Question

A used truck and a snowplow attachment are worth \(\$ 7200,\) the truck being worth 7 times as much as the plow. Find the value of each.

Short answer

The snowplow attachment is worth \$900, and the truck is worth \$6300.

Step-by-step solution

Identify the Variables

Assign variables to the unknown quantities. Let's call the value of the snowplow attachment 'p' and the value of the truck 't'.

Set Up the Equations Based on the Given Information

According to the problem, the truck is worth 7 times as much as the plow, which gives us the equation: \( t = 7p \). Additionally, together they are worth \$7200, which gives us the second equation: \( t + p = 7200 \)

Substitute and Solve for One Variable

Substitute the value of 't' from the first equation into the second equation: \( 7p + p = 7200 \). Simplify and solve for 'p'.

Solve for the Other Variable

Now that we have the value of 'p', we can substitute it back into the equation \( t = 7p \) to find the value of 't'.

Key concepts

System of Linear Equations

A system of linear equations is a set of two or more linear equations that use the same variables. The solution to this system is a set of values that satisfies all equations simultaneously. These systems can be solved using various methods, including graphing, substitution, elimination, or matrix operations. In our exercise, we are dealing with a system of two linear equations:

• For the value of the truck and snowplow attachment:
  \( t = 7p \) (Equation 1)
• For their combined value:
  \( t + p = 7200 \) (Equation 2)

To find a solution, we need to seek values for \( t \) (the truck's value) and \( p \) (the plow's value) that work for both equations.

Variable Substitution

Variable substitution is a technique to solve a system of equations where you express one variable in terms of another and then replace this expression in another equation. This method often simplifies solving because it reduces the system of equations to one with a single variable. In our scenario,

• Our first equation, \( t = 7p \), allows us to express \( t \) in terms of \( p \).
• We then substitute this expression for \( t \) into the second equation:
  \( 7p + p = 7200 \), which simplifies to \( 8p = 7200 \).
• After substitution, we are left with one linear equation with one variable, \( p \), which is much easier to solve.

Such clear substitution steps make it manageable to follow the path to the solution.

Solving Linear Equations

Solving linear equations is all about finding the value of the variable that makes the equation true. These equations can range from very simple to complex, but the principle remains the same: isolate the variable on one side of the equation. Here's how we tackled the equation from our problem:

• After substitution, we simplified the equation to \( 8p = 7200 \).
• Divide both sides by 8 to isolate \( p \):
  \( \frac{8p}{8} = \frac{7200}{8} \), which simplifies to \( p = 900 \).
• Once we have the value of \( p \), we recall that \( t = 7p \) and multiply 900 by 7 to find \( t \).

This systematic approach moves you from an equation with two unknowns to finding the specific values of each unknown.

Mathematical Problem-Solving

Mathematical problem-solving is a critical skill that involves understanding the problem, devising a plan, carrying out the plan, and then looking back to check the results. It's more than just crunching numbers; it's a process of discovery and analysis. In the context of our exercise,

• Understanding involved recognizing we had a system of linear equations with two variables.
• Devising a plan included identifying the right method (variable substitution) to solve the system.
• Carrying out the plan was the step-by-step approach of substituting and solving the equations.
• Looking back would involve checking if the found values for \( t \) and \( p \) satisfy both original equations to confirm the solution's validity.

Enhancing these problem-solving skills not only assists you with algebra but also in a variety of life's logical challenges.