If x 2 + 4y 2 = 40, xy = 6 and x 2y then the value of x - 2y is:

Question

If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:

Prepp Team · Accepted Answer

Let's analyze the given algebraic problem. We are provided with two equations and an inequality:

Equation 1: \(x^2 + 4y^2 = 40\)
Equation 2: \(xy = 6\)
Inequality: \(x > 2y\)

Our goal is to find the value of the expression \(x - 2y\).

Finding the Value of x - 2y using Algebraic Identities

We can relate the expression \(x - 2y\) to the given equations by considering the square of \(x - 2y\). Let's expand \((x - 2y)^2\):

\((x - 2y)^2 = x^2 - 2(x)(2y) + (2y)^2\)

\((x - 2y)^2 = x^2 - 4xy + 4y^2\)

Now, we can rearrange the terms to group \(x^2\) and \(4y^2\):

\((x - 2y)^2 = (x^2 + 4y^2) - 4xy\)

Substituting Given Values into the Identity

We are given that \(x^2 + 4y^2 = 40\) and \(xy = 6\). We can substitute these values into the expanded identity:

\((x - 2y)^2 = (40) - 4(6)\)

\((x - 2y)^2 = 40 - 24\)

\((x - 2y)^2 = 16\)

Determining the Sign of x - 2y

To find the value of \(x - 2y\), we need to take the square root of both sides of the equation:

\(x - 2y = \pm \sqrt{16}\)

\(x - 2y = \pm 4\)

This gives us two possible values for \(x - 2y\): \(+4\) or \(-4\). To determine the correct value, we use the given inequality:

\(x > 2y\)

This inequality can be rewritten as:

\(x - 2y > 0\)

Since \(x - 2y\) must be greater than 0, the value of \(x - 2y\) must be positive.

Therefore, \(x - 2y = 4\).

Let's check the options provided:

Option	Value
1	1
2	3
3	2
4	4

The calculated value \(x - 2y = 4\) matches Option 4.

Step-by-Step Solution Summary

Start with the expression to be evaluated: \(x - 2y\).
Consider the square of the expression: \((x - 2y)^2\).
Expand the square using the identity: \((a-b)^2 = a^2 - 2ab + b^2\). This gives \((x - 2y)^2 = x^2 - 4xy + 4y^2\).
Rearrange the terms to group \(x^2\) and \(4y^2\): \((x - 2y)^2 = (x^2 + 4y^2) - 4xy\).
Substitute the given values \(x^2 + 4y^2 = 40\) and \(xy = 6\) into the equation: \((x - 2y)^2 = 40 - 4(6)\).
Simplify the equation: \((x - 2y)^2 = 40 - 24 = 16\).
Take the square root of both sides: \(x - 2y = \pm \sqrt{16} = \pm 4\).
Use the given inequality \(x > 2y\), which implies \(x - 2y > 0\), to select the positive value.
The value of \(x - 2y\) is 4.

Revision Table: Key Concepts

Concept	Description	Application in Problem
Algebraic Identity	A mathematical equation that is true for all possible values of the variables it contains. Example: \((a-b)^2 = a^2 - 2ab + b^2\).	Used to relate the target expression \((x-2y)^2\) to the given expressions \(x^2 + 4y^2\) and \(xy\).
Solving Equations	Finding the value(s) of variables that satisfy an equation.	Used to find the value of \((x-2y)^2\) and subsequently \(x-2y\).
Inequalities	Mathematical statements that compare two expressions using symbols like >, <, ≥, ≤.	Used to determine the correct sign (positive or negative) for the square root result.

Additional Information: Solving Systems of Equations

This problem could also potentially be solved by finding the actual values of x and y first. We have the system of equations:

\(x^2 + 4y^2 = 40\)
\(xy = 6\)

From equation 2, we can express \(x\) in terms of \(y\) (or vice versa): \(x = \frac{6}{y}\). Note that \(y\) cannot be zero since \(xy=6\).

Substitute this into equation 1:

\(\left(\frac{6}{y}\right)^2 + 4y^2 = 40\)

\(\frac{36}{y^2} + 4y^2 = 40\)

Multiply by \(y^2\) to clear the denominator (assuming \(y^2 \neq 0\)):

\(36 + 4y^4 = 40y^2\)

Rearrange into a quadratic in \(y^2\):

\(4y^4 - 40y^2 + 36 = 0\)

Divide by 4:

\(y^4 - 10y^2 + 9 = 0\)

Let \(z = y^2\). The equation becomes a quadratic in \(z\):

\(z^2 - 10z + 9 = 0\)

Factor the quadratic:

\((z - 1)(z - 9) = 0\)

So, \(z = 1\) or \(z = 9\).

Since \(z = y^2\), we have \(y^2 = 1\) or \(y^2 = 9\).

Case 1: \(y^2 = 1 \implies y = \pm 1\).

If \(y = 1\), then \(x = \frac{6}{y} = \frac{6}{1} = 6\). Check inequality \(x > 2y\): \(6 > 2(1)\), \(6 > 2\). This solution \((x, y) = (6, 1)\) is valid. \(x - 2y = 6 - 2(1) = 6 - 2 = 4\).
If \(y = -1\), then \(x = \frac{6}{y} = \frac{6}{-1} = -6\). Check inequality \(x > 2y\): \(-6 > 2(-1)\), \(-6 > -2\). This is false. So \((x, y) = (-6, -1)\) is not valid.

Case 2: \(y^2 = 9 \implies y = \pm 3\).

If \(y = 3\), then \(x = \frac{6}{y} = \frac{6}{3} = 2\). Check inequality \(x > 2y\): \(2 > 2(3)\), \(2 > 6\). This is false. So \((x, y) = (2, 3)\) is not valid.
If \(y = -3\), then \(x = \frac{6}{y} = \frac{6}{-3} = -2\). Check inequality \(x > 2y\): \(-2 > 2(-3)\), \(-2 > -6\). This is true. So \((x, y) = (-2, -3)\) is also a valid solution pair that satisfies all conditions. Let's calculate \(x - 2y\) for this pair: \(x - 2y = -2 - 2(-3) = -2 + 6 = 4\).

Both valid pairs \((6, 1)\) and \((-2, -3)\) result in \(x - 2y = 4\). This confirms the result obtained using the algebraic identity method, which was much quicker for this specific problem.