Problem 15 - Precalculus with Limits

Question

Solving a System by Elimination In Exercises \(13-30,\) solve the system by the method of elimination and check any solutions algebraically. $$ \left\\{\begin{array}{l}{5 x+3 y=6} \\ {3 x-y=5}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1.5\) and \(y = -0.5\).

Step 1: Multiply Equations to Make Coefficients of y's Equal

Multiply the second equation by 3 and the first equation by 1 to make the coefficient of \(y\) in both equations the same:\n1*(5x + 3y) = 1*6 becomes 5x + 3y = 6,\n 3*(3x - y) = 3*5 becomes 9x - 3y = 15.

Step 2: Add the Two New Equations to Eliminate y

Add the two equations obtained in step 1 to eliminate variable \(y\):\n (5x + 3y) + (9x - 3y) = (6+15).\n This simplifies to 14x = 21.

Step 3: Solve for x

Solve the equation from step 2 for \(x\):\n 14x = 21,\n \(x = 21 / 14 = 1.5\).

Step 4: Substitute x's Value to Find y's Value

Substitute the value of \(x\) obtained in step 3 into the original second equation to find the value of \(y\):\n 3*(1.5) - y = 5,\n 4.5 - y = 5,\n \(y = 4.5 - 5 = -0.5\).

Step 5: Verification

To check if these values of \(x\) and \(y\) are correct, substitute them into the original equations:\n For 5x + 3y = 6, substituting for \(x\) and \(y\) gives 7.5 - 1.5 = 6.\n For 3x - y = 5, substituting for \(x\) and \(y\) gives 4.5 + 0.5 = 5.\n Since both equations are correct, \(x = 1.5\) and \(y = -0.5\) are indeed solutions to the system.

Key Concepts

The Elimination Method

The elimination method is a popular technique for solving systems of linear equations. It involves transforming the system so that one of the variables cancels out, allowing you to solve for the other. Here's how it works in this exercise:

• First, the goal is to have the coefficients of one of the variables be equal in both equations, so you can eliminate it by addition or subtraction.
• In our problem, we multiplied the entire second equation by 3 to ensure that the coefficients of the variable \(y\) were equal and opposite (\(3y\) and \(-3y\)).
• We didn't need to change the first equation because multiplying by 1 keeps it the same. This setup allows us to add the equations to eliminate \(y\), leaving a single equation with only \(x\).

After the transformation, adding the equations leads to eliminating the \(y\) variable entirely. This leaves a simpler equation involving just \(x\), making it easier to solve. The elimination method is powerful because it reduces complexity and focuses on one variable at a time.

Algebraic Verification

Algebraic verification is a crucial step in solving systems of equations, as it confirms the accuracy of your solution. After solving for \(x\) and \(y\) using the elimination method, it's essential to check these solutions against the original equations. Here's how to verify:

• Take the values of \(x\) and \(y\) that were found and substitute them back into both original equations.
• If the left-hand side of each equation equals the right-hand side, the solution is validated and correctly satisfies both equations.

For our specific problem, by substituting \(x = 1.5\) and \(y = -0.5\) into the first equation, \(5x + 3y = 6\), we get \(7.5 - 1.5 = 6\), which holds true. Similarly, in the second equation \(3x - y = 5\), the values yield \(4.5 + 0.5 = 5\), confirming the correctness of the solution. Performing these checks ensures you're not missing any calculation errors and that the answers are exact.

Solving Linear Systems

Solving linear systems involves finding values of variables that satisfy multiple equations simultaneously. Linear systems can represent real-world problems with constraints requiring a solution that suits all scenarios. The primary methods for solving include substitution, elimination, and graphical approaches.
The elimination method, as demonstrated, is effective for systems with straightforward setups where canceling a variable is feasible. By reducing the number of equations at each step, it simplifies the process substantially.
Besides elimination, the substitution method involves solving one equation for a single variable and substituting this expression into the other equation. It is handy when one variable has a coefficient of one. The graphical method involves plotting both equations on a graph and identifying where they intersect, which is the solution.
To choose the best method, consider the system's structure and coefficients. Each technique has its pros and cons, and the ideal choice depends on the specific problem context. Solving linear systems is versatile, and mastering these methods can greatly aid in tackling various mathematical challenges.