#$&* course Mth 177 9/2/13 11:15am Question: `q005. Solve by elimination the system of equations 2x + 3 y = 9, 4 x + 5 y = 5.
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Given Solution: `aWe start by writing the system in the form 2x + 3 y = 9 4 x + 5 y = 5. We then multiply the first equation by -2, leaving the second equation alone, to obtain -4x -6 y = -18 4 x + 5 y = 5. We add the two equations to obtain 0 x - 1 y = -13, which we multiply by -1 to get y = 13. We then substitute y = 13 back into the first original equation to get 2x + 3 * 13 = 9, which becomes 2x + 39 = 9. Adding -39 to both sides we get 2x = -30, the dividing by 2 we finally obtain {}x = -15. We check our solution x = -15, y = 13 in the second equation: 4 * -15 + 5 * 13 = -60 + 65 = 5, which agrees with the equation and confirms our result. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Solve by elimination the system of equations 2x + 3 y - z = 7, 4 x - y + z = 3, 5 x + y - 2z = 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x+3y-z=7 -2*(2x+3y-z=7) -4x-6y+2z=-14 -7y+3z=-11 4x-y+z=3 4x-y+z=3 4x-y+z=3 5x+y-2z=5 -5/2*(2x+3y-z=7) -5x-15/2y-5/2z=-35/2 -13y+z=-25 5x+y-2z=5 5x+y-2z=5 -7y+3z=-11 -7y+3z=-11 -7y+3z=-11 32y=64 y=2 -13y+z=-25 -3*(-13y+z=-25) 39y-3z=75 -13(2)+z=-25 z=1 2x+3(2)-(1)=7 2x=2 x=1 Soulution is X=1, y=2, z=1 confidence rating #$&*: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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!