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Mathematics 11 Online
OpenStudy (anonymous):

For the system of linear equations below, use graphing to determine in which quadrant the solution lies.

OpenStudy (anonymous):

x – y = 3 8x + 5y = -2

OpenStudy (triciaal):

linear equations by graphing draw the line for each function and find (by reading the graph) where they meet

OpenStudy (anonymous):

i not good with that

OpenStudy (triciaal):

for the first line when x = 0 what is y

OpenStudy (triciaal):

math just takes practice

OpenStudy (anonymous):

2

OpenStudy (anonymous):

why don't you solve it and figure out what quadrant you are in once you have the solution?

OpenStudy (triciaal):

x - y = 3 when x =0 0 = 3 + y 0-3 =3 - 3 + y -3 = y (0, -3) gives you one point on this line

OpenStudy (triciaal):

same line x - y = 3 when y = 0 what is x?

OpenStudy (triciaal):

show your work

OpenStudy (anonymous):

3

OpenStudy (anonymous):

CAN U GIVE ME THE AWNSER PLZ ITS PST MY BEAD TIME AND MY MOM IS MAD

OpenStudy (anonymous):

HELLO.....

OpenStudy (triciaal):

note that you can use other values of x and y to plot the graph. you just need 2 points to draw a line. for the second equation do the same thing find 2 points that satisfy the equation and draw the line. the solution is where the lines meet

OpenStudy (anonymous):

q2

OpenStudy (anonymous):

is that is

OpenStudy (anonymous):

i got it quadrent 1

OpenStudy (anonymous):

@DangerousJesse can u help me

OpenStudy (dangerousjesse):

Sure.. Your question?

OpenStudy (anonymous):

For the system of linear equations below, use graphing to determine in which quadrant the solution lies.

OpenStudy (anonymous):

x – y = 3 8x + 5y = -2

OpenStudy (anonymous):

i think its quadrent 1

OpenStudy (anonymous):

i dont mean to rush u butt can u hurry i have to go to bed

OpenStudy (anonymous):

hellooooooooooooooooooooooooooooooooooooooooooooooooooooooo

OpenStudy (anonymous):

:''(

OpenStudy (dangerousjesse):

Solve the following system: \[x-y = 3 \]\[8 x+5 y = -2 \]Swap equation 1 with equation 2: \[8 x+5 y = -2 \]\[x-y = 3 \]Subtract \(1/8 × (equation 1)\) from equation 2: \[8 x+5 y = -2 \]\[0 x-(13 y)/8 = 13/4 \]Multiply equation 2 by 8/13: \[8 x+5 y = -2 \]\[0 x-y = 2 \]Multiply equation 2 by -1: \[8 x+5 y = -2 \]\[0 x+y = -2 \]Subtract \(5 × (equation 2)\) from equation 1: \[8 x+0 y = 8 \]\[0 x+y = -2 \]Divide equation 1 by 8: \[x+0 y = 1 \]\[0 x+y = -2 \]

OpenStudy (dangerousjesse):

I was typing, sorry :)

OpenStudy (dangerousjesse):

Gather the last little bit and plot it to find your answer

OpenStudy (anonymous):

its ok whats the awnser i have to go to bed my pernrts are pissed

OpenStudy (anonymous):

awnser plzzzzzzzzzzzzzzzzzzzz

OpenStudy (anonymous):

:'''{

OpenStudy (anonymous):

plzz

OpenStudy (dangerousjesse):

What's \(0*y\)?

OpenStudy (anonymous):

0

OpenStudy (anonymous):

what the awnser plz

OpenStudy (dangerousjesse):

So \(x=1\)

OpenStudy (anonymous):

plzzzzzzzzzzzzzzzzzzzz...............

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

0,3

OpenStudy (dangerousjesse):

And \(x*0\)?

OpenStudy (anonymous):

0

OpenStudy (dangerousjesse):

So \(y=-2\) :)

OpenStudy (dangerousjesse):

(1,-2)

OpenStudy (anonymous):

in which quadrent

OpenStudy (dangerousjesse):

The fourth quadrant

OpenStudy (anonymous):

r u sure

OpenStudy (anonymous):

ok thank u love u bye

OpenStudy (dangerousjesse):

|dw:1408415611316:dw|No problem haha.

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