Problem 19 - Intermediate Algebra

Question

For Problems \(1-26\), solve each system by using the substitution method. (Objective 1) $$ \left(\begin{array}{r} 3 x+y=2 \\ 11 x-3 y=5 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(\left(\frac{11}{20}, \frac{7}{20}\right)\).

Step 1: Solve for One Variable

Let's start by solving the first equation for one of the variables. We have the equation \(3x + y = 2\). Solving for \(y\), we get \(y = 2 - 3x\).

Step 2: Substitute Into the Second Equation

Now that we have \(y\) in terms of \(x\), substitute \(y = 2 - 3x\) into the second equation \(11x - 3y = 5\). This gives us: \[ 11x - 3(2 - 3x) = 5 \].

Step 3: Simplify and Solve for x

Distribute and simplify the equation from Step 2: \[ 11x - 6 + 9x = 5 \] becomes \[ 20x - 6 = 5 \]. Solve for \(x\) by adding 6 to both sides: \(20x = 11\). Then divide both sides by 20: \(x = \frac{11}{20}\).

Step 4: Substitute Back to Find y

Use the value of \(x\) to find \(y\). Substitute \(x = \frac{11}{20}\) into the equation \(y = 2 - 3x\): \[ y = 2 - 3\left(\frac{11}{20}\right) = 2 - \frac{33}{20} = \frac{40}{20} - \frac{33}{20} = \frac{7}{20} \].

Step 5: Write the Solution as an Ordered Pair

The values of \(x\) and \(y\) found are \(x = \frac{11}{20}\) and \(y = \frac{7}{20}\). The solution to the system of equations is the ordered pair \(\left(\frac{11}{20}, \frac{7}{20}\right)\).

Key Concepts

Substitution Method

The substitution method is a strategy used in solving systems of equations where one of the equations is solved for one variable in terms of the others. This expression is then substituted into the other equation. In our example, we started with the equation \(3x + y = 2\). By isolating \(y\), we found that \(y=2-3x\).

• This makes it easy to substitute \(y\) in the second equation.
• It transforms a system of equations into a single equation in one variable.

After finding \(y\) in terms of \(x\), you substitute \(y = 2 - 3x\) into the other equation \(11x - 3y = 5\). This substitution reduces the problem to solving a single equation, making it easier to handle.
The substitution method is particularly helpful when one of the equations can be easily manipulated to express one variable in terms of another. It's a clear and logical approach, especially useful when dealing with simpler coefficients.

Solving Linear Equations

The process of solving linear equations involves finding values for the variables that make all of the equations true simultaneously.
For our system, after substitution, the equation becomes \[11x - 3(2 - 3x) = 5\].
To solve this linear equation:- Distribute the \(-3\) across the terms within the parentheses, yielding \(-6 + 9x\).- Combine like terms, resulting in \(20x - 6\).- Adjust the equation to solve for \(x\): add 6 to both sides to get \(20x = 11\) and then divide by 20 to find \(x = \frac{11}{20}\).All of these steps are foundational skills in algebra, where you perform operations on both sides of the equation to isolate the variable. Ensuring you carry out the same operation on each side maintains the balance of the equation.

Ordered Pairs

Once you have found the values for both variables, these are expressed as an ordered pair.
The solution to a system of equations is typically written as an ordered pair \((x, y)\).
- In our solution, we determined \(x = \frac{11}{20}\) and \(y = \frac{7}{20}\).- Therefore, the solution is the ordered pair \(\left(\frac{11}{20}, \frac{7}{20}\right)\).An ordered pair represents a point in the coordinate plane where both equations intersect. This means that both \((x, y)\) values satisfy each original equation in the system. An ordered pair pinpoints the exact solution, giving us a tangible understanding of where these lines meet graphically. It's a concise way to denote the outcome for both variables at once, grounding the abstract algebraic process in a more visual, coordinate-based context.