Answer :
Sure, let's solve the system of equations step-by-step:
We have the following system of equations:
[tex]\[ \left\{ \begin{array}{l} 1. \quad 2 \sqrt{x-1} + \frac{3}{y} = \frac{23}{2} \\ 2. \quad 3 \sqrt{x-1} - \frac{2}{y} = 1 \end{array} \right. \][/tex]
Step 1: Let [tex]\( z = \sqrt{x - 1} \)[/tex].
This substitution simplifies our system:
[tex]\[ \left\{ \begin{array}{l} 1. \quad 2z + \frac{3}{y} = \frac{23}{2} \\ 2. \quad 3z - \frac{2}{y} = 1 \end{array} \right. \][/tex]
Step 2: Solve for [tex]\(\frac{1}{y}\)[/tex] from the second equation.
From the second equation:
[tex]\[ 3z - \frac{2}{y} = 1 \Rightarrow \frac{2}{y} = 3z - 1 \Rightarrow \frac{1}{y} = \frac{3z - 1}{2} \][/tex]
Step 3: Substitute [tex]\(\frac{1}{y}\)[/tex] into the first equation.
Substitute [tex]\(\frac{1}{y} = \frac{3z - 1}{2}\)[/tex] into the first equation:
[tex]\[ 2z + 3 \left(\frac{3z - 1}{2}\right) = \frac{23}{2} \][/tex]
Simplify:
[tex]\[ 2z + \frac{9z - 3}{2} = \frac{23}{2} \Rightarrow 2z + \frac{9z - 3}{2} = \frac{23}{2} \][/tex]
Step 4: Clear the fraction by multiplying through by 2.
[tex]\[ 2(2z) + 2 \left(\frac{9z - 3}{2}\right) = 2 \left(\frac{23}{2}\right) \Rightarrow 4z + 9z - 3 = 23 \Rightarrow 13z - 3 = 23 \][/tex]
Step 5: Solve for [tex]\( z \)[/tex].
[tex]\[ 13z - 3 = 23 \Rightarrow 13z = 26 \Rightarrow z = 2 \][/tex]
Step 6: Substitute [tex]\( z = 2 \)[/tex] back to find [tex]\(\frac{1}{y}\)[/tex].
From [tex]\(\frac{1}{y} = \frac{3z - 1}{2}\)[/tex]:
[tex]\[ \frac{1}{y} = \frac{3(2) - 1}{2} = \frac{6 - 1}{2} = \frac{5}{2} \Rightarrow y = \frac{2}{5} \][/tex]
Thus, [tex]\( y = 0.4 \)[/tex].
Step 7: Find x using [tex]\( z = 2 \)[/tex] and [tex]\( z = \sqrt{x-1} \)[/tex].
[tex]\[ z = \sqrt{x-1} \Rightarrow 2 = \sqrt{x-1} \Rightarrow 4 = x - 1 \Rightarrow x = 5 \][/tex]
Conclusion:
The solution to the system of equations is:
[tex]\[ (x, y) = (5, 0.4) \][/tex]
We have the following system of equations:
[tex]\[ \left\{ \begin{array}{l} 1. \quad 2 \sqrt{x-1} + \frac{3}{y} = \frac{23}{2} \\ 2. \quad 3 \sqrt{x-1} - \frac{2}{y} = 1 \end{array} \right. \][/tex]
Step 1: Let [tex]\( z = \sqrt{x - 1} \)[/tex].
This substitution simplifies our system:
[tex]\[ \left\{ \begin{array}{l} 1. \quad 2z + \frac{3}{y} = \frac{23}{2} \\ 2. \quad 3z - \frac{2}{y} = 1 \end{array} \right. \][/tex]
Step 2: Solve for [tex]\(\frac{1}{y}\)[/tex] from the second equation.
From the second equation:
[tex]\[ 3z - \frac{2}{y} = 1 \Rightarrow \frac{2}{y} = 3z - 1 \Rightarrow \frac{1}{y} = \frac{3z - 1}{2} \][/tex]
Step 3: Substitute [tex]\(\frac{1}{y}\)[/tex] into the first equation.
Substitute [tex]\(\frac{1}{y} = \frac{3z - 1}{2}\)[/tex] into the first equation:
[tex]\[ 2z + 3 \left(\frac{3z - 1}{2}\right) = \frac{23}{2} \][/tex]
Simplify:
[tex]\[ 2z + \frac{9z - 3}{2} = \frac{23}{2} \Rightarrow 2z + \frac{9z - 3}{2} = \frac{23}{2} \][/tex]
Step 4: Clear the fraction by multiplying through by 2.
[tex]\[ 2(2z) + 2 \left(\frac{9z - 3}{2}\right) = 2 \left(\frac{23}{2}\right) \Rightarrow 4z + 9z - 3 = 23 \Rightarrow 13z - 3 = 23 \][/tex]
Step 5: Solve for [tex]\( z \)[/tex].
[tex]\[ 13z - 3 = 23 \Rightarrow 13z = 26 \Rightarrow z = 2 \][/tex]
Step 6: Substitute [tex]\( z = 2 \)[/tex] back to find [tex]\(\frac{1}{y}\)[/tex].
From [tex]\(\frac{1}{y} = \frac{3z - 1}{2}\)[/tex]:
[tex]\[ \frac{1}{y} = \frac{3(2) - 1}{2} = \frac{6 - 1}{2} = \frac{5}{2} \Rightarrow y = \frac{2}{5} \][/tex]
Thus, [tex]\( y = 0.4 \)[/tex].
Step 7: Find x using [tex]\( z = 2 \)[/tex] and [tex]\( z = \sqrt{x-1} \)[/tex].
[tex]\[ z = \sqrt{x-1} \Rightarrow 2 = \sqrt{x-1} \Rightarrow 4 = x - 1 \Rightarrow x = 5 \][/tex]
Conclusion:
The solution to the system of equations is:
[tex]\[ (x, y) = (5, 0.4) \][/tex]
