+589 ind $\frac{x}{y}$ if \begin{align*} 2\sqrt{x} + \frac1y &= 13 \\ 8\sqrt{x} - \frac 3y &= 3. \end{align*}
+9479 We are given that:
\(2\sqrt{x}+\frac1y=13\)
We can multiply both sides of this equation by 4 to get the following equation:
\(\color{purple}8\sqrt{x}+\frac4y=52\)
We are also given that:
\(8\sqrt{x}-\frac3y=3\)
From this equation, we can subtract \(\color{purple}8\sqrt{x}+\frac4y\) from the left side and \(\color{purple}52\) from the right side. We can do this because the amounts are equal, and so we are taking the exact same amount from each side, and the resulting equation is still true.
\(8\sqrt{x}-\frac3y-{\color{purple}(8\sqrt{x}+\frac4y)}\ =\ 3-{\color{purple}52}\\~\\ 8\sqrt{x}-\frac3y-8\sqrt{x}-\frac4y\ =\ -49\\~\\ -\frac3y-\frac4y\ =\ -49\\~\\ -\frac{7}{y}\ =\ -49\\~\\ -7\ =\ -49y\qquad\text{and}\quad y\neq0\\~\\ \frac{1}{7}\ =\ y\)
Now that we know what y is, we can find x by plugging in 1/7 for y into either equation.
\(2\sqrt{x}+\dfrac{1}{\frac17}\ =\ 13\\~\\ 2\sqrt{x}+7\ =\ 13\\~\\ 2\sqrt{x}\ =\ 6\\~\\ \sqrt{x}\ =\ 3\\~\\ x\ =\ 3^2\\~\\ x\ =\ 9\)
Now that we know y is 1/7 and x is 9, we can find x/y:
\(\dfrac{x}{y}\ =\ \dfrac{9}{\frac17}\ =\ 9\div\frac17\ =\ 9\cdot\frac71\ =\ 63\)
