document.write( "Question 111728: Use elimination to solve: 3x-y=15
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Algebra.Com's Answer #81513 by jim_thompson5910(35256)\"\" \"About 
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Solved by pluggable solver: Solving a System of Linear Equations by Elimination/Addition

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\n" ); document.write( " Lets start with the given system of linear equations
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\n" ); document.write( " \"3%2Ax-1%2Ay=15\"
\n" ); document.write( " \"5%2Ax%2B1%2Ay=9\"
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\n" ); document.write( " In order to solve for one variable, we must eliminate the other variable. So if we wanted to solve for y, we would have to eliminate x (or vice versa).
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\n" ); document.write( " So lets eliminate x. In order to do that, we need to have both x coefficients that are equal but have opposite signs (for instance 2 and -2 are equal but have opposite signs). This way they will add to zero.
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\n" ); document.write( " So to make the x coefficients equal but opposite, we need to multiply both x coefficients by some number to get them to an equal number. So if we wanted to get 3 and 5 to some equal number, we could try to get them to the LCM.
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\n" ); document.write( " Since the LCM of 3 and 5 is 15, we need to multiply both sides of the top equation by 5 and multiply both sides of the bottom equation by -3 like this:
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\n" ); document.write( " \"5%2A%283%2Ax-1%2Ay%29=%2815%29%2A5\" Multiply the top equation (both sides) by 5
\n" ); document.write( " \"-3%2A%285%2Ax%2B1%2Ay%29=%289%29%2A-3\" Multiply the bottom equation (both sides) by -3
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\n" ); document.write( " So after multiplying we get this:
\n" ); document.write( " \"15%2Ax-5%2Ay=75\"
\n" ); document.write( " \"-15%2Ax-3%2Ay=-27\"
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\n" ); document.write( " Notice how 15 and -15 add to zero (ie \"15%2B-15=0\")
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\n" ); document.write( " Now add the equations together. In order to add 2 equations, group like terms and combine them
\n" ); document.write( " \"%2815%2Ax-15%2Ax%29-5%2Ay-3%2Ay%29=75-27\"
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\n" ); document.write( " \"%2815-15%29%2Ax-5-3%29y=75-27\"
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\n" ); document.write( " \"cross%2815%2B-15%29%2Ax%2B%28-5-3%29%2Ay=75-27\" Notice the x coefficients add to zero and cancel out. This means we've eliminated x altogether.
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\n" ); document.write( " So after adding and canceling out the x terms we're left with:
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\n" ); document.write( " \"-8%2Ay=48\"
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\n" ); document.write( " \"y=48%2F-8\" Divide both sides by \"-8\" to solve for y
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\n" ); document.write( " \"y=-6\" Reduce
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\n" ); document.write( " Now plug this answer into the top equation \"3%2Ax-1%2Ay=15\" to solve for x
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\n" ); document.write( " \"3%2Ax-1%28-6%29=15\" Plug in \"y=-6\"
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\n" ); document.write( " \"3%2Ax%2B6=15\" Multiply
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\n" ); document.write( " \"3%2Ax=15-6\" Subtract \"6\" from both sides
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\n" ); document.write( " \"3%2Ax=9\" Combine the terms on the right side
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\n" ); document.write( " \"cross%28%281%2F3%29%283%29%29%2Ax=%289%29%281%2F3%29\" Multiply both sides by \"1%2F3\". This will cancel out \"3\" on the left side.
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\n" ); document.write( " \"x=3\" Multiply the terms on the right side
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\n" ); document.write( " So our answer is
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\n" ); document.write( " \"x=3\", \"y=-6\"
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\n" ); document.write( " which also looks like
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\n" ); document.write( " (\"3\", \"-6\")
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\n" ); document.write( " Notice if we graph the equations (if you need help with graphing, check out this solver)
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\n" ); document.write( " \"3%2Ax-1%2Ay=15\"
\n" ); document.write( " \"5%2Ax%2B1%2Ay=9\"
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\n" ); document.write( " we get
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\n" ); document.write( " graph of \"3%2Ax-1%2Ay=15\" (red) \"5%2Ax%2B1%2Ay=9\" (green) (hint: you may have to solve for y to graph these) and the intersection of the lines (blue circle).
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\n" ); document.write( " and we can see that the two equations intersect at (\"3\",\"-6\"). This verifies our answer.
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