SOLUTION: given 3x+5y=8;find the x and y intercept

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Question 118251: given 3x+5y=8;find the x and y intercept
Found 2 solutions by jim_thompson5910, MathLover1:
Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!

3%2Ax%2B5%2Ay=8 Start with the given equation

Let's find the x-intercept

To find the x-intercept, let y=0 and solve for x:
3%2Ax%2B5%2A%280%29=8 Plug in y=0

3%2Ax=8 Simplify

x=8%2F3 Divide both sides by 3



So the x-intercept is (note: the x-intercept will always have a y-coordinate equal to zero)



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3%2Ax%2B5%2Ay=8 Start with the given equation

Now let's find the y-intercept

To find the y-intercept, let x=0 and solve for y:
3%2A%280%29%2B5%2Ay=8 Plug in x=0

5%2Ay=8 Simplify

x=8%2F5 Divide both sides by 5




So the y-intercept is (note: the y-intercept will always have a x-coordinate equal to zero)

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So we have these intercepts:
x-intercept:

y-intercept:


Answer by MathLover1(3376) About Me  (Show Source):
You can put this solution on YOUR website!
3x%2B5y=8
x-intercept:
The x-intercept is where the graph crosses the x-+axis; just ask yourself what value is+y+always+going+to+be on the x-intercept? No matter where you are on the x-axis,+y%92s value is 0, that is a constant.
So, we will put y=0 into given equation:
3x%2B5%2A0+=+8
3x+=+8
x+=+8%2F3
x+=+2%282%2F3%29
or
x+=+2.66........so, the x-intecept is (2.66,0)

y-intercept
it is+x%92s value that is 0; any where you would cross the y-axis,+x%92s value is always 0

3%2A0%2B5y=+8
5y=+8
y+=+8%2F5
y+=+1%283%2F5%29
or
y+=+1.6........so, the y-intecept is (0,1.6)
Graph:

Solved by pluggable solver: Graphing Linear Equations


3%2Ax%2B5%2Ay=8Start with the given equation



5%2Ay=8-3%2Ax Subtract 3%2Ax from both sides

y=%281%2F5%29%288-3%2Ax%29 Multiply both sides by 1%2F5

y=%281%2F5%29%288%29-%281%2F5%29%283%29x%29 Distribute 1%2F5

y=8%2F5-%283%2F5%29x Multiply

y=%28-3%2F5%29%2Ax%2B8%2F5 Rearrange the terms

y=%28-3%2F5%29%2Ax%2B8%2F5 Reduce any fractions

So the equation is now in slope-intercept form (y=mx%2Bb) where m=-3%2F5 (the slope) and b=8%2F5 (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-9

y=%28-3%2F5%29%2A%28-9%29%2B8%2F5

y=27%2F5%2B8%2F5 Multiply

y=35%2F5 Add

y=7 Reduce

So here's one point (-9,7)



drawing%28+600%2C+600%2C+-19%2C6%2C+-6%2C+17%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.1%2C+1.5+%29+%29%0D%0A++%29

Now lets find another point

Plug in x=-4

y=%28-3%2F5%29%2A%28-4%29%2B8%2F5

y=12%2F5%2B8%2F5 Multiply

y=20%2F5 Add

y=4 Reduce

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


drawing%28+600%2C+600%2C+-19%2C6%2C+-6%2C+17%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++blue%28+circle%28+-4%2C4%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C4%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.1%2C+1.5+%29+%29%0D%0A++%29


Now draw a line through these points

drawing%28+600%2C+600%2C+-19%2C6%2C+-6%2C+17%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-19-10%2C-0.6%2A%28-19-10%29%2B1.6%2C6%2B10%2C-0.6%2A%286%2B10%29%2B1.6%29%2C%0D%0A++++blue%28+circle%28+-4%2C4%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-4%2C4%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+-9%2C7%2C+.1%2C+1.5+%29+%29%0D%0A++%29 So this is the graph of y=%28-3%2F5%29%2Ax%2B8%2F5 through the points (-9,7) and (-4,4)


So from the graph we can see that the slope is -3%2F5 (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 5 units to get to the next point), the y-intercept is (0,1.6) ,or (0,8%2F5), and the x-intercept is (2.66666666666667,0) ,or (8%2F3,0) . So all of this information verifies our graph.


We could graph this equation another way. Since b=8%2F5 this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,8%2F5).


So we have one point (0,8%2F5)


drawing%28+600%2C+600%2C+-10%2C15%2C+-11.4%2C+11.6%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.1%2C+1.5+%29+%29%0D%0A++%29



Now since the slope is -3%2F5, this means that in order to go from point to point we can use the slope to do so. So starting at (0,8%2F5), we can go down 3 units

drawing%28++600%2C+600%2C+-10%2C15%2C+-11.4%2C+11.6%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+1.6%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%0D%0A++%0D%0A++%29
and to the right 5 units to get to our next point

drawing%28++600%2C+600%2C+-10%2C15%2C+-11.4%2C+11.6%2C%0D%0A++++grid%281%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+5%2C-1.4%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+5%2C-1.4%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+1.6%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%285%2F2%29%2C+-1.4%2C+5%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%2F5%29%2Ax%2B8%2F5


drawing%28+600%2C+600%2C+-10%2C15%2C+-11.4%2C+11.6%2C%0D%0A++++grid%28+1+%29%2C%0D%0A++++line%28-10-10%2C-0.6%2A%28-10-10%29%2B1.6%2C15%2B10%2C-0.6%2A%2815%2B10%29%2B1.6%29%2C%0D%0A++++blue%28+circle%28+5%2C-1.4%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+5%2C-1.4%2C+.1%2C+1.5+%29+%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.15%2C+1.5+%29%29%2C%0D%0A++++blue%28+circle%28+0%2C1.6%2C+.1%2C+1.5+%29+%29%2C%0D%0A++blue%28+arc%28+0%2C+1.6%2B%28-3%2F2%29%2C+2%2C+-3%2C+90%2C+270+%29+%29%2C%0D%0A++blue%28+arc%28+%285%2F2%29%2C+-1.4%2C+5%2C+2%2C+0%2C+180+%29+%29%0D%0A++%29 So this is the graph of y=%28-3%2F5%29%2Ax%2B8%2F5 through the points (0,1.6) and (5,-1.4)