Problem 6 - Precalculus: Functions and Graphs

Question

Solve the system. $$\left\\{\begin{array}{l} 9 u+2 v=0 \\ 3 u-5 v=17 \end{array}\right.$$

Step-by-Step Solution

Answer

\( u = \frac{2}{3}, v = -3 \)

Step 1: Express One Variable in Terms of the Other

From the first equation, solve for \( v \) in terms of \( u \), which is \( 9u + 2v = 0 \).Rearrange to get:\[ 2v = -9u \]Divide by 2:\[ v = -\frac{9}{2}u \]

Step 2: Substitute into the Second Equation

Substitute \( v = -\frac{9}{2}u \) into the second equation \( 3u - 5v = 17 \).Replace \( v \) with \( -\frac{9}{2}u \):\[ 3u - 5\left(-\frac{9}{2}u\right) = 17 \]

Step 3: Simplify and Solve for \( u \)

Simplify the expression:\[ 3u + \frac{45}{2}u = 17 \]Combine like terms:\[ 3u + 22.5u = 17 \]This adds to:\[ 25.5u = 17 \]Solve for \( u \):\[ u = \frac{17}{25.5} = \frac{2}{3} \]

Step 4: Solve for \( v \) Using \( u = \frac{2}{3} \)

Substitute \( u = \frac{2}{3} \) back into the expression for \( v \):\[ v = -\frac{9}{2}\left(\frac{2}{3}\right) \]Simplify:\[ v = -\frac{9 \times 2}{2 \times 3} \]\[ v = -3 \]

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations, especially linear systems. This method involves solving one of the equations for one variable and then substituting that expression into the other equation. This reduces the system from two equations with two variables to a single equation with one variable, making it easier to solve.

To understand the process better, consider our original system of equations:

• 9u + 2v = 0
• 3u - 5v = 17

In Step 1, we began by isolating one of the variables. From the first equation, we expressed \( v \) in terms of \( u \) by rearranging it to get \( v = -\frac{9}{2}u \).

Once we had this expression, we substituted it back into the second equation. This substitution replaces \( v \) and allows us to focus on solving for \( u \). Using the substitution method helps simplify a complex problem into manageable steps.

Solving Equations

After performing the substitution, we now have a single equation with one unknown, which is easier to handle. Our task is to solve this simplified equation by isolating the unknown variable.
Starting with the expression from substitution:

• 3u + \( \frac{45}{2}u = 17 \)

We simplify by combining like terms. In this case, \( 3u \) and \( \frac{45}{2}u \) add up to yield \( 25.5u \). This reduces our problem to the linear equation \( 25.5u = 17 \).

To solve for \( u \), we simply divide both sides of the equation by 25.5, giving us \( u = \frac{17}{25.5} \) or simplified to \( u = \frac{2}{3} \). This methodical procedure highlights the systematic nature of solving equations once substitution has simplified the system.

Algebraic Expressions

Algebraic expressions are the building blocks of equations and systems used in algebra. These expressions contain numbers, variables, and arithmetic operations. Understanding how to manipulate them is essential for solving algebra problems.

In our exercise, once we solved for \( u \), we substituted it back into the expression we found for \( v \), which was \( v = -\frac{9}{2}u \). Substituting \( u = \frac{2}{3} \) into this expression involves performing arithmetic simplifications:

• \( v = -\frac{9}{2} \left( \frac{2}{3} \right) \)

Carrying out the multiplication and simplification gives us \( v = -3 \).

Understanding how to manipulate and substitute algebraic expressions correctly is crucial for reaching the correct solutions in a system of linear equations. Each step involves clear algebraic manipulations, ensuring the integrity of the solutions derived from these systems.