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2015-12-12T11:54:29+08:00

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This is an example of number problems. We have to translate the word problem into algebraic expressions.

Let x = the first number
      16 - x = the second number

The sum of two the square of  each number is 146 can be written as this
      x^{2} + (16-x)^{2} = 146

Let's solve for x
      x^{2} + 256 - 32x +  x^{2} = 146
     2 x^{2}  - 32x + 110 = 0
     ( \frac{1}{2})(2 x^{2}  - 32x + 110 = 0)
      x^{2}  - 16x + 55 = 0
     (x - 11)(x - 5) = 0
     x - 11 = 0              x - 5 = 0
     x = 11                    x = 5

Therefore, the two numbers are 11 & 5.

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2015-12-12T16:48:56+08:00
Lets represent:
x=first number
y=second number

Since the algebraic expression "sum", means you add them and the answer is 16
x+y=16

The sum of their square is 146 means if you add their square, the answer is 146.
x²+y²=146

You now have:
x+y=16
x²+y²=146

First find the value of y in the first equation:
x+y=16
y=16-x

Substitute to the second equation
x²+(16-x)²=146
x²+256-32x+x²=146
Add like terms
2x²-32x+256=146
Divide both sides by 2
x²-16x+128=73
Subtract both sides by 73
x²-16x+55=0
Factor out:
(x-11)(x-5)=0
x-11=0 x-5=0
x=11    x=5

Now find the two numbers:
a)if x=11
x+y=16
11+y=16
y=5

b)if x=5
x+y=16
5+y=16
y=11

Hope this helps =)
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