We need help with the following problem. The sum of the numbers is 13, the product is 36. Find the numbers. The answer is obvious, and we can write the individual equations, but we cannot figure out what process to use (elimination, etc) to work out the problem. Thank you for your help.
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Prentice Hall Algebra 1, 9.1, #67
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Write two equations:
x + y = 13
x*y = 36
Solve the first equation for x: x = 13  y
Substitute into the second equation: (13y)*y = 36
Simplify the new quadratic equation, and set equal to zero: y^2  13y +36 = 0
Factor the quadratic: (y  9)(y  4) = 0
two values of y make this true: y=9 or y=4
Subst y = 9 into original 1st equation and solve for x (4,9)
Subst y = 4 into original equation and solve for x (9,4)
The two numbers that satisfy the problem are 4 and 9 (which you knew!)
We do these "simple" problems to master the methods, so that we can later solve problems that don't have obvious answers. (Just in case your son asks!)Cindy Davis
Science and Math teacher at Highlands Latin School  Indianapolis
ds24 college graduate: working, reading, writing
ds23 college graduate: 1st year med school
dd21 college junior: Nursing

Originally posted by trob View PostThank you, Cindy! I found the solutions manual from another thread on here. Does anyone have a link for the algebra II solution manual? I looked on Amazon, but I'm not sure it is for the correct text.
HTH!Michael
Memoria Press
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