Connect with knowledgeable individuals and find the best answers at IDNLearn.com. Our community is ready to provide in-depth answers and practical solutions to any questions you may have.
Sagot :
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].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. IDNLearn.com is your reliable source for answers. We appreciate your visit and look forward to assisting you again soon.
