One positive integer is 1 less than twice another. The sum of their squares is 106. Find the integers.

1 answer:

8 0

They give us 2 pieces to the puzzle. Both are positive numbers...x and y.
1.) 1 number is 1 less than twice another number. (x = 2y -1)...and
2.) the sum of their squares is 106. (x^2 + y^2 = 106).

substitute the value for x into the second equation.

(2y-1)^2 + y^2 = 106
(2y-1) (2y-1) + y^2 = 106 (use distributive property)
4y^2 - 2y - 2y + 1 + y^2 = 106 (subtract 106 from both sides)
4y^2 - 2y - 2y + 1 + y^2 - 106 = 106 - 106 (combine like terms)
5y^2 - 4y - 105 = 0 (factor)
(y-5) (5y-21) = 0 (set to 0)
y - 5 = 0
y = 5

substitute the 5 into the equation for y (x = 2(5) - 1)
x = 9 if we square 9, we get 81.
subtracted from 106 we have 25...the square root of 25 is 5.

our answers are 5 and 9.

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The Corrected Problem is :

2:x and 12:18 identify the value of x that makes each pair of ratios equivalent .

Solution:

If a pair of Ratios are equivalent then we can write

$\frac{2}{x}=\frac{12}{18}\\ \\ \text{Simplify we get}\\ \\ 12x=36\\ \\ \text{Divide both the sides by 12 we get}\ \\ \\ \frac{12x}{12}=\frac{36}{12}\\ \\ x=3\\ \\ \text{Hence the required value of x is 3}\\$