Problem 14 - 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. $$ \left\\{\begin{array}{r}{3 x-5 y=8} \\ {2 x+5 y=22}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 6\) and \(y = 2\).

Step 1: Add the two equations

The two equations are designed in such a way that by simply adding them you will eliminate \(y\). Hence, \(3x - 5y + 2x + 5y = 8 + 22\) simplifying that gives \(5x = 30\).

Step 2: Solve for \(x\)

To find the value of \(x\), divide both sides of the equation \(5x = 30\) by 5. Hence, \(x = 30 / 5 = 6\).

Step 3: Substitute \(x\) into one of the original equations

Substitute \(x = 6\) into the first equation. This gives \(3 * 6 - 5y = 8\), which simplifies into \(18 - 5y = 8\).

Step 4: Solve for \(y\)

Rearrange the equation to solve for \(y\). This gives \(-5y = 8 - 18\) or \(-5y = -10\). By dividing both sides of \(-5y = -10\) by -5, we find that \(y = 2\).

Step 5: Check the solution algebraically

Substitute \(x = 6\) and \(y = 2\) into both original equations to check if the left side equals the right side of the equations.

Key Concepts

Solving Systems of Equations

When faced with a system of equations, we are presented with two or more equations that have common variables. The goal is to find the set of values for these variables that satisfy all equations simultaneously. The elimination method is a popular technique used to achieve this.
It involves manipulating the equations so that adding or subtracting them eliminates one of the variables, allowing us to solve for the remaining variable.
In the given exercise, we work with a system consisting of:

• \(3x - 5y = 8\)
• \(2x + 5y = 22\)

By adding these equations, the \(y\) terms cancel out, leading us to a simpler equation in terms of \(x\) only. From there, we can solve for \(x\) and subsequently substitute back to find \(y\). This step-by-step approach is useful as it reduces a two-variable problem to a single-variable problem, making the solution process straightforward.

Algebraic Verification

Once we've calculated potential solutions for the variables, it's crucial to verify our work. Algebraic verification ensures that the values we found actually satisfy the original system of equations.
To do this, we substitute the solution back into each of the initial equations:

• Substitute \(x = 6\) and \(y = 2\) into \(3x - 5y = 8\), resulting in \(3(6) - 5(2) = 18 - 10 = 8\), which is true.
• Substitute the same values into \(2x + 5y = 22\), giving \(2(6) + 5(2) = 12 + 10 = 22\), another true statement.
  

Both checks confirm that our solutions \((x, y) = (6, 2)\) satisfy the system. Verification is a vital step because it acts as an error-checking mechanism, ensuring our solutions are correct and align with the given equations.

Substitution Method

The substitution method serves as an alternative to elimination for solving systems of equations. It involves solving one equation for one variable and substituting this expression into another equation.
In contrast to elimination, which focuses on reducing equations by removing variables through combination, substitution allows us to manipulate one equation in terms of another.
Let’s illustrate the potential application using our system of equations: If we solve the first equation \(3x - 5y = 8\) for \(x\), we get \(x = \frac{8 + 5y}{3}\). Then, substitute this expression into \(2x + 5y = 22\) and simplify to find \(y\).
This approach is particularly handy when one equation can be easily solved for a particular variable, leading to simpler arithmetic and insights into the relationship between variables.