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Mathematics 10 Online
OpenStudy (joshawad):

need help with solving simultaneous equations by graphing

OpenStudy (anonymous):

just plot the lines. where the lines meet, that point is the solution.

OpenStudy (joshawad):

but they are giving me funny instructions

OpenStudy (anonymous):

like what?

OpenStudy (joshawad):

A: Y=3x-2 b: y=2x+3

OpenStudy (joshawad):

the point I came up with was (6,0) which is wrong

OpenStudy (joshawad):

can you help me to do it and not just give me the answer

OpenStudy (anonymous):

x=5, y=13

OpenStudy (joshawad):

how

OpenStudy (anonymous):

wait

OpenStudy (anonymous):

for eq 1, find three points that fit the equation. how? just plot random values for x, and find y. But use small numbers; starting from 0 or 1 always helps. do the same for second. just plot those lines based on those oints. now, find where the lines meet. if needed, just prolong the line straight. Where they meet, find the corresponding value of x and y. that is the solution.

OpenStudy (joshawad):

y=13? my graoh on the paper only goes to 6. 12 if u count the negatives

OpenStudy (anonymous):

if ur graph does not have upto 13, there are two options 1. only represent one quarter,( where both x and y are positive) in ur graph. 2. decrease the units u are using in ur graph

OpenStudy (joshawad):

I don't exactly understand what you are saying. :/ this is what they said to do in my video and why I am stuck.

OpenStudy (joshawad):

so just pick random numbers from the air?

OpenStudy (anonymous):

which part don't u understand?

OpenStudy (joshawad):

seems illogical

OpenStudy (anonymous):

no it does not. i mean, that is what a line is. it can have any values of x, and as long as u calculate y from it, the point (x,y) is within the line. Besides u need some way to represent the line in the graph. how else would u represent a line other than plotting points and joining them? and how do u get the points? just find out some points that fit the line, and plot them straightly.

OpenStudy (joshawad):

you lost me at any values of x you should know I have ADHD so trying to teach me this will be a little hard

OpenStudy (anonymous):

so lets start from the begining.

OpenStudy (joshawad):

ok

OpenStudy (amorfide):

If you are given an equation such as Y=3x-2 The x value can be anything on the graph as it is a straight line and will go on forever and ever and ever. So to plot the straight line between specific points, we substitute numbers for x (because it can go from -infinite to infinite so let us start with substituting x=0, x=1, x=2 when x =0 y=3(0)-2 continue this with values of x, then do the same with your other equation. connect the points to form two straight lines, and they will intersect at specific points, that is your solution

OpenStudy (anonymous):

you are given two equations. your task is to find a point, which means a value for x, and a value for y, that both satisfies the given equations, and u need to do it by plotting the graphs of the two equations given. clear till here?

OpenStudy (phi):

to plot a line, you need only 2 points. to find two points: pick x=0, and find y pick y=1 and find y plot those points and draw a line through them

OpenStudy (joshawad):

@mjbmanny yes I got that part!

OpenStudy (anonymous):

well, that is true phi, but in case of some lines that don,t pass through x=0, or y=0, then picking 0 and 1 can't be best.

OpenStudy (phi):

plot those points and draw a line through them, and make the line as long as necessary ...

OpenStudy (anonymous):

allright, josh, now u got to find some points that satisfy the equations. why? because u need to plot the equation on the graph as a line. and , as phi just said, picking two points for each line is enough.But, to be really certain, it is better to find out three points.

OpenStudy (joshawad):

phi im n ot all that bright u know if I don't Instantly know what 5x5 is it takes me 60 sec to figure it out

OpenStudy (joshawad):

3?

OpenStudy (phi):

Here is what the first line looks like

OpenStudy (joshawad):

Y = 2/1x -2

OpenStudy (joshawad):

in fraction form

OpenStudy (anonymous):

allright, josh, now how do u find the points that i mentioned before? u see, every point in a line satisfies the equation representing the line. which means, if u put on x, and u get some y. then if u replace the values of variables x and y in the equation, two sides of the eqaution become equal. so, picking points in a line is not awkard at all.

OpenStudy (joshawad):

but don't we only pick 1 number?

OpenStudy (joshawad):

for all the equations

OpenStudy (anonymous):

u pick one number and then put that number in place of the variable u picked for. then calculate the value of another varible, find it out. then the nuber u picked for first variable, and the answe u got after placing the number in place of the variable u picked for, that constitutes a point. U need two such points for each line to be able to draw a line.

OpenStudy (joshawad):

I thought u said 3

OpenStudy (anonymous):

yes, 3 is better. but two is enough

OpenStudy (joshawad):

what about the 3rd point

OpenStudy (anonymous):

i said that because in some cases, u need to elongate the lines a lot to be able to meet the two lines. So, taking 3 points reduces the chance of error. how? well if u have three points u are more likely to extend the line straight( which is most important). the point is , u need at least two points to draw line. but u can as well pick 3,4,5, or whatever points u want if u have time to waste.

OpenStudy (joshawad):

so how do I know when I have picked the right numbers

OpenStudy (anonymous):

just pick two points as a rule,. two is enough. think it this way. what is a line? its a series of points. so, how many points would u need to represent a line? just two. because remaining points can be found out just by elongating the line straight. But can u say something about a line by using just one point? no. u need at least two. if u are not understanding this forget three and why 3 is better, just think two.

OpenStudy (joshawad):

how do I know when I have picked the correct points

OpenStudy (anonymous):

u see every point is correct if it fits the equation. which is to say for suppose an example equation x+3y=6, i can pick 100, for the value of x,and put x=100, in the equation. then the corresponding y would be -(93/6). this point is also a correct point, because it is a point in the line. i got -(93/6) by putting x =100. Since I got this from the equation itself, there is no doubt it fits the equation. But, as u discovered ur graph doesnot have enough space for 100. and the calculations are also more difficult. so, lets pick a smaller value of x, say x=0. then the corresponding value of y would be y=6/3=2. And for another point of the line lets pick x=1( and this shows why sometimes this can be tricky). so the corresponding value of y woudl be (5/3). but as it is difficult to plot a decimal, lets pick another x. how about 3? say x=3. then y=1. so the two points of the line are (0,2) and (3, 1). Now, u got to do the same for both the equations given to you itn the above question. and plot the points for each line; join the corresponding points of a line, and see where two lines cross( if necessary, extend the lines ,straightly!; and the value of x, and y u get of the point of crossing is ur answer. and that is the point that satisfies both equations.

OpenStudy (joshawad):

what is x+3y=6 what format is that and how do I change it

OpenStudy (joshawad):

how did u get 100 from the equation

OpenStudy (joshawad):

nope nevermind u lost me again how did u get 93/6

OpenStudy (anonymous):

it was an example equation. just as the two you posted in ur question. I picked 100 from random to illustrate that theoretically, u can take any value of x. but it is more practical to take smaller values

OpenStudy (joshawad):

ok?

OpenStudy (anonymous):

lets get back to x+3y=6.

OpenStudy (anonymous):

so, i need two points that fit this equation got it?

OpenStudy (joshawad):

yes but let me see if I can do this ok?

OpenStudy (anonymous):

yes, i thinkk u can understand better if u practice right here.

OpenStudy (joshawad):

X-Y=2 and X+3Y=6 are my two equations

OpenStudy (anonymous):

ok

OpenStudy (joshawad):

I need fix the equations right?

OpenStudy (anonymous):

yes, u need to find wo points from both these equations. lets start with first; x-y=2

OpenStudy (joshawad):

so Y=x-2 right

OpenStudy (anonymous):

absolutely

OpenStudy (joshawad):

so x=1 to start and I need another point from this equation?

OpenStudy (anonymous):

yes. replace the x by 1, and see what y turns out to be.

OpenStudy (joshawad):

that would make y 1

OpenStudy (texaschic101):

y = 3x - 2 slope is 3 y intercept is (0,-2) x intercept is (2/3,0) now plot your points and draw a line to connect them y = 2x + 3 slope is 2 y intercept is (0,3) x intercept is (-3/2,0) for graphing purposes x int is (-1 1/2,0) plot your points and connect them with a line. where you lines cross each other is your solution

OpenStudy (anonymous):

no, it would make y -1.

OpenStudy (joshawad):

oh ok so it would not fit

OpenStudy (anonymous):

why not? the first point of ur equation y=x-2, is (1,-1). now find similar another using different x

OpenStudy (anonymous):

pick any from 0, 2, 3...

OpenStudy (joshawad):

? how did u get that

OpenStudy (anonymous):

what? 0, 2, 3.....

OpenStudy (joshawad):

yes

OpenStudy (anonymous):

well u can guess anything( thoretically, u can even include fractioins or negative integerts). But, i picked 0, 2, 3 because; they are small and positive, which will help me by makint the calculation short. and why i left 1? because i already used it. if i take 1 again, i would get the same point, and i wont have the necessary two points. all right, u pick anyhting u want.

OpenStudy (joshawad):

3y=x+6

OpenStudy (joshawad):

I need a negative integer

OpenStudy (joshawad):

so -2

OpenStudy (joshawad):

so y is 4

OpenStudy (joshawad):

soooo...

OpenStudy (joshawad):

-2,4

OpenStudy (joshawad):

its a straight line

OpenStudy (joshawad):

yes?

OpenStudy (joshawad):

@mjbmanny

OpenStudy (anonymous):

josh, there was some technical problem. now its fine.

OpenStudy (joshawad):

ok

OpenStudy (joshawad):

its a really long thing so a mod was probly alerted cause I couldn't either

OpenStudy (texaschic101):

2x + 3 = 3x - 2 2x - 3x = -2 -3 -x = -5 x = 5 y = 2x + 3 y = 2(5) + 3 y = 10 + 3 y = 13 your lines should cross at (5,13)

OpenStudy (joshawad):

thanks for no giving me the answer

OpenStudy (joshawad):

all that work for nothing

OpenStudy (anonymous):

first, u got to calculate another point for the earlier equation. u only got one. u need two. lets do that and move to another equation. for 1st equation, u used x=1,and got y=-1, and the point was thus(1,-1). now u need to find another point. for ur first equation y=x-2.

OpenStudy (joshawad):

so the next one is Y=2x+-2 and y=1/3x+5

OpenStudy (joshawad):

please don't post the answer

OpenStudy (anonymous):

josh, i couldnot understand what u posted. lets think this as clearly as the sunlight.

OpenStudy (joshawad):

Y=2x+-2 and y=1/3x+5

OpenStudy (anonymous):

and what value of x u chose? are u posting two new equations?

OpenStudy (joshawad):

yes

OpenStudy (anonymous):

why? u arent done with earlier ones.

OpenStudy (joshawad):

texas just posted the whole equation

OpenStudy (anonymous):

if u still insist, lets start afresh withe these two. the important thing is u understand what u are doing. texas solved the equaions you posted in the begining algebrically,a nd got the answer. the same way I did to produce the answer earlier. we are trying to solve equations graphically. arent we?

OpenStudy (joshawad):

y=2x+-2 2x= not 0 or 1 that might not make any sense so lets try 2

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