Answer :
Sure, let's solve these equations step-by-step using the Monge method.
### Given Equations:
1. [tex]\( 3s + (rt - s^2) = 2 \)[/tex]
2. [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]
### Equation (i): [tex]\( 3s + (rt - s^2) = 2 \)[/tex]
Let's start by solving for [tex]\( s \)[/tex] in the first equation:
[tex]\[ 3s + rt - s^2 = 2 \][/tex]
Rearrange the terms to form a quadratic equation:
[tex]\[ s^2 - 3s + 2 - rt = 0 \][/tex]
Now, this is a standard quadratic equation in [tex]\( s \)[/tex] of the form [tex]\( As^2 + Bs + C = 0 \)[/tex], where:
[tex]\[ A = 1 \][/tex]
[tex]\[ B = -3 \][/tex]
[tex]\[ C = 2 - rt \][/tex]
The solutions to the quadratic equation [tex]\( As^2 + Bs + C = 0 \)[/tex] are given by the quadratic formula:
[tex]\[ s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Substitute the values of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]:
[tex]\[ s = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (2 - rt)}}{2 \cdot 1} \][/tex]
[tex]\[ s = \frac{3 \pm \sqrt{9 - 8 + 4rt}}{2} \][/tex]
[tex]\[ s = \frac{3 \pm \sqrt{1 + 4rt}}{2} \][/tex]
Therefore, the solutions for [tex]\( s \)[/tex] in the first equation are:
[tex]\[ s_1 = \frac{3 + \sqrt{1 + 4rt}}{2} \][/tex]
[tex]\[ s_2 = \frac{3 - \sqrt{1 + 4rt}}{2} \][/tex]
Given specific values [tex]\( r = 1 \)[/tex] and [tex]\( t = 1 \)[/tex]:
[tex]\[ s_1 = \frac{3 + \sqrt{1 + 4 \cdot 1 \cdot 1}}{2} = \frac{3 + \sqrt{5}}{2} \approx 2.618 \][/tex]
[tex]\[ s_2 = \frac{3 - \sqrt{1 + 4 \cdot 1 \cdot 1}}{2} = \frac{3 - \sqrt{5}}{2} \approx 0.382 \][/tex]
### Equation (ii): [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]
Next, let's solve for [tex]\( s \)[/tex] in the second equation:
[tex]\[ 3s - 2(rt - s^2) = 2 \][/tex]
Rearrange the terms to form a quadratic equation:
[tex]\[ 3s - 2rt + 2s^2 = 2 \][/tex]
[tex]\[ 2s^2 + 3s - 2rt = 2 \][/tex]
[tex]\[ 2s^2 + 3s - 2(rt + 1) = 0 \][/tex]
This is a quadratic equation in [tex]\( s \)[/tex] of the form [tex]\( As^2 + Bs + C = 0 \)[/tex], where:
[tex]\[ A = 2 \][/tex]
[tex]\[ B = 3 \][/tex]
[tex]\[ C = -2(rt + 1) \][/tex]
The solutions to the quadratic equation [tex]\( As^2 + Bs + C = 0 \)[/tex] are given by:
[tex]\[ s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Substitute the values of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]:
[tex]\[ s = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2(rt + 1))}}{2 \cdot 2} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{9 + 16(rt + 1)}}{4} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{9 + 16rt + 16}}{4} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{25 + 16rt}}{4} \][/tex]
Therefore, the solutions for [tex]\( s \)[/tex] in the second equation are:
[tex]\[ s_1 = \frac{-3 + \sqrt{25 + 16rt}}{4} \][/tex]
[tex]\[ s_2 = \frac{-3 - \sqrt{25 + 16rt}}{4} \][/tex]
Given the specific values [tex]\( r = 1 \)[/tex] and [tex]\( t = 1 \)[/tex]:
[tex]\[ s_1 = \frac{-3 + \sqrt{25 + 16 \cdot 1 \cdot 1}}{4} = \frac{-3 + \sqrt{41}}{4} \approx 0.851 \][/tex]
[tex]\[ s_2 = \frac{-3 - \sqrt{25 + 16 \cdot 1 \cdot 1}}{4} = \frac{-3 - \sqrt{41}}{4} \approx -2.351 \][/tex]
### Summary of Solutions
For the given equations, the solutions for [tex]\( s \)[/tex] are:
1. [tex]\( 3s + (rt - s^2) = 2 \)[/tex]:
[tex]\( s_1 \approx 2.618 \)[/tex]
[tex]\( s_2 \approx 0.382 \)[/tex]
2. [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]:
[tex]\( s_1 \approx 0.851 \)[/tex]
[tex]\( s_2 \approx -2.351 \)[/tex]
### Given Equations:
1. [tex]\( 3s + (rt - s^2) = 2 \)[/tex]
2. [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]
### Equation (i): [tex]\( 3s + (rt - s^2) = 2 \)[/tex]
Let's start by solving for [tex]\( s \)[/tex] in the first equation:
[tex]\[ 3s + rt - s^2 = 2 \][/tex]
Rearrange the terms to form a quadratic equation:
[tex]\[ s^2 - 3s + 2 - rt = 0 \][/tex]
Now, this is a standard quadratic equation in [tex]\( s \)[/tex] of the form [tex]\( As^2 + Bs + C = 0 \)[/tex], where:
[tex]\[ A = 1 \][/tex]
[tex]\[ B = -3 \][/tex]
[tex]\[ C = 2 - rt \][/tex]
The solutions to the quadratic equation [tex]\( As^2 + Bs + C = 0 \)[/tex] are given by the quadratic formula:
[tex]\[ s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Substitute the values of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]:
[tex]\[ s = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (2 - rt)}}{2 \cdot 1} \][/tex]
[tex]\[ s = \frac{3 \pm \sqrt{9 - 8 + 4rt}}{2} \][/tex]
[tex]\[ s = \frac{3 \pm \sqrt{1 + 4rt}}{2} \][/tex]
Therefore, the solutions for [tex]\( s \)[/tex] in the first equation are:
[tex]\[ s_1 = \frac{3 + \sqrt{1 + 4rt}}{2} \][/tex]
[tex]\[ s_2 = \frac{3 - \sqrt{1 + 4rt}}{2} \][/tex]
Given specific values [tex]\( r = 1 \)[/tex] and [tex]\( t = 1 \)[/tex]:
[tex]\[ s_1 = \frac{3 + \sqrt{1 + 4 \cdot 1 \cdot 1}}{2} = \frac{3 + \sqrt{5}}{2} \approx 2.618 \][/tex]
[tex]\[ s_2 = \frac{3 - \sqrt{1 + 4 \cdot 1 \cdot 1}}{2} = \frac{3 - \sqrt{5}}{2} \approx 0.382 \][/tex]
### Equation (ii): [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]
Next, let's solve for [tex]\( s \)[/tex] in the second equation:
[tex]\[ 3s - 2(rt - s^2) = 2 \][/tex]
Rearrange the terms to form a quadratic equation:
[tex]\[ 3s - 2rt + 2s^2 = 2 \][/tex]
[tex]\[ 2s^2 + 3s - 2rt = 2 \][/tex]
[tex]\[ 2s^2 + 3s - 2(rt + 1) = 0 \][/tex]
This is a quadratic equation in [tex]\( s \)[/tex] of the form [tex]\( As^2 + Bs + C = 0 \)[/tex], where:
[tex]\[ A = 2 \][/tex]
[tex]\[ B = 3 \][/tex]
[tex]\[ C = -2(rt + 1) \][/tex]
The solutions to the quadratic equation [tex]\( As^2 + Bs + C = 0 \)[/tex] are given by:
[tex]\[ s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \][/tex]
Substitute the values of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]:
[tex]\[ s = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2(rt + 1))}}{2 \cdot 2} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{9 + 16(rt + 1)}}{4} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{9 + 16rt + 16}}{4} \][/tex]
[tex]\[ s = \frac{-3 \pm \sqrt{25 + 16rt}}{4} \][/tex]
Therefore, the solutions for [tex]\( s \)[/tex] in the second equation are:
[tex]\[ s_1 = \frac{-3 + \sqrt{25 + 16rt}}{4} \][/tex]
[tex]\[ s_2 = \frac{-3 - \sqrt{25 + 16rt}}{4} \][/tex]
Given the specific values [tex]\( r = 1 \)[/tex] and [tex]\( t = 1 \)[/tex]:
[tex]\[ s_1 = \frac{-3 + \sqrt{25 + 16 \cdot 1 \cdot 1}}{4} = \frac{-3 + \sqrt{41}}{4} \approx 0.851 \][/tex]
[tex]\[ s_2 = \frac{-3 - \sqrt{25 + 16 \cdot 1 \cdot 1}}{4} = \frac{-3 - \sqrt{41}}{4} \approx -2.351 \][/tex]
### Summary of Solutions
For the given equations, the solutions for [tex]\( s \)[/tex] are:
1. [tex]\( 3s + (rt - s^2) = 2 \)[/tex]:
[tex]\( s_1 \approx 2.618 \)[/tex]
[tex]\( s_2 \approx 0.382 \)[/tex]
2. [tex]\( 3s - 2(rt - s^2) = 2 \)[/tex]:
[tex]\( s_1 \approx 0.851 \)[/tex]
[tex]\( s_2 \approx -2.351 \)[/tex]
