document.write( "Question 607392: An electronics company produces three models of stereo speakers, models A, B, and C, and can deliver them by truck, van or station wagon. A truck holds 2 boxes of model A, 1 of model B, and 3 of model C. A van holds 1 box of model A, 3 boxes of model B, and 2 boxes of model C. A station wagon holds 1 box of model A, 3 boxes of model B, and 1 box of model C. If 15 boxes of model A, 20 boxes of model B and 22 boxes of model C are to be delivered, how many vehicles of each type should be used so that all operate at full capacity?\r
\n" ); document.write( "\n" ); document.write( "I need help with setting up the equations for this problem. This is a matrix problem.\r
\n" ); document.write( "\n" ); document.write( "Is it something like this?
\n" ); document.write( "This is how I first translated it..\r
\n" ); document.write( "\n" ); document.write( "2x + 1y + 3z = truck
\n" ); document.write( "1x + 3y + 2z = van
\n" ); document.write( "1x + 3y + 1z = station wagon\r
\n" ); document.write( "\n" ); document.write( "Then I thought to maybe combine Boxes A, B, and C together within an equation:\r
\n" ); document.write( "\n" ); document.write( "Boxes A) 2x + 1y + 1z = 15
\n" ); document.write( "Boxes B) 1x + 3y + 3z = 20
\n" ); document.write( "Boxes C) 3x + 2y + 1z = 22\r
\n" ); document.write( "\n" ); document.write( "I'm a little lost.
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Algebra.Com's Answer #382665 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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\n" ); document.write( "Good Work on the Second Set-Up: x=#Trucks, y =#vans and z=#station wagons
\n" ); document.write( "2x + 1y + 1z = 15
\n" ); document.write( "1x + 3y + 3z = 20
\n" ); document.write( "3x + 2y + 1z = 22 (5, 2, 3)
\n" ); document.write( "x = 5, y=2 and z = 3
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

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\n" ); document.write( " \"system%282%2Ax%2B1%2Ay%2B1%2Az=15%2C1%2Ax%2B3%2Ay%2B3%2Az=20%2C3%2Ax%2B2%2Ay%2B1%2Az=22%29\"
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\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29\". This is the matrix formed by the coefficients of the given system of equations.
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\n" ); document.write( " Take note that the right hand values of the system are \"15\", \"20\", and \"22\" and they are highlighted here:
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\n" ); document.write( " These values are important as they will be used to replace the columns of the matrix A.
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\n" ); document.write( " Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=-5\". 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.
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\n" ); document.write( " Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
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\n" ); document.write( " 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).
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\n" ); document.write( " Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=-25\". 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.
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\n" ); document.write( " 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=%28-25%29%2F%28-5%29=5\"
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\n" ); document.write( " So the first solution is \"x=5\"
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\n" ); document.write( " We'll follow the same basic idea to find the other two solutions. Let's reset by letting \"A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29\" again (this is the coefficient matrix).
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\n" ); document.write( " 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).
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\n" ); document.write( " Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=-10\".
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\n" ); document.write( " 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=%28-10%29%2F%28-5%29=2\"
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\n" ); document.write( " So the second solution is \"y=2\"
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\n" ); document.write( " Let's reset again by letting \"A=%28matrix%283%2C3%2C2%2C1%2C1%2C1%2C3%2C3%2C3%2C2%2C1%29%29\" which is the coefficient matrix.
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\n" ); document.write( " 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\"
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\n" ); document.write( " Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=-15\".
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\n" ); document.write( " 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=%28-15%29%2F%28-5%29=3\"
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\n" ); document.write( " So the third solution is \"z=3\"
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\n" ); document.write( " Final Answer:
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\n" ); document.write( " So the three solutions are \"x=5\", \"y=2\", and \"z=3\" giving the ordered triple (5, 2, 3)
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\n" ); document.write( " 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.
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