Respuesta :
Given:
a.) A positive integer is 5 less than another.
b.) The sum of the reciprocal of a smaller and twice the reciprocal of the larger is 11/14.
From the given, let's generate the equation each of the givens represents.
Let,
x = The lesser number
y = The larger number
a.) A positive integer is 5 less than another.
[tex]\text{ x = y - 5}[/tex]b.) The sum of the reciprocal of a smaller and twice the reciprocal of the larger is 11/14.
[tex]\frac{1}{x}+2(\frac{1}{y})\text{ = }\frac{11}{14}[/tex][tex]\frac{1}{x}+\text{ }\frac{2}{y}=\frac{11}{14}[/tex]Step 1: Let's substitute x = y - 5 to the other equation to be able to find x.
[tex]\frac{1}{x}+\text{ }\frac{2}{y}=\frac{11}{14}\text{ }\rightarrow\frac{1}{y-5}+\text{ }\frac{2}{y}=\frac{11}{14}[/tex]Step 2: Let's simplify the equation, let's transform the two addends into a fraction with a common denominator. The common denominator will be y(y-5). We get,
[tex]\frac{1}{y-5}+\text{ }\frac{2}{y}=\frac{11}{14}\text{ }\rightarrow\text{ }\frac{(1)(y)}{y(y-5)}\text{ + }\frac{(2)(y-5)}{y(y-5)}\text{ = }\frac{11}{14}[/tex][tex]\frac{y\text{ + 2y - 10}}{y^2-5y}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ (14)(}y\text{ + 2y - 10) = (11)(}y^2-5y)[/tex][tex]14y+28y-140=11y^2\text{ - 55y}[/tex][tex]42y-140=11y^2\text{ - 55y }\rightarrow11y^2\text{ - 55y - 42y + 140 = 0 }\rightarrow11y^2\text{ -97y + 140 = 0}[/tex]Step 2: Let's find the solution to the quadratic equation.
[tex]\text{y = }\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2-4ac}}{2a}[/tex]From the given equation, a = 11, b = -97 and c = 140. Let's plug in the values.
[tex]\text{ y = }\frac{-b\text{ }\pm\text{ }\sqrt[]{b^2-4ac}}{2a}\text{ }\rightarrow\text{ y = }\frac{-(-97)\pm\text{ }\sqrt[]{(-97)^2-4(11)(140)}}{2(11)}[/tex][tex]\text{ y = }\frac{97\pm\text{ }\sqrt[]{9,409^{}-6,160}}{22}\text{ }\rightarrow\text{ y = }\frac{97\pm\text{ }\sqrt[]{3249}}{22}[/tex][tex]\text{ y = }\frac{97\text{ }\pm\text{ 57}}{22}[/tex][tex]y_1\text{ = }\frac{97\text{ + 57}}{22}\text{ = }\frac{154}{22}\text{ = 7}[/tex][tex]\text{ y}_2\text{ =}\frac{97\text{ - 57}}{22}=\frac{40}{22}\text{ = }\frac{20}{11}[/tex]Step 3: Let's substitute 7 and 20/11 to the equation if which among them is real and correct.
a.) At y = 7, x = y - 5 = 7 - 5 = 2
[tex]\frac{1}{x}+\frac{2}{y}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{1}{2}+\frac{2}{7}=\text{ }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{7}{14}+\frac{4}{14}\text{ = }\frac{11}{14}[/tex][tex]\frac{11}{14}\text{ = }\frac{11}{14}[/tex]b.) At y = 20/11, x = x - 5 = 20/11 - 5 = 20/11 - 55/11 = -35/11
[tex]\frac{1}{x}+\frac{2}{y}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{1}{-\frac{35}{11}}+\frac{2}{\frac{20}{11}}=\text{ }\frac{11}{14}\text{ }\rightarrow\text{ -}\frac{11}{35}\text{ + }\frac{2(11)}{20}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ -}\frac{11}{35}\text{ + }\frac{22}{20}\text{ = }\frac{11}{14}[/tex][tex]\frac{(22)(35)}{700}\text{ - }\frac{(11)(20)}{700}=\text{ }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{770\text{ - 220}}{700}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{550}{700}\text{ = }\frac{11}{14}[/tex][tex]\frac{\frac{550}{50}}{\frac{700}{50}}\text{ = }\frac{11}{14}\text{ }\rightarrow\text{ }\frac{11}{14}\text{ = }\frac{11}{14}[/tex]The integers that will satisfy the criteria.
Pair 1: Lesser number = 2 and the Larger number = 7
The second set of pairs is not to be considered integers because it has a concurrent decimal equivalent. A fraction can only be called an integer if it can be simplified into a whole number.
