document.write( "Question 76284: Hi, I know I could get one question like this if someone could explain it to me - Three times the smaller of 2 numbers increased by 5 times the larger is 229. Four times the smaller decreased by 25 equals 3 times the larger. Find the numbers.
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document.write( "I started out with - x=bigger number, and y=smaller number
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document.write( "3y+5x=229 (first equation)
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document.write( "4y-25=3x (second equation)
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document.write( "How do you go on? \n" );
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Algebra.Com's Answer #54744 by wodkit02(9)![]() ![]() ![]() You can put this solution on YOUR website! Once you have 2 simultaneous equations, you can solve for a variable in one equation by using a modified version of the other.\r \n" ); document.write( "\n" ); document.write( "For this particular question, your equations are correct. So we have \n" ); document.write( "3y+5x=229 \n" ); document.write( "4y-25=3x\r \n" ); document.write( "\n" ); document.write( "I'm going to start by putting the 2nd equation in terms of x, so when I solve for x I get: \n" ); document.write( "4y-25=3x \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Now I can plug in that x value into the 1st equation, so I can solve for y. \n" ); document.write( "3y+5x=229 \n" ); document.write( "3y+5* \n" ); document.write( "3y+ \n" ); document.write( " \n" ); document.write( "9y+20y-125=687...add like terms to get \n" ); document.write( "29y-125=687...add 125 to each side to get \n" ); document.write( "29y=812...finally, we can divide to get \n" ); document.write( "y=28\r \n" ); document.write( "\n" ); document.write( "Back to the 2nd equation, we can now plug in a number for y, and then solve for x: \n" ); document.write( "4y-25=3x \n" ); document.write( "4(28)-25=3x \n" ); document.write( "112-25=3x \n" ); document.write( "87=3x \n" ); document.write( "29=x\r \n" ); document.write( "\n" ); document.write( "So the numbers are 28 and 29.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |