Can you solve this Algebraic word problem?

Question

Suppose the sum of two numbers is 20, and the sum of their squares is 232. Find the numbers.
Please help!

Answer 1

Start with 2 equations:

a + b = 20
a^2 + b^2 = 232

Solve one equation for one variable (I usually choose the easiest):

a = 20 - b

Now substitute that answer into the other equation:

(20 - b)^2 + b^2 = 232

Solve for the remaining variable:

(-b + 20)^2 + b^2 = 232
b^2 - 40b + 400 + b^2 = 232 (solve parenthetical expression)
2b^2 - 40b + 400 = 232 (combine like monomials)
2b^2 - 40b + 168 = 0 (subtract 232 from each side)
b^2 - 20b + 84 = 0 (divide each side by 2)
(b - 6) (b - 14) = 0 (factor)
b = {6, 14} (solve for each polynomial)

You can confirm by substituting each possible value for "b" into each equation and solving for "a".

Hope the explanation was more helpful than just the answer.

Answer 2

14 and 6

Answer 3

14 and 6

Answer 4

14 and 6

Answer 5

First, make two equations to represent the wording using a and b:
a+b=20
a^2+b^2=232

now solve for one of the numbers in the first equation and then substitute it into the second:
b=20-a
a^2+(20-a)^2=232

now solve for a:
a^2+400-40a+a^2=232

conbine like terms:
2a^2-40a+400=232

Factor out 2:
2(a^2-20+200)=232

divide both sides by 2:
a^2-20+200=116

subtract 116 from both sides:
a^2-20+84=0

now factor:
(a-6) (a-14)=

a-6=0 a-14=0
a=6 a=14

now you have the two numbers. you can plug them into the first equation to check:
a+b=20

6+b=20
b=14

14+b=20
b=6

either way you write it, the two numbers are 6 and 14

Answer 6

14 & 6

Answer 7

14 and 6. Now answer my question.

Answer 8

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