Problem 49 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

(a) \( x=\sqrt{10}, y=10 \); (b) \( x=3, y=1 \); (c) Solve \( 3y^2 + 9y - 120 = 0 \) for \(y\); (d) \( x=3, y=-1 \).

Step 1: Simplify the System (a)

We are given two equations: \( \log x + \log y = \frac{3}{2} \) and \( 2 \log x - \log y = 0 \). Begin by adding these equations. Adding them gives:\[ (\log x + \log y) + (2 \log x - \log y) = \frac{3}{2} \]This simplifies to \( 3 \log x = \frac{3}{2} \).

Step 2: Solve for \(x\) (a)

From \( 3 \log x = \frac{3}{2} \), divide both sides by 3 to isolate \( \log x \):\[ \log x = \frac{1}{2} \]This implies \( x = 10^{\frac{1}{2}} = \sqrt{10} \).

Step 3: Find \(y\) using \( x \) (a)

Substitute \( \log x = \frac{1}{2} \) back into the first equation: \( \log y = \frac{3}{2} - \log x = \frac{3}{2} - \frac{1}{2} = 1 \). Thus, \( y = 10^1 = 10 \).

Step 1: Introduce Substitution (b)

Introduce substitution by letting \( a = 2^x \) and \( b = 2^y \). The system becomes:\( a + b = 10 \) and \( a^2 + b^2 = 68 \).

Step 2: Solve Simultaneous Equations (b)

Use the identity \( (a+b)^2 = a^2 + 2ab + b^2 \). Substitute:\[ 10^2 = 68 + 2ab \]\[ 100 = 68 + 2ab \]Solve for \( ab \): \[ 2ab = 32 \Rightarrow ab = 16 \].

Step 3: Use Quadratic Formula (b)

The values of \( a \) and \( b \) are roots of the equation: \( t^2 - 10t + 16 = 0 \).Use quadratic formula: \( t = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)\[ t = \frac{10 \pm \sqrt{36}}{2} \]\( t = 8 \) or \( t = 2 \). Thus, \( 2^x = 8 \) and \( 2^y = 2 \).

Step 4: Solve for \(x\) and \(y\) (b)

Since \( 2^x = 8 \), then \( x = 3 \). Similarly, \( 2^y = 2 \), so \( y = 1 \).

Step 1: Use Difference of Cubes (c)

Given equations are \( x - y = 3 \) and \( x^3 - y^3 = 387 \). Use the factorization: \[ x^3 - y^3 = (x-y)(x^2 + xy + y^2) \]Substituting \( x - y = 3 \), get:\[ 3(x^2 + xy + y^2) = 387 \].

Step 2: Solve for \(xy\) and Equations (c)

Divide by 3:\[ x^2 + xy + y^2 = 129 \]Substitute \( x = y + 3 \):\( (y+3)^2 + (y+3)y + y^2 = 129 \).

Step 3: Simplify and Solve (c)

Expand and simplify:\( 3y^2 + 6y + 9 + y^2 + 3y = 129 \)\( 3y^2 + 9y + 9 = 129 \)Simplifying gives:\( 3y^2 + 9y - 120 = 0 \)Solve the quadratic equation for \( y \).

Step 1: Combine and Factor Equations (d)

Given \( x^2 + xy = 1 \) and \( xy + y^2 = 3 \). Add both equations:\[ x^2 + 2xy + y^2 = 4 \].

Step 2: Factor Quadratic Form (d)

The equation represents \( (x+y)^2 = 4 \). This implies \( x+y = 2 \) or \( x+y = -2 \).

Step 3: Solve for \(x\) and \(y\) (d)

Substitute \( x = 2-y \) into original equations to find \( x = 2-y \) and after calculations, \( y = -1 \). Thus, \( x = 3 \).

Key Concepts

Logarithmic Equations

Logarithmic equations involve the logarithm functions and are essential in many mathematical problems. Simplifying a system of logarithmic equations often requires combining or transforming logarithmic expressions. For instance, in system (a), given the equations \( \log x + \log y = \frac{3}{2} \) and \( 2 \log x - \log y = 0 \), we use properties of logarithms such as the addition rule: \( \log a + \log b = \log(ab) \). Adding these equations can help isolate terms, enabling further simplification.

Once simplified, solving for a variable involves using the inverse of the logarithm, which is the exponential function. For example, \( \log x = \frac{1}{2} \) implies \( x = 10^{\frac{1}{2}} = \sqrt{10} \). This transformation from a logarithmic to an exponential form makes the equations easier to handle. Understanding these properties helps in manipulating and solving logarithmic equations effectively.

Exponential Equations

Exponential equations, like those in system (b), involve expressions where variables are the exponents of constants. Such systems often require strategic substitutions to simplify and solve. In the given problem, we used substitutions \( a = 2^x \) and \( b = 2^y \), converting the system into polynomial form: \( a + b = 10 \) and \( a^2 + b^2 = 68 \).

By using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), we can find critical relationships between expressions. Solving for \( ab \) was essential for cracking this system. With equations in simplified forms, use the quadratic formula to find actual values of \( a \) and \( b \). Subsequently, we equate these to powers of 2 and solve for \( x \) and \( y \). Mastering substitutions and transformations is crucial for successfully solving exponential equations.

Polynomial Equations

Polynomial equations such as those in system (c) can appear daunting due to their degree and the number of terms. However, they can often be simplified using algebraic identities like the difference of cubes. Given \( x - y = 3 \) and \( x^3 - y^3 = 387 \), using \( x^3 - y^3 = (x-y)(x^2 + xy + y^2) \) allows factoring to simplify the problem significantly.

Substituting known differences can reduce equations to a single variable, easily solvable by traditional means. In this situation, substituting \( x = y + 3 \) into a factored identity reduced a polynomial equation to a quadratic one, making it manageable. Recognizing when and how to apply such identities is valuable in reducing complexity in polynomial systems.

Quadratic Equations

Quadratic equations are among the most common algebraic equations students encounter. They often serve as tools for simplifying more complex equations like those found in system (d). For instance, given \( x^2 + xy = 1 \) and \( xy + y^2 = 3 \), adding these equations and recognizing it represents \( (x+y)^2 = 4 \) allows factoring into simpler linear equations.

This simplification means the quadratic can be expressed in terms of its sum or difference, i.e., \( x+y = 2 \) or \( x+y = -2 \). Once one expression is found, substitute back to find individual variable values. Solving these involves completing the square or using the quadratic formula. Mastering these tools helps you find solutions to virtually any quadratic equation, applying them to various algebraic contexts efficiently.