If \(\sqrt {\frac{{\rm{x}}}{{\rm{y}}}} = \frac{{24}}{5} + \sqrt {\frac{{\rm{y}}}{{\rm{x}}}} \) and x + y = 26, then what is the value of xy?
25
Solving Equations with Square Roots and Sum
The problem asks us to find the value of \(xy\) given two equations: \(\sqrt {\frac{{\rm{x}}}{{\rm{y}}}} = \frac{{24}}{5} + \sqrt {\frac{{\rm{y}}}{{\rm{x}}}} \) and \({\rm{x}} + {\rm{y}} = 26\).
Step 1: Simplify the First Equation
The first equation involves square roots of ratios. Let's make a substitution to simplify it. Let \({\rm{a}} = \sqrt {\frac{{\rm{x}}}{{{\rm{y}}}}} \). Then, the term \(\sqrt {\frac{{\rm{y}}}{{{\rm{x}}}}} \) is the reciprocal of \({\rm{a}}\), which is \(\frac{1}{{\rm{a}}}\).
Substitute this into the first equation:
\[ {\rm{a}} = \frac{{24}}{5} + \frac{1}{{\rm{a}}} \]
Step 2: Solve for 'a'
This is a rational equation involving 'a'. To eliminate the denominators, multiply the entire equation by \(5{\rm{a}}\) (assuming \({\rm{a}} \ne 0\)).
\[ 5{\rm{a}} \times {\rm{a}} = 5{\rm{a}} \times \frac{{24}}{5} + 5{\rm{a}} \times \frac{1}{{\rm{a}}} \]
\[ 5{{\rm{a}}^2} = 24{\rm{a}} + 5 \]
Rearrange the terms to form a standard quadratic equation:
\[ 5{{\rm{a}}^2} - 24{\rm{a}} - 5 = 0 \]
We can solve this quadratic equation for 'a' by factoring. We look for two numbers that multiply to \(5 \times -5 = -25\) and add up to -24. These numbers are -25 and 1.
Rewrite the middle term:
\[ 5{{\rm{a}}^2} - 25{\rm{a}} + {\rm{a}} - 5 = 0 \]
Factor by grouping:
\[ 5{\rm{a}}({\rm{a}} - 5) + 1({\rm{a}} - 5) = 0 \]
\[ (5{\rm{a}} + 1)({\rm{a}} - 5) = 0 \]
This gives two possible solutions for 'a':
- \(5{\rm{a}} + 1 = 0 \implies 5{\rm{a}} = -1 \implies {\rm{a}} = -\frac{1}{5}\)
- \({\rm{a}} - 5 = 0 \implies {\rm{a}} = 5\)
Since \({\rm{a}} = \sqrt {\frac{{\rm{x}}}{{{\rm{y}}}}} \), 'a' must be non-negative for real values of x and y. Therefore, we choose the positive value:
\[ {\rm{a}} = 5 \]
Step 3: Relate 'a' back to x and y
Substitute the value of 'a' back into the substitution \({\rm{a}} = \sqrt {\frac{{\rm{x}}}{{{\rm{y}}}}} \):
\[ 5 = \sqrt {\frac{{\rm{x}}}{{{\rm{y}}}}} \]
Square both sides of the equation to remove the square root:
\[ {5^2} = {\left( {\sqrt {\frac{{\rm{x}}}{{{\rm{y}}}}} } \right)^2} \]
\[ 25 = \frac{{\rm{x}}}{{{\rm{y}}}} \]
This gives us a relationship between x and y:
\[ {\rm{x}} = 25{\rm{y}} \]
Step 4: Use the Second Equation to Find x and y
We have the second equation given in the problem: \({\rm{x}} + {\rm{y}} = 26\).
Now we have a system of two linear equations:
- \({\rm{x}} = 25{\rm{y}}\)
- \({\rm{x}} + {\rm{y}} = 26\)
Substitute the expression for x from the first equation into the second equation:
\[ (25{\rm{y}}) + {\rm{y}} = 26 \]
Combine like terms:
\[ 26{\rm{y}} = 26 \]
Solve for y:
\[ {\rm{y}} = \frac{{26}}{{26}} \]
\[ {\rm{y}} = 1 \]
Now substitute the value of y back into the equation \({\rm{x}} = 25{\rm{y}}\) to find x:
\[ {\rm{x}} = 25(1) \]
\[ {\rm{x}} = 25 \]
Step 5: Calculate the Value of xy
We found that \({\rm{x}} = 25\) and \({\rm{y}} = 1\). The problem asks for the value of \(xy\).
\[ {\rm{xy}} = 25 \times 1 \]
\[ {\rm{xy}} = 25 \]
Let's verify these values with the original equations:
- Equation 1: \(\sqrt {\frac{{25}}{1}} = \sqrt{25} = 5\). Also, \(\frac{{24}}{5} + \sqrt {\frac{1}{25}} = \frac{{24}}{5} + \frac{1}{5} = \frac{{25}}{5} = 5\). The first equation holds true.
- Equation 2: \(25 + 1 = 26\). The second equation holds true.
The values \({\rm{x}} = 25\) and \({\rm{y}} = 1\) are correct, and the value of \(xy\) is 25.
| Calculated Value | Result |
|---|---|
| x | 25 |
| y | 1 |
| xy | 25 |
Revision Table: Key Steps to Find xy
| Step | Description | Equation/Result |
|---|---|---|
| 1 | Substitute \(\sqrt{\frac{x}{y}}\) with 'a' in the first equation. | \({\rm{a}} = \frac{{24}}{5} + \frac{1}{{\rm{a}}}\) |
| 2 | Solve the resulting quadratic equation for 'a'. | \(5{{\rm{a}}^2} - 24{\rm{a}} - 5 = 0 \implies {\rm{a}} = 5\) (positive root) |
| 3 | Relate 'a' back to x and y. | \(5 = \sqrt{\frac{x}{y}} \implies {\rm{x}} = 25{\rm{y}}\) |
| 4 | Use the second equation \(x+y=26\) and the relation \(x=25y\) to find x and y. | \(25{\rm{y}} + {\rm{y}} = 26 \implies 26{\rm{y}} = 26 \implies {\rm{y}} = 1\) \({\rm{x}} = 25(1) \implies {\rm{x}} = 25\) |
| 5 | Calculate the product xy. | \({\rm{xy}} = 25 \times 1 = 25\) |
Additional Information on Solving Systems of Equations
A system of equations involves two or more equations with two or more variables. The goal is to find values for the variables that satisfy all equations simultaneously. Different methods can be used to solve systems of equations, depending on the type of equations.
Common methods include:
- Substitution Method: Solve one equation for one variable in terms of the other variables, and then substitute that expression into the other equation(s). This is the method we used in this problem.
- Elimination Method: Multiply equations by constants so that when the equations are added or subtracted, one or more variables are eliminated. This method is often used for systems of linear equations.
- Graphical Method: Graph each equation on the same coordinate plane. The solution(s) to the system are the point(s) where the graphs intersect. This method is less precise for non-integer solutions.
- Matrix Method: For systems of linear equations, matrices can be used (e.g., using augmented matrices, Cramer's rule, or inverse matrices).
In this problem, we transformed a system including a complex equation with square roots into a system of linear equations, which was then easily solved using substitution.
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