Problem 38 - Algebra 2

Question

Solve each system of equations by using either substitution or elimination. \(s+3 t=27\) \(2 t=19-\frac{1}{2} s\)

Step-by-Step Solution

Answer

The solution is \(s = -6\) and \(t = 11\).

Step 1: Rearrange Second Equation

Start with the second equation: \[ 2t = 19 - \frac{1}{2}s \]We need to isolate the variable \(t\). Multiply the entire equation by 2 to remove the fraction:\[ 4t = 38 - s \].Now, solve for \(s\) to find a suitable substitution:\[ s = 38 - 4t \].

Step 2: Substitute into the First Equation

Take the expression for \(s\) from Step 1 and substitute it into the first equation:\[ s + 3t = 27 \] becomes \[ (38 - 4t) + 3t = 27 \].

Step 3: Solve for t

Simplify the equation from Step 2:\[ 38 - 4t + 3t = 27 \].Combine like terms:\[ 38 - t = 27 \].Subtract 38 from both sides:\[ -t = -11 \].Multiply both sides by -1 to solve for \(t\):\[ t = 11 \].

Step 4: Solve for s

Now that we have \(t = 11\), substitute it back into the equation \(s = 38 - 4t\):\[ s = 38 - 4(11) \].Calculate:\[ s = 38 - 44 = -6 \].

Step 5: Verify the Solution

To verify the solution, substitute \(s = -6\) and \(t = 11\) back into the original equations:1. \(s + 3t = 27\):\[ -6 + 3(11) = -6 + 33 = 27 \] (holds true).2. \(2t = 19 - \frac{1}{2}s\):\[ 2(11) = 19 + 3 = 22 \] (also holds true).Both equations are satisfied. The solution is verified.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. It involves solving one of the equations for one of the variables and then substituting this expression into the other equation.
This method works well when one of the equations is easy to solve for one of the variables. Here's how you can apply the substitution method:

• Solve one of the equations for one variable. For example, in our original problem, from the equation \( 2t = 19 - \frac{1}{2}s \), we isolated \( s \) to find \( s = 38 - 4t \).
• Substitute this expression into the other equation. This helps to replace the variable in one equation, transforming it into a single-variable equation. We then substituted \( s = 38 - 4t \) into \( s + 3t = 27 \), simplifying to \( (38 - 4t) + 3t = 27 \).
• Solve this new equation for the remaining variable. This helps find the value for one variable, which can then be used to find the other.
• Finally, substitute back to find the value of the first variable.

By identifying suitable equations to isolate and substitute, this method becomes an effective tool for solving linear equations in systems.

Elimination Method

The elimination method is another strategy for solving systems of linear equations. It involves combining the equations in a way that eliminates one of the variables. To use the elimination method:

• First, arrange both equations in a standard format, making sure the variables line up vertically.
• Multiply one or both equations by a constant if necessary, so that the coefficients of one of the variables are opposites (e.g., \(x\) in one equation has a coefficient of 3, and in the other, -3).
• Add or subtract the equations to eliminate one of the variables. This will leave you with a single-variable equation that you can solve.
• Substitute the obtained value back into one of the original equations to solve for the remaining variable.

While the substitution method is about replacing variables, the elimination method focuses on removing them by combining equations. Each method has situations where it is more efficient depending on the structure of the system.

Solving Linear Equations

Solving linear equations is a fundamental skill in algebra. It involves finding the value of the variable(s) that makes the equation true. Linear equations can be solved using several methods, including substitution and elimination, as explored above. When solving linear equations, key steps generally include:

• Isolating the variable, which might involve distributing, combining like terms, and using inverse operations (such as addition/subtraction and multiplication/division).
• Performing operations equally on both sides of the equation to maintain the balance, which ensures that the equation remains valid.
• Rechecking your solution by substituting your answer back into the original equation to see if it produces a true statement.

These steps are applied in various ways depending on the complexity and format of the equations. Understanding the balance and manipulation of these equations using algebraic properties is crucial for solving them effectively.