Answer:
Let's first solve for \(x\) and \(y\) using the given equations and then substitute those values to find the value of \(9x^2 + 4y^2\).
Given equations:
1. \(4x^2 + y^2 = 40\)
2. \(xy = 6\)
From equation 2, we can isolate \(y\):
\[y = \frac{6}{x}\]
Now, substitute the value of \(y\) from equation 2 into equation 1:
\[4x^2 + \left(\frac{6}{x}\right)^2 = 40\]
Simplify the equation:
\[4x^2 + \frac{36}{x^2} = 40\]
Multiply both sides by \(x^2\) to eliminate the fraction:
\[4x^4 + 36 = 40x^2\]
Rearrange the equation:
\[4x^4 - 40x^2 + 36 = 0\]
Factor the equation:
\[x^4 - 10x^2 + 9 = 0\]
\[(x^2 - 9)(x^2 - 1) = 0\]
This gives us two possible values for \(x\):
1. \(x^2 - 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3\)
2. \(x^2 - 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\)
Now that we have the possible values of \(x\), we can find the corresponding values of \(y\) using the equation \(xy = 6\).
For \(x = 3\):
\[y = \frac{6}{3} = 2\]
For \(x = -3\):
\[y = \frac{6}{-3} = -2\]
For \(x = 1\):
\[y = 6\]
For \(x = -1\):
\[y = -6\]
Now, substitute each pair of \(x\) and \(y\) values into the expression \(9x^2 + 4y^2\) to find the value in each case:
1. For \(x = 3\) and \(y = 2\):
\[9x^2 + 4y^2 = 9(3^2) + 4(2^2) = 81 + 16 = 97\]
2. For \(x = -3\) and \(y = -2\):
\[9x^2 + 4y^2 = 9(-3^2) + 4(-2^2) = 81 + 16 = 97\]
3. For \(x = 1\) and \(y = 6\):
\[9x^2 + 4y^2 = 9(1^2) + 4(6^2) = 9 + 144 = 153\]
4. For \(x = -1\) and \(y = -6\):
\[9x^2 + 4y^2 = 9(-1^2) + 4(-6^2) = 9 + 144 = 153\]
Therefore, the possible values for \(9x^2 + 4y^2\) are 97 and 153, depending on the combinations of \(x\) and \(y\) you choose.
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Answers & Comments
Answer:
Let's first solve for \(x\) and \(y\) using the given equations and then substitute those values to find the value of \(9x^2 + 4y^2\).
Given equations:
1. \(4x^2 + y^2 = 40\)
2. \(xy = 6\)
From equation 2, we can isolate \(y\):
\[y = \frac{6}{x}\]
Now, substitute the value of \(y\) from equation 2 into equation 1:
\[4x^2 + \left(\frac{6}{x}\right)^2 = 40\]
Simplify the equation:
\[4x^2 + \frac{36}{x^2} = 40\]
Multiply both sides by \(x^2\) to eliminate the fraction:
\[4x^4 + 36 = 40x^2\]
Rearrange the equation:
\[4x^4 - 40x^2 + 36 = 0\]
Factor the equation:
\[x^4 - 10x^2 + 9 = 0\]
\[(x^2 - 9)(x^2 - 1) = 0\]
This gives us two possible values for \(x\):
1. \(x^2 - 9 = 0 \Rightarrow x^2 = 9 \Rightarrow x = \pm 3\)
2. \(x^2 - 1 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1\)
Now that we have the possible values of \(x\), we can find the corresponding values of \(y\) using the equation \(xy = 6\).
For \(x = 3\):
\[y = \frac{6}{3} = 2\]
For \(x = -3\):
\[y = \frac{6}{-3} = -2\]
For \(x = 1\):
\[y = 6\]
For \(x = -1\):
\[y = -6\]
Now, substitute each pair of \(x\) and \(y\) values into the expression \(9x^2 + 4y^2\) to find the value in each case:
1. For \(x = 3\) and \(y = 2\):
\[9x^2 + 4y^2 = 9(3^2) + 4(2^2) = 81 + 16 = 97\]
2. For \(x = -3\) and \(y = -2\):
\[9x^2 + 4y^2 = 9(-3^2) + 4(-2^2) = 81 + 16 = 97\]
3. For \(x = 1\) and \(y = 6\):
\[9x^2 + 4y^2 = 9(1^2) + 4(6^2) = 9 + 144 = 153\]
4. For \(x = -1\) and \(y = -6\):
\[9x^2 + 4y^2 = 9(-1^2) + 4(-6^2) = 9 + 144 = 153\]
Therefore, the possible values for \(9x^2 + 4y^2\) are 97 and 153, depending on the combinations of \(x\) and \(y\) you choose.