document.write( "Question 1100689: Kayanna decided to save the 5-cent, 10-cent, and 25-cent pieces she collected after purchase made while shopping with her father. After a period of days she had a total of 246 coins. The money value of the coins was $25.50. If the amount of 5-cent pieces was 40 less than twice the amount of 10-cent pieces.
\n" ); document.write( "a. Derive a system of three equations showing the information given.
\n" ); document.write( "b. Use Cramer's Rule to find how many of each type of coins she had.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Please assist
\n" ); document.write( "

Algebra.Com's Answer #715218 by richwmiller(17219)\"\" \"About 
You can put this solution on YOUR website!
f+d+t=246
\n" ); document.write( ".05f+.10d+.25t=25.50\r
\n" ); document.write( "\n" ); document.write( "f=2*d-40\r
\n" ); document.write( "\n" ); document.write( "f-2*d+0*t=-40
\n" ); document.write( "f+d+t=246
\n" ); document.write( ".05f+.10d+.25t=25.50
\n" ); document.write( "1,-2,0,-40
\n" ); document.write( "1,1,1,246
\n" ); document.write( ".05, .10,.25,25.50\r
\n" ); document.write( "\n" ); document.write( "f=120 nickels, d=80 dimes, and t=46 quarters
\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( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " First let \"A=%28matrix%283%2C3%2C1%2C1%2C1%2C.05%2C.10%2C.25%2C1%2C-2%2C0%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 \"246\", \"25.5\", and \"-40\" 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=0.55\". 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=66\". 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=%2866%29%2F%280.55%29=120\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the first solution is \"x=120\"
\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%2C.05%2C.10%2C.25%2C1%2C-2%2C0%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=44\".
\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=%2844%29%2F%280.55%29=80\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the second solution is \"y=80\"
\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%2C.05%2C.10%2C.25%2C1%2C-2%2C0%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=25.3\".
\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=%2825.3%29%2F%280.55%29=46\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( " So the third solution is \"z=46\"
\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=120\", \"y=80\", and \"z=46\" giving the ordered triple (120, 80, 46)
\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" );