I have no clue how to solve this

Solve by graphing

y=x+2

y=4-x

Printable View

- Jan 25th 2006, 07:35 AMjosh_csolving a linear system by graphing
I have no clue how to solve this

Solve by graphing

y=x+2

y=4-x - Jan 25th 2006, 08:05 AMCaptainBlackQuote:

Originally Posted by**josh_c**

The solution is the point at which the two lines cross, as this is

a point where the x and y values satisfy both equations, and hence

is the required solution.

RonL - Jan 25th 2006, 08:06 AMSkyWatcher
hi

you've got to take a nice piece of paper and trace the lines (i dont know if its a proper expression) that are related to your two equation

The solution should be the coordonates (x,y) of the intersection of those two lines so handle your rule corectly and dont shiver!

You dont know how to trace a "line" related to an linear equation?

no problemos:

if you know to trace a line when youve got two points of the line i will tell you:

ax+b=y is the general form of a linear equation

so when you fix x you should know how to calculate y (if you dont give me your account number and your telephon number i will show you a trick)

so you choose two x for which the point (x,y) (or (x,ax+b) if you preffer) are easy to plot on you piece of paper and there you go

By - Jan 25th 2006, 08:20 AMjosh_cQuote:

Originally Posted by**CaptainBlack**

- Jan 25th 2006, 09:03 AMCaptainBlackQuote:

Originally Posted by**josh_c**

call this and a y-coordinate call this .

Then the solution is and .

To check that this is a solution plug these values back into the equations

and you should find that they balance.

RonL - Jan 25th 2006, 09:04 AMjosh_cQuote:

Originally Posted by**CaptainBlack**

- Jan 25th 2006, 09:11 AMSkyWatcher
re re hi

so you want to understand what you are doing (take care this is no politicaly correct)

When you plot a line (satisfying a linear equation) you plot on your paper

all (a part indeed) the point which coordonates satisfy the equation

why that? i have to think of it a few minuts to it (may be more)

So you plot all the point satisfying equation 1

and all the plot satisfying equation 2

So if there is a point that satisfy the two equation it is the intersection of this two line you surely will understand why (you just have to say this point ...ect)

if you know some litle basis of geometry you know that two unparralell line have only one intersection and that two parrallel line are the same line or have no intersection

for the more tricky question wait a few minute

or watch your boock

re re By! - Jan 25th 2006, 09:18 AMjosh_cQuote:

Originally Posted by**SkyWatcher**

- Jan 25th 2006, 09:30 AMCaptainBlackQuote:

Originally Posted by**josh_c**

- Jan 25th 2006, 10:02 AMjosh_cQuote:

Originally Posted by**CaptainBlack**

- Jan 25th 2006, 10:20 AMCaptainBlackQuote:

Originally Posted by**josh_c**

RonL - Jan 25th 2006, 10:29 AMjosh_cQuote:

Originally Posted by**CaptainBlack**

- Jan 25th 2006, 10:33 AMCaptainBlackQuote:

Originally Posted by**josh_c**

else on the other side.

So 2x-y+9=0, can be rewritten:

y=2x+9,

and similarly x+y-3=0 can be rewritten:

y=3-x

RonL - Jan 25th 2006, 12:26 PMThePerfectHackerQuote:

Originally Posted by**josh_c**

Here are the graphs for what you said,