Systems of Eqns

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course MTH 177

9/2 2:30 am

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you

do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

014. Determinants; Systems of Equations and their Geometrical Interpretation

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Question: `q005. Solve by elimination the system of equations 2x + 3 y = 9, 4 x + 5 y = 5.

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Your solution:

2x+3y=9

4x+5y=5

First multiply 2x+3y=9 by -2 the result is -4x-6y=-18 then add this to 4x+5y=5 to get the result of 0x-y=-13. solving for y gives you y=13

then substitute y=13 into 2x+3y=9 which is 2x+3(13)=9 simplify and you get 2x=30 solve for x and you get x=15

confidence rating #$&*: 3

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Given Solution:

`aWe start by writing the system in the form

2x + 3 y = 9

4 x + 5 y = 5.

We then multiply the first equation by -2, leaving the second equation alone, to obtain

-4x -6 y = -18

4 x + 5 y = 5.

We add the two equations to obtain

0 x - 1 y = -13, which we multiply by -1 to get

y = 13.

We then substitute y = 13 back into the first original equation to get

2x + 3 * 13 = 9, which becomes

2x + 39 = 9. Adding -39 to both sides we get

2x = -30, the dividing by 2 we finally obtain

{}x = -15.

We check our solution x = -15, y = 13 in the second equation: 4 * -15 + 5 * 13 = -60 + 65 = 5, which agrees with the equation and confirms our result.

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Self-critique (if necessary): OK

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Self-critique Rating: 3

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Question: `q006. Solve by elimination the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5.

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Your solution:

2x+3y-z=7

4x-y+z=3

5x+y-2z=5

fist we multiply 2x+3y-z=7 by -2 to get -4x-6y+2z=-14 then add this to 4x-y+z=3 and get -7y+3z=-11

then we multiply 2x+3y-z=7 by (-5/2) to get -5x-(15/2)y+(5/2)z=(-35/2), then add this to 5x+y-2z=5 to get (-13/2)y+(1/2)z=(-25/2)

then multiply (-13/2)y+(1/2)z=(-25/2) by the common denominator of 2 to get -13y+z=-25. now we have successfully eliminated the x from all equations

-7y+3z=-11

-13y+z=-25

now multiply -13y+z=-25 by -3 to get 39y-3z=75 add this to -7y+3z=-11 to get 32y=64. solve for y to get y=2

substitute y=2 into -7y+3z=-11 to get -7(2)+3z=-11 which simplifies to 3z=3 then solving for z you get z=1

now substitute y=2 and z=1 into the original 2x+3y-z=7 to get 2x+3(2)-(1)=7. simplify to get 2x=2. solve for x to get x=1

The final result is x=1 y=2 and z=1

confidence rating #$&*: 3

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Given Solution:

`aWe start by writing the system in the form

2x + 3 y - z = 7

4 x - y + z = 3

5 x + y - 2z = 5.

We multiply the first equation by -2 to get -4x - 6y + 2z = -14, which we add to the second equation 4x - y + z = 3, resulting in solution

-7y + 3z = -11.

Then we multiply the first equation by -5/2 to get -5 x - 15/2 y + 5/2 z = -35/2, which we add to the third equation 5x + y - 2z = 5 to obtain

-13/2 y + 1/2 z = -25/2, which we multiply by the common denominator 2 to get

-13 y + z = -25.

We have therefore eliminated x from the second and third equations. We form a system consisting of these two equations:

-7y + 3z = -11

-13y + z = -25.

We can choose to eliminate z from these equations, multiplying the second by -3 to getthe system

-7y + 3z = -11

39y - 3z = 75.

Adding these equations we obtain

32 y = 64, which has solution y = 2.

Substituting y = 2 into either of the two equations in this system, choosing the first we get

-7 * 2 + 3 z = -11 which simplifies to

-14 + 3z = -11 then to

3z = 3 and

z = 1.

Substituting z = 1 and y = 2 back into any equation of the original system, choosing the second we get

4 x - 2 + 1 = 3 so that

4x - 1 = 3 and

4x = 4 with solution

x = 1

we have a complete solution to the original system:

x=1

y=2

z=1.

We can check this solution by substituting these values into each of the three equations; the result will verify the solution.

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Self-critique (if necessary):ok

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Self-critique Rating:3

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