assignment 4

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course Mth 177

9/2/13 11:15am

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 -2*(2x+3y=9) -4x-6y=-18 -y=-13 y=13

4x+5y=5 4x+5y=5 4x+5y=5

2x+3(13)=9 2x=-30 x=30

SOulution is x=30, y=13

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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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 -2*(2x+3y-z=7) -4x-6y+2z=-14 -7y+3z=-11

4x-y+z=3 4x-y+z=3 4x-y+z=3

5x+y-2z=5

-5/2*(2x+3y-z=7) -5x-15/2y-5/2z=-35/2 -13y+z=-25

5x+y-2z=5 5x+y-2z=5

-7y+3z=-11 -7y+3z=-11 -7y+3z=-11 32y=64 y=2

-13y+z=-25 -3*(-13y+z=-25) 39y-3z=75

-13(2)+z=-25 z=1

2x+3(2)-(1)=7 2x=2 x=1

Soulution is X=1, y=2, 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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