SOLUTION: please i have a problem with the following equation and don't know how to go about it and the deadline for me to submit this answer is early tommorow 1) Solve the linear equat

Algebra ->  Linear-equations -> SOLUTION: please i have a problem with the following equation and don't know how to go about it and the deadline for me to submit this answer is early tommorow 1) Solve the linear equat      Log On


   

Question 841766: please i have a problem with the following equation and don't know how to go about it and the deadline for me to submit this answer is early tommorow
1) Solve the linear equation 2x+3y=1, 5x+7y=3.
a)x=2, y=1
b)x=4, y=-2
c)x=2, y=-2
d)x=2, y=-1
2)Solve the linear equations 2x+4y=10 and 3x+6y=15.
x=5-2a,y=a
x=5-2a,y=4
x=5,y=a
x=-2a,y=a
3)Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for c

3
1
5
7
4)Solve the linear equation : 2x+3y=3, x-2y=5 and 3x+2y=7.

x=2 and y=-1
x=3 and y=1
x=3 and y=-1
x=1 and y=-1
5)Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find c

4
5
9
10
6)Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for b

9
-3
5
-4
7)Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find b

4
-5
-3
5
8)Solve the set of linear equations by Guassian elimination method : a+2b+3c=5, 3a-b+2c=8, 4a-6b-4c=-2. Find a

1
4
5
-1
9)Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Solve for a

2
4
7
3
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

1) d)x=2, y=-1 , 2)x=5-2a,y=a    |Found by substituting values into EQs given

Couple other Exs:

5) 

I.a+2b+3c=5 

II.3a-b+2c=8 

III.4a-6b-4c=-2

Solve for c :  (Eliminate one of the other two (picking a to eliminate)

I.a+2b+3c=5 

ll.3a-b+2c=8

-3a-6b-9c=-15 

3a-b+2c=8

 -7b -7c = -7,  b + c = 1,  b+=highlight%28+1-c%29

 

II.3a-b +2c=8 

III.4a-6b-4c=-2

12a-4b +8c=32 

-12a+18b+12c=6

      14b + 20c = 48

      14(1-c) + 20c = 48   |Substitute 1-c  for b

               6c = 24

                c+=+highlight%284%29  

6)a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9  Find b

a=2,+b=highlight%28-3%29, and c=5




 
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables




  

  

  system%281%2Ax%2B3%2Ay%2B2%2Az=3%2C2%2Ax%2B-1%2Ay%2B-3%2Az=-8%2C5%2Ax%2B2%2Ay%2B1%2Az=9%29

  

  

  

  First let A=%28matrix%283%2C3%2C1%2C3%2C2%2C2%2C-1%2C-3%2C5%2C2%2C1%29%29. This is the matrix formed by the coefficients of the given system of equations.

  

  

  Take note that the right hand values of the system are 3, -8, and 9 and they are highlighted here: 

  

  

  

  

  These values are important as they will be used to replace the columns of the matrix A.

  

  

  

  

  Now let's calculate the the determinant of the matrix A to get abs%28A%29=-28. To save space, I'm not showing the calculations for the determinant. However, if you need help with calculating the determinant of the matrix A, check out this solver.

  

  

  

  Notation note: abs%28A%29 denotes the determinant of the matrix A.

  

  

  

  ---------------------------------------------------------

  

  

  

  Now replace the first column of A (that corresponds to the variable 'x') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bx%5D (since we're replacing the 'x' column so to speak).

  

  

  

  

  

  

  Now compute the determinant of A%5Bx%5D to get abs%28A%5Bx%5D%29=-56. Again, as a space saver, I didn't include the calculations of the determinant. Check out this solver to see how to find this determinant.

  

  

  

  To find the first solution, simply divide the determinant of A%5Bx%5D by the determinant of A to get: x=%28abs%28A%5Bx%5D%29%29%2F%28abs%28A%29%29=%28-56%29%2F%28-28%29=2

  

  

  

  So the first solution is x=2

  

  

  

  

  ---------------------------------------------------------

  

  

  We'll follow the same basic idea to find the other two solutions. Let's reset by letting A=%28matrix%283%2C3%2C1%2C3%2C2%2C2%2C-1%2C-3%2C5%2C2%2C1%29%29 again (this is the coefficient matrix).

  

  

  

  

  Now replace the second column of A (that corresponds to the variable 'y') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5By%5D (since we're replacing the 'y' column in a way).

  

  

  

  

  

  

  Now compute the determinant of A%5By%5D to get abs%28A%5By%5D%29=84.

  

  

  

  To find the second solution, divide the determinant of A%5By%5D by the determinant of A to get: y=%28abs%28A%5By%5D%29%29%2F%28abs%28A%29%29=%2884%29%2F%28-28%29=-3

  

  

  

  So the second solution is y=-3

  

  

  

  

  ---------------------------------------------------------

  

  

  

  

  

  Let's reset again by letting A=%28matrix%283%2C3%2C1%2C3%2C2%2C2%2C-1%2C-3%2C5%2C2%2C1%29%29 which is the coefficient matrix.

  

  

  

  Replace the third column of A (that corresponds to the variable 'z') with the values that form the right hand side of the system of equations. We will denote this new matrix A%5Bz%5D 

  

  

  

  

  

  

  Now compute the determinant of A%5Bz%5D to get abs%28A%5Bz%5D%29=-140.

  

  

  

  To find the third solution, divide the determinant of A%5Bz%5D by the determinant of A to get: z=%28abs%28A%5Bz%5D%29%29%2F%28abs%28A%29%29=%28-140%29%2F%28-28%29=5

  

  

  

  So the third solution is z=5

  

  

  

  

  ====================================================================================

  

  Final Answer:

  

  

  

  

  So the three solutions are x=2, y=-3, and z=5 giving the ordered triple (2, -3, 5)

  

  

  

  

  Note: there is a lot of work that is hidden in finding the determinants. Take a look at this 3x3 Determinant Solver to see how to get each determinant.