Problem 29 - Algebra 1

Question

Use the substitution method to solve the linear system. $$\begin{aligned} &3 w-2 u=12\\\ &w-u=60 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the system of equations are \(u = -168\) and \(w = -108\).

Step 1: Express One Variable in Terms of the Other

From the second equation \(w-u=60\), isolate \(w\) to get \(w=u+60\).

Step 2: Substitute for w in the First Equation

We can now substitute \(u+60\) for \(w\) in the first equation to give us \(3(u+60) - 2u = 12\). Simplify this equation by distributing and combining like-terms to get \(3u + 180 - 2u = 12\). This simplifies to \(u + 180 = 12\).

Step 3: Solve for u

Subtract \(180\) from both sides to get \(u = -168\).

Step 4: Substitute u in Second Equation

Substitute \(u = -168\) into the second equation \(w = u + 60\). Doing so gives us \(w = -168 + 60\).

Step 5: Solve for w

Calculate the value of \(w\) which gives \(w = -108\).

Key Concepts

Substitution Method

The substitution method is a fundamental technique used in linear algebra to solve systems of linear equations. The idea is to solve one equation for one variable and then substitute this expression into the other equation(s). This allows you to reduce the system to a single equation in one unknown, which is often easier to solve.

Here’s how it works:

• Choose an equation and isolate one variable. This usually means expressing one variable in terms of the others.
• Substitute this expression into the other equation(s).
• Solve the resulting equation for the remaining variable.
• Once you have the value, back-substitute it to find the value of the other variable(s).

In our exercise, we first isolated the variable \(w\) in the equation \(w-u=60\), getting \(w = u+60\). This expression for \(w\) was then substituted into the first equation to solve for \(u\) and eventually \(w\).

Remember, the substitution method is very effective for systems where it is easy to express one variable in terms of others. It is particularly useful for solving systems by hand.

Solving Equations

Solving equations is the process of finding values for variables that make the equation true. In the context of linear equations, this involves finding values that satisfy all equations simultaneously.

In solving linear equations through substitution, once you've reduced the system to a single equation with one variable, you're left with a simple linear equation. Solving this involves straightforward algebraic manipulation:

• Simplifying expressions by combining like terms.
• Isolating the variable on one side of the equation through addition, subtraction, multiplication, or division.

For our system, when we substituted \(w = u+60\) into \(3w-2u=12\), we simplified the equation to \(u + 180 = 12\). By subtracting \(180\) from each side, we found \(u = -168\).

From here, solving for the second variable \(w\) involved substituting back into the expression \(w = u+60\), leading us to find that \(w = -108\). This highlights the importance of substitution in transforming and simplifying equations to make them solvable.

Linear Algebra

Linear algebra is a branch of mathematics focusing on vectors, matrices, and systems of linear equations. It is a powerful tool used to understand the structure and behavior of mathematical models across various fields. One of the primary functions of linear algebra is to provide methods for solving systems of linear equations.

A system of linear equations is made up of two or more equations with multiple variables. Each equation describes a line, and the solution to the system is the point(s) where these lines intersect.

Key aspects of linear algebra related to our exercise include:

• Variables and Equations: Linear equations involve terms that are either constants or the product of a constant and a single variable.
• Systems of Equations: A collection of two or more linear equations involving the same set of variables.

When we use methods like substitution, we are leveraging the principles of linear algebra to transform and ultimately solve these systems. Successfully solving our system, and determining \(w\) and \(u\), required us to understand each equation's role in the broader context of the system, and how they work together to reveal a solution.

Linear algebra is foundational for many scientific and engineering disciplines, as it provides the tools to model and solve real-world problems systematically and efficiently.