Problem 51 - Algebra 2

Question

Solve each equation. \(\left[\begin{array}{l}{x+3 y} \\ {2 x-y}\end{array}\right]=\left[\begin{array}{r}{-22} \\ {19}\end{array}\right]\)

Step-by-Step Solution

Answer

Solution: \(x = 5\), \(y = -9\).

Step 1: Understand the System of Equations

The matrix equation given is \(\begin{bmatrix} x + 3y \ 2x - y \end{bmatrix} = \begin{bmatrix} -22 \ 19 \end{bmatrix}\). This can be split into two separate equations:1. \(x + 3y = -22\)2. \(2x - y = 19\).

Step 2: Solve for One Variable

Let's solve the first equation for \(x\): \(x = -22 - 3y\). This expresses \(x\) in terms of \(y\).

Step 3: Substitute into the Second Equation

Substitute \(x = -22 - 3y\) into the second equation \(2(-22 - 3y) - y = 19\). Simplify to get:\(-44 - 6y - y = 19\),which simplifies further to\(-44 - 7y = 19\).

Step 4: Solve for \(y\)

Solve \(-44 - 7y = 19\) for \(y\):1. Add 44 to both sides to get: \(-7y = 63\),2. Divide both sides by -7 to find: \(y = -9\).

Step 5: Substitute back to find \(x\)

Use \(y = -9\) in the expression \(x = -22 - 3y\):\(x = -22 - 3(-9) = -22 + 27 = 5\).Thus, \(x = 5\).

Step 6: Verify the Solution

Substitute \(x = 5\) and \(y = -9\) back into the original equations:1. \(5 + 3(-9) = -22\) checks out,2. \(2(5) - (-9) = 19\) checks out.Both equations are satisfied, so the solution is correct.

Key Concepts

Matrix Equations

Matrix equations are a powerful way to represent and solve systems of equations. Instead of writing equations separately, you can bundle them into a matrix. This helps in visualizing and calculating the solutions more efficiently. In our example, the matrix equation is given by: \[ \begin{bmatrix} x + 3y \ 2x - y \end{bmatrix} = \begin{bmatrix} -22 \ 19 \end{bmatrix} \]This visual representation holds two equations in one package:

• The first row corresponds to the equation: \(x + 3y = -22\)
• The second row represents: \(2x - y = 19\)

By transforming these into a matrix, it gives us a clear understanding of parallel equations needing solutions for the variables \(x\) and \(y\). When working with matrix equations, it's crucial to carefully interpret each element to ensure accuracy.

Algebraic Substitution

Algebraic substitution is a method used for solving systems of equations by expressing one variable in terms of another. It works by taking one equation from a system and solving it to find an expression for one variable, which is then substituted into the other equation. This way, you simplify the problem to find the solution more easily.For our system, we've started by taking the first equation \(x + 3y = -22\) and isolated \(x\):\[ x = -22 - 3y \]This step lets you express \(x\) in terms of \(y\), which is crucial for the next phase, where \(x\) is plugged into the second equation \(2x - y = 19\). Here's the substitution part:\[ 2(-22 - 3y) - y = 19 \]With this approach, you reduce a two-variable equation down to one, which is simpler to solve. Algebraic substitution is often used because it breaks down complex systems into easier-to-handle parts.

Variable Isolation

Variable isolation is a critical technique in solving equations, where you focus on getting one variable on one side of the equation. This simplifies analysis and computation. In solving our given system, after substitution, we obtain a single variable expression:From the equation:\[ -44 - 7y = 19 \]We aimed to isolate \(y\). Performing algebraic operations:

• Add 44 to both sides: \(-7y = 63\)
• Then, divide each side by -7: \(y = -9\)

Once \(y\) is calculated, we substitute \(y\) back to get \(x\) as well:\[ x = -22 - 3(-9) = -22 + 27 = 5 \]Using variable isolation like this not only provides a systematic pathway to solutions but ensures you maintain a structured approach in verifying correctness. This method is foundational in algebra, ensuring students can approach similar problems with confidence.