document.write( "Question 651018: how do i determine whether (1,-5,10) is a solution of the following system.\r
\n" ); document.write( "\n" ); document.write( "x+y+z=6
\n" ); document.write( "3x-y-z=-2
\n" ); document.write( "2x-y+4z=37
\n" ); document.write( "

Algebra.Com's Answer #407537 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "to determine whether , plug in values for \"x=1\", \"y=-5\" and \"z=10\". I you get that left side of equation is equal to right side, than (1,-5,10) is a solution of the following system\r
\n" ); document.write( "\n" ); document.write( "\"x%2By%2Bz=6\"..->..\"1-5%2B10=6\"->..\"11-5=6\"->..\"6=6\"\r
\n" ); document.write( "\n" ); document.write( "\"3x-y-z=-2\"..->..\"3%2A1-%28-5%29-10=-2\"->..\"3%2B5-10=-2\"->..\r
\n" ); document.write( "\n" ); document.write( "\"8-10=-2\"->..\"-2=-2\"\r
\n" ); document.write( "\n" ); document.write( "\"2x-y%2B4z=37\"..->..\"2%2A1-%28-5%29%2B4%2A10=37\".->..\"2%2B5%2B40=37\"->..\"47=37\"...not true\r
\n" ); document.write( "\n" ); document.write( "so, the ordered triple (\"1\", \"-5\", \"10\") is NOT a solution of this system\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "here is a solution:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "
Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 3 variables

\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " \"system%281%2Ax%2B1%2Ay%2B1%2Az=6%2C3%2Ax%2B-1%2Ay%2B-1%2Az=-2%2C2%2Ax%2B-1%2Ay%2B4%2Az=37%29\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%29%29\". This is the matrix formed by the coefficients of the given system of equations.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Take note that the right hand values of the system are \"6\", \"-2\", and \"37\" and they are highlighted here:
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " These values are important as they will be used to replace the columns of the matrix A.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Now let's calculate the the determinant of the matrix A to get \"abs%28A%29=-20\". 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.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Notation note: \"abs%28A%29\" denotes the determinant of the matrix A.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " ---------------------------------------------------------
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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).
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=-20\". 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.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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=%28-20%29%2F%28-20%29=1\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the first solution is \"x=1\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " ---------------------------------------------------------
\n" ); document.write( "
\n" ); document.write( "
\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%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%29%29\" again (this is the coefficient matrix).
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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).
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=60\".
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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=%2860%29%2F%28-20%29=-3\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the second solution is \"y=-3\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " ---------------------------------------------------------
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Let's reset again by letting \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C3%2C-1%2C-1%2C2%2C-1%2C4%29%29\" which is the coefficient matrix.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " Now compute the determinant of \"A%5Bz%5D\" to get \"abs%28A%5Bz%5D%29=-160\".
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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=%28-160%29%2F%28-20%29=8\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the third solution is \"z=8\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " ====================================================================================
\n" ); document.write( "
\n" ); document.write( " Final Answer:
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the three solutions are \"x=1\", \"y=-3\", and \"z=8\" giving the ordered triple (1, -3, 8)
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\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.
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "

\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "as you can see, the three solutions are \"x=1\", \"y=-3\", and \"z=8\" giving the ordered triple (\"1\", \"-3\", \"8\")\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "
\n" );