document.write( "Question 3346: the sum of two integers is ten. three times the larger integer is three less than eight times the smaller interger. Find the intergers. i know the answer is 3 and 7 but teacher wants the algebraic expression and solution worked out and i can't do that \n" ); document.write( "

Algebra.Com's Answer #1462 by drglass(89) $\"\"$ $\"About$

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To solve problems like this start by giving names to the integers. Call the bigger integer x and the smaller one y. Now, the problem tells us that the sum of x and y is 10. We can write this mathematically as:

\n" ); document.write( " $\"x+%2B+y+=+10\"$

\n" ); document.write( "The problem also tells us that \"three times the larger integer is three less then eight times the smaller integer.\" Let's break this up a little, three times the larger means three times x $\"3x\"$ (x is the larger, right?). Next we are told that $\"3x\"$ is three less than eight times the smaller. Well, eight times the smaller is $\"8y\"$ and three less than $\"8y\"$ is $\"8y+-+3\"$ . The problem says $\"3x\"$ is $\"8y+-+3\"$ or, $\"3x+=+8y+-+3\"$ .

\n" ); document.write( "This gives us two equations:

\n" ); document.write( "

$\"x+%2B+y+=+10\"$ and

$\"3x+=+8y+-+3\"$

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\n" ); document.write( "\n" ); document.write( "If we subtract y from both sides of the first equation, we get $\"x+=+10+-+y\"$

\n" ); document.write( "Now substitute x in the second equation with $\"10+-+y\"$ and we get $\"3%2810+-+y%29+=+8y+-+3\"$

\n" ); document.write( "multiply 3 into the first part of the equation to get $\"30+-+3y+=+8y+-+3\"$

\n" ); document.write( "add 3 to both sides to get $\"33+-+3y+=+8y\"$

\n" ); document.write( "add 3y to both sides to get $\"33+=+11y\"$

\n" ); document.write( "finally, divide both sides by 11 to get $\"y+=+3\"$ .

\n" ); document.write( "Since we know $\"x+%2B+3+=+10\"$ , we know $\"x+=+7\"$ \n" ); document.write( "

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