Answer :
To solve the problem where we need to determine the value of [tex]\(3x + 5y\)[/tex] given the system of equations:
1. [tex]\(9x^2 + 25y^2 = 181\)[/tex]
2. [tex]\(xy = -6\)[/tex]
we can proceed with the following steps:
### Step 1: Substitution Strategy
To solve the system of equations, consider solving for one variable in terms of the other from the second equation and substituting it into the first equation.
From [tex]\(xy = -6\)[/tex]:
[tex]\[ y = -\frac{6}{x} \][/tex]
### Step 2: Substitute [tex]\(y\)[/tex] into the First Equation
Substitute [tex]\( y = -\frac{6}{x} \)[/tex] into the first equation [tex]\(9x^2 + 25y^2 = 181\)[/tex]:
[tex]\[ 9x^2 + 25\left(-\frac{6}{x}\right)^2 = 181 \][/tex]
[tex]\[ 9x^2 + 25\frac{36}{x^2} = 181 \][/tex]
[tex]\[ 9x^4 + 25 \cdot 36 = 181x^2 \][/tex]
[tex]\[ 9x^4 + 900 = 181x^2 \][/tex]
### Step 3: Rearrange the Equation
Rearrange the equation to form a standard polynomial:
[tex]\[ 9x^4 - 181x^2 + 900 = 0 \][/tex]
Let [tex]\( z = x^2 \)[/tex]. Substitute [tex]\(z\)[/tex] into the polynomial equation:
[tex]\[ 9z^2 - 181z + 900 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Solve for [tex]\(z\)[/tex] using the quadratic formula [tex]\( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with [tex]\( a = 9 \)[/tex], [tex]\( b = -181 \)[/tex], and [tex]\( c = 900 \)[/tex]:
[tex]\[ z = \frac{181 \pm \sqrt{(-181)^2 - 4 \cdot 9 \cdot 900}}{2 \cdot 9} \][/tex]
[tex]\[ z = \frac{181 \pm \sqrt{32761 - 32400}}{18} \][/tex]
[tex]\[ z = \frac{181 \pm \sqrt{361}}{18} \][/tex]
[tex]\[ z = \frac{181 \pm 19}{18} \][/tex]
This gives us two solutions for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{200}{18} = \frac{100}{9} \quad \text{and} \quad z = \frac{162}{18} = 9 \][/tex]
So, the possible values for [tex]\(x^2\)[/tex] are [tex]\( \frac{100}{9} \)[/tex] and [tex]\( 9 \)[/tex].
### Step 5: Find [tex]\(x\)[/tex] and Correspondingly [tex]\(y\)[/tex]
For each [tex]\(z\)[/tex]:
[tex]\[ \text{If } x^2 = \frac{100}{9} \Rightarrow x = \pm \frac{10}{3} \][/tex]
[tex]\[ \text{If } x = \frac{10}{3}, y = -\frac{6}{\frac{10}{3}} = -\frac{9}{5} \][/tex]
[tex]\[ \text{If } x = -\frac{10}{3}, y = -\frac{6}{-\frac{10}{3}} = \frac{9}{5} \][/tex]
[tex]\[ \text{If } x^2 = 9 \Rightarrow x = \pm 3 \][/tex]
[tex]\[ \text{If } x = 3, y = -\frac{6}{3} = -2 \][/tex]
[tex]\[ \text{If } x = -3, y = -\frac{6}{-3} = 2 \][/tex]
Thus, the pairs [tex]\((x, y)\)[/tex] that satisfy both equations are:
1. [tex]\(\left(\frac{10}{3}, -\frac{9}{5}\right)\)[/tex]
2. [tex]\(\left(-\frac{10}{3}, \frac{9}{5}\right)\)[/tex]
3. [tex]\((3, -2)\)[/tex]
4. [tex]\((-3, 2)\)[/tex]
### Step 6: Calculate [tex]\(3x + 5y\)[/tex]
We now compute [tex]\(3x + 5y\)[/tex] for each pair:
For [tex]\(\left(\frac{10}{3}, -\frac{9}{5}\right)\)[/tex]:
[tex]\[ 3 \left(\frac{10}{3}\right) + 5 \left(-\frac{9}{5}\right) = 10 - 9 = 1 \][/tex]
For [tex]\(\left(-\frac{10}{3}, \frac{9}{5}\right)\)[/tex]:
[tex]\[ 3 \left(-\frac{10}{3}\right) + 5 \left(\frac{9}{5}\right) = -10 + 9 = -1 \][/tex]
For [tex]\((3, -2)\)[/tex]:
[tex]\[ 3(3) + 5(-2) = 9 - 10 = -1 \][/tex]
For [tex]\((-3, 2)\)[/tex]:
[tex]\[ 3(-3) + 5(2) = -9 + 10 = 1 \][/tex]
Therefore, the values of [tex]\(3x + 5y\)[/tex] given the constraints are [tex]\(-1\)[/tex] and [tex]\(1\)[/tex].
1. [tex]\(9x^2 + 25y^2 = 181\)[/tex]
2. [tex]\(xy = -6\)[/tex]
we can proceed with the following steps:
### Step 1: Substitution Strategy
To solve the system of equations, consider solving for one variable in terms of the other from the second equation and substituting it into the first equation.
From [tex]\(xy = -6\)[/tex]:
[tex]\[ y = -\frac{6}{x} \][/tex]
### Step 2: Substitute [tex]\(y\)[/tex] into the First Equation
Substitute [tex]\( y = -\frac{6}{x} \)[/tex] into the first equation [tex]\(9x^2 + 25y^2 = 181\)[/tex]:
[tex]\[ 9x^2 + 25\left(-\frac{6}{x}\right)^2 = 181 \][/tex]
[tex]\[ 9x^2 + 25\frac{36}{x^2} = 181 \][/tex]
[tex]\[ 9x^4 + 25 \cdot 36 = 181x^2 \][/tex]
[tex]\[ 9x^4 + 900 = 181x^2 \][/tex]
### Step 3: Rearrange the Equation
Rearrange the equation to form a standard polynomial:
[tex]\[ 9x^4 - 181x^2 + 900 = 0 \][/tex]
Let [tex]\( z = x^2 \)[/tex]. Substitute [tex]\(z\)[/tex] into the polynomial equation:
[tex]\[ 9z^2 - 181z + 900 = 0 \][/tex]
### Step 4: Solve the Quadratic Equation
Solve for [tex]\(z\)[/tex] using the quadratic formula [tex]\( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] with [tex]\( a = 9 \)[/tex], [tex]\( b = -181 \)[/tex], and [tex]\( c = 900 \)[/tex]:
[tex]\[ z = \frac{181 \pm \sqrt{(-181)^2 - 4 \cdot 9 \cdot 900}}{2 \cdot 9} \][/tex]
[tex]\[ z = \frac{181 \pm \sqrt{32761 - 32400}}{18} \][/tex]
[tex]\[ z = \frac{181 \pm \sqrt{361}}{18} \][/tex]
[tex]\[ z = \frac{181 \pm 19}{18} \][/tex]
This gives us two solutions for [tex]\(z\)[/tex]:
[tex]\[ z = \frac{200}{18} = \frac{100}{9} \quad \text{and} \quad z = \frac{162}{18} = 9 \][/tex]
So, the possible values for [tex]\(x^2\)[/tex] are [tex]\( \frac{100}{9} \)[/tex] and [tex]\( 9 \)[/tex].
### Step 5: Find [tex]\(x\)[/tex] and Correspondingly [tex]\(y\)[/tex]
For each [tex]\(z\)[/tex]:
[tex]\[ \text{If } x^2 = \frac{100}{9} \Rightarrow x = \pm \frac{10}{3} \][/tex]
[tex]\[ \text{If } x = \frac{10}{3}, y = -\frac{6}{\frac{10}{3}} = -\frac{9}{5} \][/tex]
[tex]\[ \text{If } x = -\frac{10}{3}, y = -\frac{6}{-\frac{10}{3}} = \frac{9}{5} \][/tex]
[tex]\[ \text{If } x^2 = 9 \Rightarrow x = \pm 3 \][/tex]
[tex]\[ \text{If } x = 3, y = -\frac{6}{3} = -2 \][/tex]
[tex]\[ \text{If } x = -3, y = -\frac{6}{-3} = 2 \][/tex]
Thus, the pairs [tex]\((x, y)\)[/tex] that satisfy both equations are:
1. [tex]\(\left(\frac{10}{3}, -\frac{9}{5}\right)\)[/tex]
2. [tex]\(\left(-\frac{10}{3}, \frac{9}{5}\right)\)[/tex]
3. [tex]\((3, -2)\)[/tex]
4. [tex]\((-3, 2)\)[/tex]
### Step 6: Calculate [tex]\(3x + 5y\)[/tex]
We now compute [tex]\(3x + 5y\)[/tex] for each pair:
For [tex]\(\left(\frac{10}{3}, -\frac{9}{5}\right)\)[/tex]:
[tex]\[ 3 \left(\frac{10}{3}\right) + 5 \left(-\frac{9}{5}\right) = 10 - 9 = 1 \][/tex]
For [tex]\(\left(-\frac{10}{3}, \frac{9}{5}\right)\)[/tex]:
[tex]\[ 3 \left(-\frac{10}{3}\right) + 5 \left(\frac{9}{5}\right) = -10 + 9 = -1 \][/tex]
For [tex]\((3, -2)\)[/tex]:
[tex]\[ 3(3) + 5(-2) = 9 - 10 = -1 \][/tex]
For [tex]\((-3, 2)\)[/tex]:
[tex]\[ 3(-3) + 5(2) = -9 + 10 = 1 \][/tex]
Therefore, the values of [tex]\(3x + 5y\)[/tex] given the constraints are [tex]\(-1\)[/tex] and [tex]\(1\)[/tex].
