Basic algebra math help again

[Nocturnal]

Ok help me thanks!

Find the x- and y-intercepts of each graph; then draw the graphs.

2x + y = 11

x + 3y = 18

My book says to Let x = 0; then y = 11

Then, let y = 0; then x = 11/2

HOW IN THE WORLD DO THEY GET THE 0 AND 11? AND THEN THE 11/2?

 

[gopunk]

2(0)+y=11
y = 11

2x + 0 = 11
x = 11/2

these are the x and y intercepts for the first eq... sometimes it helps to plot a few points before plotting hte whole thing

 

[Nocturnal]

How did they get the 0 and 11 to begin with? And then the 11/2 to begin with? This is an example straight out of my math book. I tried reading and reading and reading and I finally came to the conclusion that I'm not going to figure it out without asking for help.

 

[Nocturnal]

Ok I understand that now. But, how did they get the numbers 0 and 11 to begin with? Do you just randomly pick any number and hope that it works?

 

[johnjohn320]

Originally posted by: Nocturnal
Ok I understand that now. But, how did they get the numbers 0 and 11 to begin with? Do you just randomly pick any number and hope that it works?

No, when you're trying to find y you substitute 0 in for x. When you're finding x, you substitute a 0 in for y.

 

[illusion88]

Originally posted by: johnjohn320

Originally posted by: Nocturnal
Ok I understand that now. But, how did they get the numbers 0 and 11 to begin with? Do you just randomly pick any number and hope that it works?

No, when you're trying to find y you substitute 0 in for x. When you're finding x, you substitute a 0 in for y.

bingo.

 

[Nocturnal]

Determine the solution to each system of inequalities:

x + y > 2

x - y < 2

How would I work a problem like that?

 

[johnjohn320]

Originally posted by: Nocturnal
Determine the solution to each system of inequalities:

x + y > 2

x - y < 2

How would I work a problem like that?

Maybe I'm remembering wrong, but aren't these the ones you use elimination for?

 

[Nocturnal]

I'm solving systems of linear inequalities.

 

[gopunk]

Originally posted by: johnjohn320

Originally posted by: Nocturnal
Determine the solution to each system of inequalities:

x + y > 2

x - y < 2

How would I work a problem like that?

Maybe I'm remembering wrong, but aren't these the ones you use elimination for?

i could have sworn you do it with absolute values... but i forget specficis..

 

[johnjohn320]

Originally posted by: Nocturnal
I'm solving systems of linear inequalities.

Yeah, I'm thinking of solving systems of linear equations, I don't think it works the same.

 

[jamesave]

draw the linear line where X+Y = 2.

Realize that the line basically divide into two areas:.. one are represent >2, the other one represent <2

Then draw the linear line where X-Y = 2

Determine the area where X-Y < 2 (similar to the first step)

Combine the area where both are common, so there you go.

EDIT: grammar

 

[Yossarian]

by definition the x-intercept is the x value where the line crosses the x axis, or where y=0. vice versa for the y-intercept. all you're doing is plugging zeros in and finding the other variable's value from the equation. 2 points = straight line.

 

[fornax]

Determine the solution to each system of inequalities:

x + y > 2

x - y < 2

How would I work a problem like that?

I assume you have to show the solution graphically? If so, first imagine you have two EQUATIONS (not inequalities):

x + y = 2
x - y = 2

Each of these defines a line, and the solution to the first inequality are all points in the plane above this line (i.e. all points for which x + y > 2). The solution to the second inequality are all point above the line x - y = 2. The solution to the SYSTEM of the two inequalities is the quarter of the plane above the instersection of these two line (just graph it, it's easier to see than to explain).

If you have to write the solution analytically, it's harder and looks ugly.

 

[her209]

Stop fvcking around and pay attention in class.

The reasons you set X=0 and solve for Y is because on the graph, you are trying to find the Y-intercept (where the graph crosses the Y-axis). To find the Y-intercept, you know that there is a point that lies on the Y-axis at (X=0,Y).

Now to find the X intercept (where the graph crosses the X-axis), do the same except set Y=0 and solve for X because the point that lies on the X-axis is (X,Y=0)

Get it!?

 

[rgwalt]

A "y-intercept" is the value of y when the line intercepts (or crosses) the y-axis. At this point, x=0. So, to find a y-intercept, you set x=0 and solve for y.

A "x-intercept" is the value of x when the line intercepts (or crosses) the x-axis. At this point, y=0. So, to find a x-intercept, you set y=0 and solve for x.

Ryan