# math help

• Dec 6th 2006, 02:48 PM
fancyface
math help
solve by addition method indicate whether it is independent, inconsistent or depnedent,

x+3(y-1)=11
2(x-y+8y=28 :confused:
• Dec 6th 2006, 04:39 PM
topsquark
Quote:

Originally Posted by fancyface
solve by addition method indicate whether it is independent, inconsistent or depnedent,

x+3(y-1)=11
2(x-y+8y=28 :confused:

Need a little help with that last equation. You're missing a ")."

-Dan
• Dec 7th 2006, 04:16 AM
topsquark
From a PM from fancyface:

This is the message:
x+3(y-1)=11
2(x-y)+8y=28

-Dan
• Dec 7th 2006, 04:19 AM
topsquark
Quote:

Originally Posted by topsquark
From a PM from fancyface:

This is the message:
x+3(y-1)=11
2(x-y)+8y=28

-Dan

I'm going to simplify these equations a touch:
$\displaystyle x + 3y = 14$
$\displaystyle 2x + 6y = 28$

Note that if we multiply the top equation by 2 we get:
$\displaystyle 2x + 6y = 28$
which is just the same as the second equation.

This means we have an infinite number of solutions to x and y, all lying on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$. You'd have to look up the definitions of dependent and independent. But I can say that the system is NOT inconsistent since we have at least one solution.

-Dan
• Dec 7th 2006, 08:41 AM
AfterShock
Quote:

Originally Posted by topsquark
I'm going to simplify these equations a touch:
$\displaystyle x + 3y = 14$
$\displaystyle 2x + 6y = 28$

Note that if we multiply the top equation by 2 we get:
$\displaystyle 2x + 6y = 28$
which is just the same as the second equation.

This means we have an infinite number of solutions to x and y, all lying on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$. You'd have to look up the definitions of dependent and independent. But I can say that the system is NOT inconsistent since we have at least one solution.

-Dan

x = 14
y = 0

Note the gausselim (echelon form) is x + 3y = 14 // -5y = 0;

I agree with topsquark when he says that the system is not inconsistent (and thus consistent), since we have at least one solution. In fact, we have exactly one solution. Setting up the matrix of coefficients for the above and then augmenting it, we can clearly see when row reducing it that there are no pivots in the augmented column, thus it is consistent. Further, we know it is linearly INDEPENDENT since there are no free variables and the columns span R^2.
• Dec 7th 2006, 11:27 AM
topsquark
Quote:

Originally Posted by AfterShock
In fact, we have exactly one solution.

As I mentioned previously we have an infinite number of solutions. All points on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$ will work.

-Dan
• Dec 7th 2006, 11:36 AM
AfterShock
Quote:

Originally Posted by topsquark
As I mentioned previously we have an infinite number of solutions. All points on the line $\displaystyle y = -\frac{1}{3}x + \frac{14}{3}$ will work.

-Dan

Indeed, I believe I was looking at the homogeneous system.