document.write( "Question 999769: In cramers rule and elimination and substitution
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\n" ); document.write( "Y=-5/2x+4
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Algebra.Com's Answer #617355 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
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Solved by pluggable solver: Using Cramer's Rule to Solve Systems with 2 variables

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\n" ); document.write( " \"system%280%2Ax%2B1%2Ay=-1%2C2.5%2Ax%2B1%2Ay=4%29\"
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\n" ); document.write( " First let \"A=%28matrix%282%2C2%2C0%2C1%2C2.5%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 \"-1\" and \"4\" which are highlighted here:
\n" ); document.write( " \"system%280%2Ax%2B1%2Ay=highlight%28-1%29%2C2.5%2Ax%2B1%2Ay=highlight%284%29%29\"
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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=%280%29%281%29-%281%29%282.5%29=-2.5\". Remember that the determinant of the 2x2 matrix \"A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29\" is \"abs%28A%29=ad-bc\". If you need help with calculating the determinant of any two by two matrices, then 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( " \"A%5Bx%5D=%28matrix%282%2C2%2Chighlight%28-1%29%2C1%2Chighlight%284%29%2C1%29%29\"
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\n" ); document.write( " Now compute the determinant of \"A%5Bx%5D\" to get \"abs%28A%5Bx%5D%29=%28-1%29%281%29-%281%29%284%29=-5\". Once again, remember that the determinant of the 2x2 matrix \"A=%28matrix%282%2C2%2Ca%2Cb%2Cc%2Cd%29%29\" is \"abs%28A%29=ad-bc\"
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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=%28-5%29%2F%28-2.5%29=2\"
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\n" ); document.write( " So the first solution is \"x=2\"
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\n" ); document.write( " We'll follow the same basic idea to find the other solution. Let's reset by letting \"A=%28matrix%282%2C2%2C0%2C1%2C2.5%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( " \"A%5Bx%5D=%28matrix%282%2C2%2C0%2Chighlight%28-1%29%2C2.5%2Chighlight%284%29%29%29\"
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\n" ); document.write( " Now compute the determinant of \"A%5By%5D\" to get \"abs%28A%5By%5D%29=%280%29%284%29-%28-1%29%282.5%29=2.5\".
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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=%282.5%29%2F%28-2.5%29=-1\"
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\n" ); document.write( " So the second solution is \"y=-1\"
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\n" ); document.write( " Final Answer:
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\n" ); document.write( " So the solutions are \"x=2\" and \"y=-1\" giving the ordered pair (2, -1)
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\n" ); document.write( " Once again, Cramer's Rule is dependent on determinants. Take a look at this 2x2 Determinant Solver if you need more practice with determinants.
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\n" ); document.write( "\n" ); document.write( "solutions:
\n" ); document.write( "\"x=2\"
\n" ); document.write( "\"y=-1+\"\r
\n" ); document.write( "\n" ); document.write( "2.
\n" ); document.write( "elimination\r
\n" ); document.write( "\n" ); document.write( "\"y=-1+\".....eq.1
\n" ); document.write( "\"%285%2F2%29x%2By=4\"....eq.2
\n" ); document.write( "-----------------------------------subtract eq.1 from eq.2\r
\n" ); document.write( "\n" ); document.write( "\"%285%2F2%29x%2By-y=4-%28-1%29\"
\n" ); document.write( "\"%285%2F2%29x=4%2B1\"
\n" ); document.write( "\"5x%2F2=5\"
\n" ); document.write( "\"5x=5%2A2\"
\n" ); document.write( "\"x=%28cross%285%29%2A2%29%2Fcross%285%29\"
\n" ); document.write( "\"x=2\"\r
\n" ); document.write( "\n" ); document.write( "solutions:
\n" ); document.write( "\"x=2\"
\n" ); document.write( "\"y=-1+\"\r
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\n" ); document.write( "\n" ); document.write( "3.substitution \r
\n" ); document.write( "\n" ); document.write( "\"y=-1+\".....eq.1
\n" ); document.write( "\"%285%2F2%29x%2By=4\"....eq.2
\n" ); document.write( "-----------------------------------substitute \"-1\" for \"y\" in eq.2 annd solve for \"x\"\r
\n" ); document.write( "\n" ); document.write( "\"%285%2F2%29x%2B%28-1%29=4\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "\"%285%2F2%29x-1=4\"
\n" ); document.write( "\"5x%2F2=4%2B1\"
\n" ); document.write( "\"5x=5%2A2\"
\n" ); document.write( "\"x=%28cross%285%29%2A2%29%2Fcross%285%29\"
\n" ); document.write( "\"x=2\"\r
\n" ); document.write( "\n" ); document.write( "solutions:
\n" ); document.write( "\"x=2\"
\n" ); document.write( "\"y=-1+\"\r
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