another algebra problem (linear)

[coder1]

Ok, here is the equation:

x-3y=0
5x-y=-14

I came up with 3,1

But it appears to be wrong.

Also does anyone know of a math forum where people can discuss algebra.

Thanks

 

[minendo]

nevermind

 

[ryzmah]

set x=3y and sub into bottom equation
14y = -14
y = -1

x- 3(-1) = 0
x=-3

you probably forgot a negative

 

[Soccer55]

-3,-1 works. Maybe you just forgot the negative sign on the 14 in the 2nd equation?

Since it's a system with only 2 equations, you can check my answer by manipulating the first equation to get x=something or y=something, then substituting that into the 2nd equation. Now you have an equation with only 1 variable. Solve for that variable, and then use that value in the first equation to get the value of your other variable. Hopefully this helps and doesn't confuse you too much.....

-Tom

 

[notfred]

x - 3y = 0
add 3y to both sides

x = 3y

Substitute in the second equation.
5(3y)-y = -14
Multiply
15y-y = -14
14y = -14
y = -1

Then use -1 in the original problem

x-3(-1) = 0
Multiply
x+3 = 0
subtract
x = -3

y = -1, x = -3

 

[notfred]

BTW, this isn't really linear algebra 🙂

 

[coder1]

hrmm, trying to figure out why I keep getting y=1

After I multiple the coeffeicents I then get

5x-y=-14
5x-15y=0

I then get

-14y=-14



Where am I going wrong?

Thanks again for all of the replies.

 

[notfred]

Originally posted by: coder1
hrmm, trying to figure out why I keep getting y=1

After I multiple the coeffeicents I then get

5x-y=-14
5x-15y=0

I then get

-14y=-14



Where am I going wrong?

Thanks again for all of the replies.

5x - y is jsut (5x) - y, it is NOT 5(x-y) . Only the x is multiplied by 5, not the y.

 

[Haircut]

Originally posted by: coder1
hrmm, trying to figure out why I keep getting y=1

After I multiple the coeffeicents I then get

5x-y=-14
5x-15y=0

I then get

-14y=-14

Where am I going wrong?
Thanks again for all of the replies.

5x - y = -14
5x - 15y = 0

Thus -15y - (-y) = 0 - (-14)
-14y = 14
y = -1

 

[dighn]

After I multiple the coeffeicents I then get

5x-y=-14
5x-15y=0

then get

-14y=-14

-1 - (-15) = 14, not -14

i hate making this kind of errors. so annoying, esp some how the mind is fixated temporarily and can't see the error

 

[coder1]

I see it now. That seems to always throw me off. Thanks so much for the help guys.

 

[coder1]

Ok here's another one. I feel this has no solution.

x+3y=4
2x+6y=1


I have used the subsitution method as well as multiplying the coefficients. I keep getting a 8=1 or someothing along that line. All the variables cancel each other out.

 

[Yossarian]

those equations seem to be false.

 

[coder1]

Pipboy, that seems right to me. Thanks for the help.

 

[notfred]

Originally posted by: coder1
Ok here's another one. I feel this has no solution.

x+3y=4
2x+6y=1


I have used the subsitution method as well as multiplying the coefficients. I keep getting a 8=1 or someothing along that line. All the variables cancel each other out.

Yeah, that doesn't seem to work.

 

[Haircut]

Originally posted by: coder1
Ok here's another one. I feel this has no solution.

x+3y=4
2x+6y=1


I have used the subsitution method as well as multiplying the coefficients. I keep getting a 8=1 or someothing along that line. All the variables cancel each other out.

Yeah, that doesn't work out.

You can rewrite the equations as follows

3y = -x + 4
6y = -2x + 1

y = -x/3 + 4/3
y = -x/3 + 1/6

Thus the two lines have the same gradient so must be parallel and therefore can't meet.

 

[Yax]

Interesting. Here's an old one for you to think about:

___
.99 = 1

The proof was provided to me long ago in 10th grade algebra class. I thought it was interesting.

Edit: Note for the less informed, the bar over the 99 means that 9 repeats to infinity.