document.write( "Question 132472This question is from textbook Algebra 1
\n" ); document.write( ": 3b+2a=2 and -2b+a=8 \n" ); document.write( "
Algebra.Com's Answer #96825 by jim_thompson5910(35256)\"\" \"About 
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\n" ); document.write( "\n" ); document.write( "Start with the given system of equations:\r
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\n" ); document.write( "\n" ); document.write( "\"system%283b%2B2a=2%2C-2b%2Ba=8%29\"\r
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\n" ); document.write( "\n" ); document.write( "Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for a.\r
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\n" ); document.write( "\n" ); document.write( "So let's isolate a in the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"3b%2B2a=2\" Start with the first equation\r
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\n" ); document.write( "\n" ); document.write( "\"2a=2-3b\" Subtract \"3b\" from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"2a=-3b%2B2\" Rearrange the equation\r
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\n" ); document.write( "\n" ); document.write( "\"a=%28-3b%2B2%29%2F%282%29\" Divide both sides by \"2\"\r
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\n" ); document.write( "\n" ); document.write( "\"a=%28%28-3%29%2F%282%29%29b%2B%282%29%2F%282%29\" Break up the fraction\r
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\n" ); document.write( "\n" ); document.write( "\"a=%28-3%2F2%29b%2B1\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "Since \"a=%28-3%2F2%29b%2B1\", we can now replace each \"a\" in the second equation with \"%28-3%2F2%29b%2B1\" to solve for \"b\"\r
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\n" ); document.write( "\n" ); document.write( "\"-2b%2Bhighlight%28%28%28-3%2F2%29b%2B1%29%29=8\" Plug in \"a=%28-3%2F2%29b%2B1\" into the first equation. In other words, replace each \"a\" with \"%28-3%2F2%29b%2B1\". Notice we've eliminated the \"a\" variables. So we now have a simple equation with one unknown.\r
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\n" ); document.write( "\n" ); document.write( "\"%282%29%28-2b-%283%2F2%29b%2B1%29=%282%29%288%29\" Multiply both sides by the LCM of 2. This will eliminate the fractions (note: if you need help with finding the LCM, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "\"-4b-3b%2B2=16\" Distribute and multiply the LCM to each side\r
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\n" ); document.write( "\n" ); document.write( "\"-7b%2B2=16\" Combine like terms on the left side\r
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\n" ); document.write( "\n" ); document.write( "\"-7b=16-2\"Subtract 2 from both sides\r
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\n" ); document.write( "\n" ); document.write( "\"-7b=14\" Combine like terms on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"b=%2814%29%2F%28-7%29\" Divide both sides by -7 to isolate b\r
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\n" ); document.write( "\n" ); document.write( "\"b=-2\" Divide\r
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\n" ); document.write( "\n" ); document.write( "-----------------First Answer------------------------------\r
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\n" ); document.write( "\n" ); document.write( "So the first part of our answer is: \"b=-2\"\r
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\n" ); document.write( "\n" ); document.write( "Since we know that \"b=-2\" we can plug it into the equation \"a=%28-3%2F2%29b%2B1\" (remember we previously solved for \"a\" in the first equation).\r
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\n" ); document.write( "\n" ); document.write( "\"a=%28-3%2F2%29b%2B1\" Start with the equation where \"a\" was previously isolated.\r
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\n" ); document.write( "\n" ); document.write( "\"a=%28-3%2F2%29%28-2%29%2B1\" Plug in \"b=-2\"\r
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\n" ); document.write( "\n" ); document.write( "\"a=6%2F2%2B1\" Multiply\r
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\n" ); document.write( "\n" ); document.write( "\"a=4\" Combine like terms and reduce. (note: if you need help with fractions, check out this solver)\r
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\n" ); document.write( "\n" ); document.write( "-----------------Second Answer------------------------------\r
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\n" ); document.write( "\n" ); document.write( "So the second part of our answer is: \"a=4\"\r
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\n" ); document.write( "\n" ); document.write( "-----------------Summary------------------------------\r
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\n" ); document.write( "\n" ); document.write( "So our answers are:\r
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\n" ); document.write( "\n" ); document.write( "\"b=-2\" and \"a=4\"\r
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