document.write( "Question 131817This question is from textbook
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document.write( ": Arranging Like Terms:
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document.write( "Use linear combinations to solve the system of linear equations.\r
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document.write( "3b + 2c = 46
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document.write( "5c + b = 11\r
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document.write( "Do I need to put the bottom equation into slope intercept form, y = mx + b? \n" );
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Algebra.Com's Answer #96264 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Things are out of order here because the 'b' terms aren't lined up and neither are the 'c' terms. If you put either of the equations in slope-intercept form, you are just going to make a bigger mess of things. What you need to do first is to use the commutative property of addition to swap the position of the terms on the left side of either of the equations so that 'b' in the second equation is lined up under '3b' in the first, etc.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now you are asked to solve the system by the method of linear combination, aka 'the addition method' I like to call this method the 'elimination' method because we are going to perform some operations that will eliminate one of the variables.\r \n" ); document.write( "\n" ); document.write( "What we need is a number that can be multiplied across one of the equations so that the coefficient on one of the variables in that equation becomes the additive inverse of the coefficient on that same variable in the other equation. (remember that the additive inverse is the number you can add to any number and get zero as a result. -2 and 2 are additive inverses of each other.)\r \n" ); document.write( "\n" ); document.write( "In this case, we can multiply the second equation by -3. Let's do that. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "And then add the two equations together, term by term: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "There are two ways to go from here. We can either take this value for c that we just discovered and substitute it back into either of the original equations allowing us to solve for the other variable, b, or we can go back to the original equations and find a way to eliminate the c variable, thus solving for b the same way we just solved for c. I'm going to demonstrate this second method because there is a little fly in the ointment that you need to see to learn to solve these types of problems in general.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "We could multiply the second equation by \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Adding: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Always check your answers. \n" ); document.write( "Remember we said \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( "Answer checks. \n" ); document.write( " \n" ); document.write( " |