document.write( "Question 715404: Find two integers whose product is 296 such that one of the integers is three less than five times the other integer.
\n" ); document.write( "I've tried working this out and can't figure it out! Someone please help.
\n" ); document.write( "-Kassidy \n" ); document.write( "

Algebra.Com's Answer #439413 by fcabanski(1391) $\"\"$ $\"About$

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Use the words to make equations that relate the two numbers.

\n" ); document.write( "The first equation is \"Find two integers whose product is 296 \". Product is the result of multiplication. Call the integers x and y. x*y=296.

\n" ); document.write( "The second equation is \" one of the integers is three less than five times the other integer. x = 5*y-3 (is means =, three less than means subtract three, and five times means 5*.)

\n" ); document.write( "Substitute the value the second equation gives for x (that's (5y-3)) into the first equation. Then solve it for y. xy=296 becomes (5y-3)*y = 296.

\n" ); document.write( "Distribute the y: 5y^2 -3y = 296 Subtract 296 from both sides 5y^2 -3y-296=0.

\n" ); document.write( "Find y using the quadratic equation $\"%28-b+%2B+or+-+sqrt%28b%5E2+-+2ac%29%29%2F2a\"$ where a, b and c are from the equation ax^2 +bx +c. In this problem a = 5, b=-3 and c=-296.

\n" ); document.write( " $\"%283+%2B+or+-+sqrt%285929%29%29%2F10+=+%283+%2B+or+-+77%29%2F10\"$ = 80/10 or -74/10 = 8 or -7.4

\n" ); document.write( "The problem states that both numbers are integers. So discard y=-7.4. That leaves y=8. If y = 8 you can find x: x*8=296 ---> x=296/8 = 37.
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