Problem 15 - Intermediate Algebra

Question

For Problems \(1-18\), use the elimination-by-addition method to solve each system. (Objective 1 ) $$ \left(\begin{array}{l} 8 x-3 y=13 \\ 4 x+9 y=3 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{3}{2}\) and \(y = -\frac{1}{3}\).

Step 1: Align the equations

Write the equations one under the other: 1. \(8x - 3y = 13\)2. \(4x + 9y = 3\)

Step 2: Eliminate a variable

Choose to eliminate \(x\) by making the coefficients equal. Multiply the second equation by 2:1. \(8x - 3y = 13\)2. \(8x + 18y = 6\)

Step 3: Subtract the equations

Subtract the first equation from the second equation to eliminate \(x\):\((8x + 18y) - (8x - 3y) = 6 - 13\).This simplifies to:\(21y = -7\).

Step 4: Solve for the remaining variable

Divide both sides of the equation by 21 to solve for \(y\):\[y = \frac{-7}{21} = -\frac{1}{3}\]

Step 5: Substitute back to find the other variable

Substitute \(y = -\frac{1}{3}\) back into one of the original equations, say \(8x - 3y = 13\):\(8x - 3(-\frac{1}{3}) = 13\).Solve for \(x\):\(8x + 1 = 13\) \(8x = 12\)\(x = \frac{12}{8} = \frac{3}{2}\).

Step 6: Verify the solution

Substitute \(x = \frac{3}{2}\) and \(y = -\frac{1}{3}\) back into the second equation to ensure they satisfy it:\(4(\frac{3}{2}) + 9(-\frac{1}{3}) = 3\)\(6 - 3 = 3\)This confirms the solution is correct.

Key Concepts

systems of equations

A system of equations is a set of two or more equations with the same unknowns. Solving a system typically involves finding values for the unknown variables that satisfy all the equations in the system simultaneously. For example, in our given problem, we have a system of two equations:

• Equation 1: \(8x - 3y = 13\)
• Equation 2: \(4x + 9y = 3\)

The goal is to determine the values of \(x\) and \(y\) that make both equations true.
Depending on their nature, systems of equations can be categorized as consistent or inconsistent. Consistent systems have at least one solution, while inconsistent systems have none. The system from our exercise is consistent, as there is a unique solution for \(x\) and \(y\) that satisfies both equations.
If the equations in a system have exactly the same line after simplification, they have infinitely many solutions, meaning they are dependent.

linear algebra

Linear algebra is a branch of mathematics focusing on vectors, vector spaces, and linear mappings between these spaces. It's a powerful tool for solving systems of linear equations. In the context of our exercise, we are dealing with linear equations, which means each term is either a constant or the product of a constant and a single variable.
When solving our two given equations, we use coefficients of variables and constants to form vectors and matrices to employ different methods like substitution or elimination to derive solutions.
Linear algebra heavily relies on matrix representations. Although this problem doesn't explicitly use matrices, advanced approaches to solving systems of equations often involve converting the equations into matrix form for efficient computation. Understanding these basics helps simplify the process of tackling systems, especially those with more variables or equations.

solving equations

Solving equations, particularly linear ones, often involves finding the value of the variable that makes the equation true. For our system, we employed the elimination method by focusing on one variable at a time to find the values of \(x\) and \(y\). Here's how it works:

• **Align the Equations**: Start by writing one equation beneath the other, aligning similar terms.
• **Eliminate a Variable**: Multiply one or both equations by suitable numbers so that when you add or subtract them, one variable gets eliminated.
• **Solve for Remaining Variable**: Simplify the resulting equation to find the value of the remaining variable.
• **Back Substitute**: Insert this value back into one of the original equations to find the other variable.

The solution of our system revealed \[ x = \frac{3}{2} \] and \[ y = -\frac{1}{3} \]. By substituting these values back into the equations and verifying them, we ensure the accuracy of our solution. This method is reliable and systematic, making it ideal for beginners to learn and apply effectively.