what is the easiest step to find two numbers that can give me a sum of 80 and a product of 203?

Question

I'm doing math homework with my child, and I don't remember how to do this. I need to be reminded of the simple steps to find how many of each item at different prices were sold in order to reach the given amount of money. HELP!

Answer 1

suppose x and y are two unknown numbers.
As per the data given,
x+y=80
x*y=203
so, x=80-y
x*y=203
(80-y)*y=203
80y-y^2=203
y^2-80y+203=0,by reversing the sign.
here, a=1,b=-80,c=203
now y=-b+-sqrt((-80)^2-4*1*203 divided by 2*1
y=80+sqrt(6400-4*1*203)divided by 2
y=80+sqrt(6400-4*1*203)divided by 2
y=77.4
now,
y=80-sqrt(6400-4*1*203)divided by 2
y=2.6235

Answer 2

Check your numbers again as there is no solution to the problem as stated. It might help if you present one of the problems exactly as stated. However, you would analyze it as follows:

First determine the prime factors of 203 = 7x29
7 and 29 are both prime numbers.
Now assuming you have at least 2 different items priced at 7 price units and 29 price units, and you want to know how many of each will be required so they sum to 80 price units, you have the equation:

7m+29n = 80 where you have to determine possible values of m and n

80-29 = 51 = 3x17, so that's not the answer since 51 is not divisible by 7.
80-(2x29) = 80-58 = 22 = 2x11 and that's wrong too.
80-(3x29) is negative, so we see that there is no solution.

Answer 3

if I'm not mistaken:
1. let one of the numbers equal x, the other would then be 80-x
2. (x)(80-x)=203

Answer 4

product=203
1*203
7*29
no this cannot be split to give a sum of 80 and product of 203

Answer 5

x+y=80 y=80-x
xy=203
x(80-x)=203
80x-x^2=203
x^2-80x+203=0

unfortunately, you have to use the quadratic formula and this gets messy
my calculator gives about 77.376463 and 2.623568

Answer 6

x+y=80

xy=203

x=80-y

so (80-y)y = 203

Answer 7

im really good at math

but can u write what the question asks
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trust me i can help u