A linear system may have a unique solution, no solution, or infinitely many solutions. Indicate the type of the system for th following examples by U , N , or I
7x+3y= pi 4x-6y= pi^2 2x+3y= 0 4x+6y= 0 2x+3y=1 4x+ 6y= 1
x+y=5 x+2y=10
2x-3y=5 4x-6y=10
Do you see anything here:
7x+3y= pi 4x-6y= pi^2 2x+3y= 0 4x+6y= 0 2x+3y=1 4x+ 6y= 1
x+y=5 x+2y=10
I don't see anything wrong with the following:
x+y=5 x+2y=10
Do you see anything here?
2x-3y=5 4x-6y=10
I'm soryy i think i wrote this wrong there are like 4 questions in one of the questions
1. 7x+3y= pi 4x-6y= pi^2
2. 2x+3y= 0 4x+6y= 0 2
3. x+3y=1 4x+ 6y= 1
4. x+y=5 x+2y=10
5. 2x-3y=5 4x-6y=10
I am pretty sure i can figure out 2-4 (i wanted to check my answers on those though) but i am not sure how to do the fist one
The best way to check is to eliminate one of the two variables.
1. case U
If it comes up with ax+by=c like any other equation, there is a unique solution.
2. Case I
If it comes up with 0x+0y=0, this means that one equation is the linear transformation of the other, such as
3x+2y=4
6x+4y=8
Then we conclude that there is only one equation for two unknowns, the solution of which is
x=(4-2y)/3
Any value of y with the corresponding value of x is acceptable as solution, so this is case I, infinitely many solutions.
3. case N
If, after elimination of one variable, we come up with 0x+0y=5, which is impossible, then we conclude that there is no solution. Example:
x+y=3
2x+2y=8
Post your answers for a check if you wish.
1. case U
If it comes up with x=2, or 4y=6 like any other solution, there is a unique solution.
From http://www.jiskha.com/display.cgi?id=1252725501
1. 7x+3y= pi 4x-6y= pi^2
2. 2x+3y= 0 4x+6y= 0
3. x+3y=1 4x+ 6y= 1
4. x+y=5 x+2y=10
5. 2x-3y=5 4x-6y=10
1.U 2.N 3.N 4.U 5.I
Here is the check requested:
1. correct
2 If the equations are
2x+3y= 0
4x+6y= 0
as you posted above, the answer of No Solution is not correct. Try to eliminate one variable and see to which case it belongs.
3. Try again to eliminate one of the variables and see to which case it belongs.
4. correct
5. correct
To determine the type of a linear system, we need to analyze the coefficient matrix and the augmented matrix. Let's go through each example step by step:
1) 7x + 3y = π 4x - 6y = π^2
x + y = 5 x + 2y = 10
First, let's write down the augmented matrix:
[ 7 3 | π ]
[ 4 -6 | π^2 ]
[ 1 1 | 5 ]
[ 1 2 | 10 ]
To simplify the calculations, let's apply Gaussian elimination to bring it to its row-echelon form:
[R2: 4R1 - 7R2 -> R2]
[ 7 3 | π ]
[ 0 -30 | π^2 - 4π ]
[ 1 1 | 5 ]
[ 1 2 | 10 ]
[R4: R4 - R3 -> R4]
[ 7 3 | π ]
[ 0 -30 | π^2 - 4π ]
[ 1 1 | 5 ]
[ 0 1 | 5 ]
[R1: 7R4 - 3R1 -> R1]
[ 0 0 | 2π - 19 ]
[ 0 -30 | π^2 - 4π ]
[ 1 1 | 5 ]
[ 0 1 | 5 ]
[R2: -R2/30 -> R2]
[ 0 0 | 2π - 19 ]
[ 0 1 | (4π - π^2)/30 ]
[ 1 1 | 5 ]
[ 0 1 | 5 ]
[R1: R1 - R2 -> R1]
[ 0 -1 | (2π - 19)/30 ]
[ 0 1 | (4π - π^2)/30 ]
[ 1 1 | 5 ]
[ 0 1 | 5 ]
[R3: -R3 + R1 -> R3]
[ 0 -1 | (2π - 19)/30 ]
[ 0 1 | (4π - π^2)/30 ]
[ 1 0 | 68/30 ]
[ 0 1 | 5 ]
The coefficient matrix and the augmented matrix are now in row-echelon form. Let's look at the equations:
- y = (2π - 19)/30
y = (4π - π^2)/30
These equations contradict each other, so this system has no solution. Therefore, for the first example, the type of the system is N.
2) 2x + 3y = 0 4x + 6y = 0
Let's write down the augmented matrix:
[ 2 3 | 0 ]
[ 4 6 | 0 ]
We can observe that the second row is a scalar multiple of the first row. This means that the second equation is just a multiple of the first equation. Therefore, the system has infinitely many solutions. The type of the system for the second example is I.
3) 2x + 3y = 1 4x + 6y = 1
Let's write down the augmented matrix:
[ 2 3 | 1 ]
[ 4 6 | 1 ]
We can observe that the second row is a scalar multiple of the first row. Again, this means that the second equation is just a multiple of the first equation. Therefore, the system has infinitely many solutions. The type of the system for the third example is I.
In summary, the types of the systems are as follows:
1) N (no solution)
2) I (infinitely many solutions)
3) I (infinitely many solutions)