Problem 49 - Precalculus Mathematics for Calculus

Question

Follow the hints and solve the systems. (a) \(\left\\{\begin{array}{c}\log x+\log y=\frac{3}{2} \\ 2 \log x-\log y=0\end{array}\right.\) \([\text {Hint: Add the equations.}]\) (b) \(\left\\{\begin{array}{l}2^{x}+2^{y}=10 \\\ 4^{x}+4^{y}=68\end{array}\right.\) \(\left[\text {Hint: Note that } 4^{x}=2^{2 x}=\left(2^{x}\right)^{2}\right]\) (c) \(\left\\{\begin{array}{c}x-y=3 \\ x^{3}-y^{3}=387\end{array}\right.\) [Hint: Factor the left-hand side of the second equation. \(]\) (d) \(\left\\{\begin{array}{l}x^{2}+x y=1 \\ x y+y^{2}=3\end{array}\right.\) [Hint: Add the equations, and factor the result.]

Step-by-Step Solution

Answer

Solutions: (a) (3.17, 10.05), (b) (3, 1) or (1, 3), (c) (9, 6), (d) (1,1).

Step 1: Solve System (a): Express Logarithms in Terms of Variable

Rewrite the first equation using the property of logarithms: \( \log x + \log y = \log(xy) = \frac{3}{2} \). Therefore, \( xy = 10^{\frac{3}{2}} = \sqrt{1000} \). Simplifying gives \( xy = 31.622 \).

Step 2: Substitute and Solve for y in Terms of x

From the second equation, \( 2 \log x = \log y \), hence \( y = x^2 \).

Step 3: Combine and Solve Equations

Substitute \( y = x^2 \) into \( xy = 31.622 \), producing \( x(x^2) = 31.622 \). This becomes \( x^3 = 31.622 \), so \( x = \sqrt[3]{31.622} \).

Step 4: Complete Solution for System (a)

Calculate \( x \) to find \( x \approx 3.17 \). Substitute back to find \( y = (3.17)^2 \approx 10.05 \). The solution is \( (x, y) \approx (3.17, 10.05) \).

Step 5: Solve System (b): Transform to a Single Variable

Rewrite the second equation \( 4^x + 4^y = 68 \) as \((2^x)^2 + (2^y)^2 = 68 \). Let \( a = 2^x \) and \( b = 2^y \), transforming the equations to \( a + b = 10 \) and \( a^2 + b^2 = 68 \).

Step 6: Express and Solve Using Substitution

Substitute \( b = 10 - a \) back into \( a^2 + (10-a)^2 = 68 \). Simplify the equation: \( a^2 + (100 - 20a + a^2) = 68 \), giving \( 2a^2 - 20a + 32 = 0 \). Simplify further and solve \( a^2 - 10a + 16 = 0 \).

Step 7: Solve the Quadratic Equation

Using the quadratic formula, \( a = \frac{10 \pm \sqrt{100 - 64}}{2} = \frac{10 \pm \sqrt{36}}{2} = \frac{10 \pm 6}{2} \). Thus, \( a = 8 \) or \( a = 2 \).

Step 8: Find Variable Values for System (b)

When \( a = 8 \), \( b = 2 \) and vice-versa. Back-substitute to find \( (x, y) = (3, 1) \) or \( (1, 3) \).

Step 9: Solve System (c): Simplify Using Factoring

Rewrite the second equation \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) = 387 \). Given \( x-y = 3 \), substitute and simplify: \( 3(x^2 + xy + y^2) = 387 \), giving \( x^2 + xy + y^2 = 129 \).

Step 10: Substitution from First into Second Equation

From \( x-y = 3 \), express \( x = y + 3 \) and substitute into \( x^2 + xy + y^2 = 129 \), resulting in equation \( (y+3)^2 + (y+3)y + y^2 = 129 \).

Step 11: Solve for y in System (c)

The expression simplifies to \( y^2+6y+9+y^2+3y+y^2 = 129 \) or \( 3y^2+9y-120=0 \). Use quadratic formula to solve: \( y = \frac{-9 \pm \sqrt{81+(4 \times 3 \times 120)}}{6} \).

Step 12: Compute Variables for System (c)

Solving provides \( y = 6 \) or \( y = -9.33 \). Since \( x-y=3 \), if \( y=6 \) then \( x=9 \).

Step 13: Solve System (d): Combine Equations

Add up the equations: \( x^2 + xy + xy + y^2 = 1+3 \), thus \( x^2 + 2xy + y^2 = 4 \).

Step 14: Factor the New Equation

Notice that \( x^2 + 2xy + y^2 = (x+y)^2 \). Thus, \((x+y)^2 = 4\), giving \(x+y = 2\) or \(x+y = -2\).

Step 15: Solve for the Variables for System (d)

With \( x+y=2 \), substitute back into \( x^2 + xy = 1 \); solving the system, we get \( (x,y) = (1,1) \). The factors will provide another possible set of solutions assuming valid roots.

Key Concepts

Logarithmic Equations

Logarithmic equations involve the logarithm of an unknown variable. A fundamental property of logarithms is that they convert multiplication into addition, which is very useful when solving systems of equations. For example, the equation \( \log x + \log y = \log(xy) \) simplifies the process of finding the product of \( x \) and \( y \) if the values of their logs are known.
When faced with a system of logarithmic equations, you can often combine them by applying rules such as the product rule \( \log a + \log b = \log(ab) \) and the power rule \( b \log a = \log(a^b) \).
In the provided problem, once the logarithmic equations are expressed in terms of a single variable, you can solve for the unknowns typically through substitution. This approach simplifies complex log equations into manageable algebraic forms.

Exponential Equations

Exponential equations are scenarios in which variables appear as exponents. These equations often arise in real-world applications such as population growth or radioactive decay.
In mathematical terms, when you see expressions like \( 2^x \) or \( 4^x \), it's important to recognize their potential for simplification. Knowing that \( 4^x = (2^x)^2 \) can be crucial because it allows the equation to be rewritten in terms of a single base.
By setting variables equal to a common exponent base, like substituting \( a = 2^x \), exponential systems can be translated into polynomial equations which are easier to solve. This conversion is vital for system (b) where leveraging this method helps isolate variables efficiently. Simplifying every step helps prevent errors, especially with complex calculations.

Factoring Polynomials

Factoring polynomials is a method used to solve equations where variables are raised to powers. The process involves expressing a polynomial as a product of simpler polynomials or factors. This is particularly useful when you have a polynomial equation and need to solve for the variable.
For instance, in system (c), the equation \( x^3 - y^3 = 387 \) can be expressed using the factoring formula for the difference of cubes: \( x^3 - y^3 = (x - y)(x^2 + xy + y^2) \). This not only simplifies computation but helps in aligning the equation to match known variable terms, allowing for easier manipulation.
Once factored, substitution from other equations in the system becomes straightforward. Solving the remaining simpler equations after substitution efficiently provides the required solutions.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
In contexts involving systems of equations, quadratic equations often arise as intermediate steps. Knowing how to apply the quadratic formula can help find roots efficiently. For example, in system (d), factorization led to an expression resembling a quadratic equation, allowing further simplification using the quadratic formula.
It's crucial to first bring the equation into the standard quadratic form before applying the formula. Checking the discriminant \( b^2 - 4ac \) helps determine the nature of the roots. Whether solutions are real or complex impacts the interpretation of the original problem's variables.