Question 715404
Use the words to make equations that relate the two numbers.<P>
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.<P>
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*.)<P>
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.<P>
Distribute the y:  5y^2 -3y = 296 Subtract 296 from both sides  5y^2 -3y-296=0.<P>
Find y using the quadratic equation {{{(-b + or - sqrt(b^2 - 2ac))/2a}}}  where a, b and c are from the equation ax^2 +bx +c.  In this problem a = 5, b=-3 and c=-296.<P>
{{{(3 + or - sqrt(5929))/10 = (3 + or - 77)/10}}} = 80/10 or -74/10 = 8 or -7.4<P>
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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