document.write( "Question 202668: I need help with this question,
\n" ); document.write( "{Find a number which is 150 greater than its opposite.}\r
\n" ); document.write( "\n" ); document.write( "The opposite of 150 is -150 right? so is the answer zero?
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Algebra.Com's Answer #152853 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
If the problem said: \"Find a number 'x' which is 150 greater than its opposite.\", then it might have been more obvious. Anyways, they're not referring to the opposite of 150, they're referring to the opposite of the unknown number \"x\". For ANY number \"x\", the opposite is \"-x\" (ie negative 'x')\r
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\n" ); document.write( "\n" ); document.write( "The given sentence \"a number which is 150 greater than its opposite\" translates to \"x=150%2B%28-x%29\"\r
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\n" ); document.write( "\n" ); document.write( "\"x=150%2B%28-x%29\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"x%2Bred%28x%29=150%2B%28-x%29%2Bred%28x%29\" Add 'x' to both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"x%2Bx=150%2Bcross%28%28-x%29%2Bx%29\" Cancel out the opposite (but common) terms.\r
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\n" ); document.write( "\n" ); document.write( "\"red%282%29x=150%2Bred%280x%29\" Combine like terms.\r
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\n" ); document.write( "\n" ); document.write( "\"2x=150\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"%282x%29%2Fred%282%29=150%2Fred%282%29\" Divide both sides by 2 to isolate \"x\".\r
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\n" ); document.write( "\n" ); document.write( "\"%28cross%282%29x%29%2Fcross%282%29=150%2F2\" Cancel out the common terms.\r
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\n" ); document.write( "\n" ); document.write( "\"x=150%2F2\" Simplify\r
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\n" ); document.write( "\n" ); document.write( "\"x=red%2875%29\" Reduce\r
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\n" ); document.write( "\n" ); document.write( "So the solution is \"x=75\" which means that the number 75 is 150 greater than its opposite (which is -75). If you're bold enough (or bored enough), you can draw out a number line and count the distance from 75 to -75. Sure enough, you'll find that the distance is 150.
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