Problem 22 - Algebra 1

Question

Use the substitution method to solve the linear system. $$\begin{aligned} &p+q=4\\\ &4 p+q=1 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(p = -1\) and \(q = 5\)

Step 1: Solve the first equation for p or q

From the first equation \(p + q = 4\), we will solve for p, giving us \(p = 4 - q\).

Step 2: Substitute into Second Equation

Substitute \(p = 4 - q\) into the second equation and solve for q. So the second equation \(4p + q = 1\) becomes \(4(4 - q) + q = 1\). Simplifying gives \(16 - 4q + q = 1\) which simplifies further to \(16 - 3q = 1\).

Step 3: Solve the resulting equation for q

Subtract 16 from both sides of the equation \(16 - 3q = 1\) to get \(-3q = -15\). Then divide both sides by -3 to solve for q to get \(q = 5\).

Step 4: Substitute Value of q into First Equation

Substitute \(q = 5\) into the first equation to get \(p = 4 - 5\), simplifying to \(p = -1\).

Key Concepts

Linear Equations

Linear equations are mathematical expressions that form straight lines when graphed on a coordinate plane. They often appear in the form: \[ax + by = c\], where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. In our exercise, we have two linear equations:

• \(p+q=4\)
• \(4p+q=1\)

These equations describe a line on a graph when plotted.
Linear equations are fundamental in algebra since they relate directly to everyday situations such as budgeting and planning.
The key characteristic of these equations is that they exhibit a constant rate of change, depicted by straight lines. Every increase or decrease by 1 in one variable results in a constant change in the other variable.

Systems of Equations

A system of equations is a set of multiple equations that share the same variables. In a linear system, all involved equations are linear. The main goal is to find the values of variables that simultaneously satisfy each equation within the system. Here, we have:

• Equation 1: \(p + q = 4\)
• Equation 2: \(4p + q = 1\)

Solving such systems helps us find out the point where two or more lines intersect when graphed.
In solving systems, various methods like graphing, substitution, and elimination are used. Each has its own advantage depending on the complexity of the equations.
This example uses the substitution method, which is effective when you can isolate one variable easily.

Solving Equations

Solving linear equations involves finding the value of the variable that makes the equation true. Using the substitution method means you solve one equation for one variable and then use that expression to replace the variable in another equation.
Here’s a quick recap of the steps taken:

• Step 1: First, isolate one variable. From \(p + q = 4\), we isolated \(p\), giving us \(p = 4 - q\).
• Step 2: Substitute \(p\) in the second equation \(4p + q = 1\), which becomes \(4(4 - q) + q = 1\).
• Step 3: Simplify and solve for \(q\). This gives us \(q = 5\).
• Step 4: Use this \(q\) value in the first equation to find \(p = -1\).

The substitution method is efficient for systems where equations are easily rearrangeable. Understanding this concept ensures accuracy when determining variable values in algebraic problems.