%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+4%2C0%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C3%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+3%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%284%2F2%29%2C+0%2C+4%2C+2%2C+0%2C+180+%29+%29%0D%0A++%29 So this is the graph of y=%28-3%2F4%29%2Ax%2B3 through the points (0,3) and (4,0)