document.write( "Question 134244: Solve system by substitution or addition, whichever is easier.\r
\n" ); document.write( "\n" ); document.write( "2y - x = 3
\n" ); document.write( " x = 3y - 5 \n" ); document.write( "
Algebra.Com's Answer #98167 by algebrapro18(249)\"\" \"About 
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Since one of the equations has a variable solved for already, lets use substitution.
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\n" ); document.write( "\n" ); document.write( "The steps in substitution are as follows:
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\n" ); document.write( "\n" ); document.write( "1) Solve one of the equations for a variable.
\n" ); document.write( "2) Plug in the value of that variable into the other equation and solve for the one variable left.
\n" ); document.write( "3) Plug that value into 1 and solve for the first variable.
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\n" ); document.write( "\n" ); document.write( "Since step one has already been done for us we can just jump to step to.
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\n" ); document.write( "\n" ); document.write( "2y - x = 3
\n" ); document.write( " x = 3y - 5
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\n" ); document.write( "\n" ); document.write( "Plugging x = 3y-5 into the top equation you get:
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\n" ); document.write( "\n" ); document.write( "2y -(3y-5) = 3
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\n" ); document.write( "\n" ); document.write( "now we just solve for y
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\n" ); document.write( "\n" ); document.write( "2y -(3y-5) = 3
\n" ); document.write( "2y - 3y + 5 = 3
\n" ); document.write( "-y + 5 = 3
\n" ); document.write( "-y = -2
\n" ); document.write( "y = 2
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\n" ); document.write( "\n" ); document.write( "Now we move on to step 3. Plugging the value y = 2 into the second equation we get:
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\n" ); document.write( "\n" ); document.write( "x = 3(2)-5
\n" ); document.write( "x = 6 - 5
\n" ); document.write( "x = 1
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\n" ); document.write( "\n" ); document.write( "now we are done, our solution is (1,2) or x = 1 and y =2 depending on what form we want to leave it in. \n" ); document.write( "
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