Problem 10 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether each ordered pair is a solution of the system of equations. $$\left\\{\begin{aligned} -\log _{10} x+3 &=y \\ \frac{1}{9} x+y &=\frac{28}{9} \end{aligned}\right.$$ (a) (100,1) (b) (10,2) (c) (1,3) (d) (1,1)

Step-by-Step Solution

Answer

Of the given ordered pairs, only (1,3) is a solution to the system of equations.

Step 1: Analyze the System of Equations

The provided system contains two equations: \(-\log_{10}x+3 = y\) and \(\frac{1}{9}x+y = \frac{28}{9}\). They are to be checked for certain ordered pairs (x, y). Remember that logarithm is the reverse process of exponentiation; hence \(-\log_{10} 100 = -2, -\log_{10}10 = -1, -\log_{10}1 = 0 \). Now, replace each pair into both equations to see if they yield true statements.

Step 2: Check Pair (100,1)

Substitute x=100, y=1 into the two equations. For the first equation, you get \(-\log_{10} 100 + 3 = -2 + 3 =1 \), which matches the y value of 1. For the second equation, you get \(\frac{1}{9}*100 + 1 = \frac{28}{9}\) which simplifies to \(\frac{100}{9} + 1 \neq \frac{28}{9}\), which is not true. Hence, (100,1) is not a solution.

Step 3: Check Pair (10,2)

Substitute x=10, y=2 into the two equations. For the first equation, you end up with \(-\log_{10}10 + 3 = -1 + 3 = 2 \), which matches the y value of 2. For the second equation, you get \(\frac{1}{9}*10 + 2 = \frac{28}{9}\), which simplifies to \(\frac{10}{9} + 2 \neq \frac{28}{9}\), which is untrue. Hence, (10,2) is not a solution.

Step 4: Check pair (1,3)

Substitute x=1, y=3 into the equations. For the first equation, you get \(-\log_{10}1+3 = 0 + 3 = 3\), which aligns with the y value of 3. In the second equation, you get \(\frac{1}{9}*1 + 3 = \frac{28}{9}\), which simplifies to \(\frac{1}{9} + 3 = \frac{28}{9}\). This is a true statement. Hence, (1,3) is a solution.

Step 5: Check pair (1,1)

Substitute x=1, y=1 into the equations. For the first equation, you end up with \(-\log_{10}1 + 3 = 0 + 3 = 3\), which does not match the y value. Hence, (1,1) is not a solution.

Key Concepts

Logarithmic Functions

Logarithms are a fundamental concept in algebra and calculus, often cropping up in equations where the variable is an exponent. A logarithmic function, like those in our exercise, acts as the inverse of an exponential function. In simpler terms, it helps us determine what exponent we need to get a certain result.

In the expression \(-\log_{10}x\), the logarithm calculates the power to which 10 must be raised to result in \(x\). For example:

• \(-\log_{10} 100\) yields \(-2\), since 10 squared equals 100.
• \(-\log_{10} 10\) yields \(-1\), as 10 to the power of 1 is 10.
• \(-\log_{10} 1\) equals 0, because 10 to the power of 0 equals 1.

Logarithms enable us to work through equations where the sought variable appears as a power, making them especially useful in solving systems of equations that involve exponential growth or decay.

Ordered Pairs

Ordered pairs represent solutions to systems of equations and consist of two elements: \(x\) and \(y\), usually written as \(x, y\). Each pair suggests a potential solution that could satisfy both equations in our system.

When determining if a pair such as (100,1) is a solution:

• First check both values in the system's first equation.
• Next verify the pair in the second equation of the system.
• If both equations hold true for the same pair, then it is a solution of the system.

In the exercise, only some ordered pairs will satisfy both equations simultaneously, highlighting the importance of systematic verification. This step ensures that no possible solution is overlooked.

Solution Verification

Verifying solutions in a system of equations is crucial to finding accurate results. This involves checking whether given ordered pairs make both equations true.

Here is a simple process to perform solution verification:

• Substitute the \(x\) and \(y\) values of the ordered pair into the first equation.
• Examine the outcome to see if it matches the result expected in the equation.
• Repeat this for the second equation using the same values of \(x\) and \(y\).
• If both verifications result in true statements, the ordered pair satisfies the system.

For example, the pair (1,3) was found to satisfy both equations in the given system, confirming its validity as a solution. This method ensures precision in identifying correct solutions and pairings.

Algebraic Substitution

Algebraic substitution is a straightforward yet powerful technique wherein you replace a variable with its equivalent numerical value or expression.

To apply substitution effectively:

• Identify the variable you need to substitute, often given directly in the ordered pair \((x, y)\).
• Replace \(x\) or \(y\) in the equation with its specific value from the pair.
• Solve the equation to see if it stands true with these replacements.

For instance, substituting \(x=10\) and \(y=2\) into the equation \(-\log_{10}x + 3 = y\) results in equality, verifying if it holds true for the system. Through smooth transitions between variable and numerical value, substitution facilitates checking potential solutions efficiently.