Problem 59 - Algebra 2

Question

Solve each system by using substitution or elimination. $$ \begin{array}{l}{2 s+7 t=39} \\ {5 s-t=5}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( s = 2 \), \( t = 5 \).

Step 1: Choose a Method to Solve

Start by deciding whether to use substitution or elimination to solve the system. In this case, since the coefficients of \(t\) are already opposites (7 and -1), we will use elimination.

Step 2: Eliminate One Variable

To eliminate \(t\), add the two equations directly: \( (2s + 7t) + (5s - t) = 39 + 5 \). This simplifies to \( 7s + 6t = 44 \). However, it's easier to notice that if we multiply the second equation by 7, we'll form a system that can eliminate \(t\) completely.

Step 3: Multiply Second Equation

Multiply the second equation by 7: \( 7 \times (5s - t) = 7 \times 5 \). This results in the equation: \( 35s - 7t = 35 \).

Step 4: Add the Equations

Now add the modified second equation to the first equation to eliminate \(t\): \( (2s + 7t) + (35s - 7t) = 39 + 35 \). Simplifying this gives \( 37s = 74 \).

Step 5: Solve for s

Divide both sides by 37: \( s = \frac{74}{37} = 2 \). Now we have the value for \(s\).

Step 6: Substitute and Solve for t

Substitute \( s = 2 \) back into the second original equation to solve for \(t\): \( 5(2) - t = 5 \). Simplifying gives \( 10 - t = 5 \). Solving for \(t\), we get \( t = 5 \).

Step 7: Verify the Solution

Substitute \( s = 2 \) and \( t = 5 \) back into the first original equation to verify the solution: \( 2(2) + 7(5) = 4 + 35 = 39 \). Both equations are satisfied.

Key Concepts

Elimination Method

The elimination method is a popular approach for solving systems of linear equations. The idea is to eliminate one variable at a time to simplify solving for the others. This is often achieved by adding or subtracting the equations in a way that cancels out one of the variables. To use this method effectively, follow these steps:

• Look for coefficients of a variable that are opposites or can be easily manipulated to become opposites.
• Add or subtract the equations to eliminate one variable.
• Solve the resulting equation for the remaining variable.

In our exercise, the elimination method was used because the coefficients of the variable \( t \) in the two equations were conveniently set as opposites (7 and -1). This made it easier to eliminate \( t \) by adjusting the second equation, multiplying it by 7, and thereby simplifying the problem to solve for \( s \). This approach is very systematic and can be applied to complex systems where variables have matching or easily adjusted coefficients.

Substitution Method

Another effective way to solve linear systems is the substitution method. This method involves solving one equation for one variable and then substituting that result into the other equation. This technique is particularly useful when it is easy to express one variable in terms of the other. Here’s a simple plan to use substitution:

• Solve one of the equations for one of the variables.
• Substitute this result into the other equation, replacing the chosen variable.
• Solve the new equation for the remaining variable.

This approach creates a single-variable equation, which can be much simpler to handle. In problems where systems have simpler expressions or clear relationships between variables, substitution can be quicker and easier than elimination.

Linear Equations

Linear equations are fundamental mathematical expressions where each term is either a constant or the product of a constant and a single variable. A linear equation can be in the form \( ax + by = c \). These equations produce a straight line when graphed on a coordinate plane.

Key characteristics of linear equations include:

• They have no exponents on the variables, meaning no variable is squared or cubed, etc.
• There are no products of variables—each variable stands alone.

Understanding linear equations is crucial because they are the building blocks for more complicated mathematical concepts. They appear frequently in algebra and are used to explore the relationships between changing values. In the given system, both equations were linear, which establishes that the solutions would be a point where these straight lines intersect.

System of Equations

A system of equations is a set of two or more equations with the same variables. Solving a system means finding the values of the variables that satisfy all the equations simultaneously. Systems of equations can be categorized in several ways:

• Consistent systems have at least one solution.
• Inconsistent systems have no solution.
• Dependent systems have infinitely many solutions.

In our exercise, the system of equations was solved by finding the values \( s = 2 \) and \( t = 5 \) using the elimination method. These values satisfied both equations, demonstrating that the system was consistent. Systems of equations often appear in real-world problems and can represent multiple constraints in business, physics, or other sciences, making their solution is fundamental in various fields.