Problem 29 - Understanding Elementary Algebra with Geometry

Question

In Exercises \(29-62,\) solve the system of equations using any method you choose. $$\left\\{\begin{array}{c} r+2 t=10 \\ 3 r+t=-15 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(r = -8\) and \(t = 9\) or \((-8, 9)\).

Step 1: Choose the Method

Select a method to solve the system of equations. In this case, use the substitution method.

Step 2: Solve for One Variable

Solve one of the equations for one of the variables. From the first equation, solve for \(r\): \[ r + 2t = 10 \] Subtract \(2t\) from both sides to get: \[ r = 10 - 2t \]

Step 3: Substitute and Solve for the Other Variable

Substitute \(r = 10 - 2t\) into the second equation: \[ 3r + t = -15 \] Replace \(r\) with \(10 - 2t\): \[ 3(10 - 2t) + t = -15 \] Solve the equation: \[ 30 - 6t + t = -15 \] Combine like terms: \[ 30 - 5t = -15 \] Subtract 30 from both sides: \[ -5t = -45 \] Divide by -5: \[ t = 9 \]

Step 4: Solve for the First Variable

Substitute \(t = 9\) back into the equation \(r = 10 - 2t\): \[ r = 10 - 2(9) \] Calculate: \[ r = 10 - 18 \] \[ r = -8 \]

Step 5: Solution of the System

The solution to the system of equations is \(r = -8\) and \(t = 9\). This means that the ordered pair \((-8, 9)\) satisfies both equations.

Key Concepts

Substitution Method

The substitution method is a great way to solve systems of linear equations. It involves solving one of the equations for one variable and substituting that expression into the other equation. This reduces the system to a single equation with one variable, which is easier to solve. In our example, we start by solving the first equation for \( r \):
\( r + 2t = 10 \).
We subtract \( 2t \) from both sides to isolate \( r \):
\( r = 10 - 2t \).
Now, we have expressed \( r \) in terms of \( t \). This expression for \( r \) can now be substituted into the second equation to find \( t \):
\( 3r + t = -15 \).
When you use substitution, always remember to simplify your equations step by step.

Linear Equations

Linear equations are equations of the first degree, meaning they have no variables raised to a power other than one. They typically take the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. In our exercise, we have two linear equations:
\( r + 2t = 10 \)
and
\( 3r + t = -15 \).
Each variable in a linear equation represents a straight line when plotted on a graph. The solution to a system of linear equations is the point where these lines intersect. In our case, solving the equations gives us \( r = -8 \) and \( t = 9 \), which means that the lines intersect at the point \((-8, 9)\).

Step-by-Step Solution

Breaking down the problem into clear steps can make solving equations much easier. Let's revisit the steps outlined in the solution:

• Step 1: Choose the Method. We chose the substitution method for this example.
• Step 2: Solve for One Variable. We solved the first equation for \( r \): \( r = 10 - 2t \).
• Step 3: Substitute and Solve for the Other Variable. Substituted \( r = 10 - 2t \) into the second equation and solved for \( t \): \( t = 9 \).
• Step 4: Solve for the First Variable. Used \( t = 9 \) in the expression \( r = 10 - 2t \) to find \( r = -8 \).
• Step 5: Solution of the System. The solution to the system is \( r = -8 \) and \( t = 9 \), providing the ordered pair \((-8, 9)\).

Each of these steps builds on the previous one, leading you methodically to the solution. This structured approach helps in clearly understanding the flow of the problem-solving process.