Problem 23 - Algebra 1

Question

Choose a method to solve the linear system. Explain your choice, and then solve the system. $$ \begin{aligned} &x+y=0\\\ &3 x+2 y=1 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 1\) and \(y = -1\).

Step 1: Isolate a variable in one equation

The first equation is \(x + y = 0\), which can be rearranged as \(x = -y\).

Step 2: Substitute into the second equation

Next, substitute \(x = -y\) into the second equation \(3x + 2y = 1\). This gives you \(3(-y) + 2y = 1\), simplifying to \(-y = 1\) or \(y = -1\).

Step 3: Substitute y back into the first equation

Substitute \(y = -1\) back into the first equation \(x + y = 0\), resulting in \(x + (-1) = 0\), or \(x = 1\).

Key Concepts

Substitution Method

The substitution method is a powerful technique used to solve systems of linear equations, where one equation is solved for one variable and then substituted into the other equations. This method can be particularly useful when the equations are set up in a way that makes isolating a variable straightforward.

Imagine you're at a crossroad, and the paths are your equations. By choosing one path (isolating one equation), you effectively find a route that guides you to your destination (solution). In the case of our exercise, the equation \(x + y = 0\) allows us to easily isolate \(x\), resulting in \(x = -y\). This clear path now replaces \(x\) in the second equation, simplifying our journey towards the solution.

Students may wonder why we'd pick this method over others. It's simply because it turns the complex relationship of two variables into single-variable equations, which are much easier to solve. Infusing our mathematical 'crossroad' with this method, we single out signposts that clearly mark the way forward.

Algebraic Equations

Algebraic equations are like puzzles. Each variable is a piece of the puzzle, and the equation itself is a clue on how those pieces fit together to complete the picture. In our system of equations, we are given two algebraic equations, \(x + y = 0\) and \(3x + 2y = 1\).

These equations tell us a relationship between \(x\) and \(y\): if we know the value of one, we can find the other. Solving these puzzles requires us to manipulate these relationships, by adding, subtracting, multiplying, or dividing, until we find the values that satisfy both equations simultaneously.

In the context of our exercise, we rearranged the first algebraic equation to make \(x\) the subject. This kind of manipulation gives us the power to unlock the values of variables, much like finding the right key to a locked treasure chest. Each step we take in solving an algebraic equation moves us closer to discovering the hidden treasures of \(x\) and \(y\).

Systems of Equations

A system of equations is like a team of detectives working together to solve a mystery. Each equation in the system provides a different clue. To solve the mystery—that is, to find the values of the variables that satisfy all the equations—we often need all the clues to fit together just right.

In our exercise, the system is composed of two linear equations which intersect at a point. The coordinates of this point are the solution to the system. When we're using the substitution method, it's as if we're letting the first detective (the first equation) lead us to a piece of critical evidence. That evidence (\(x\) in terms of \(y\)) is then handed over to the second detective (the second equation), who uses it to crack the case.

Understanding how to work with systems of equations is a fundamental skill in algebra. It's not just about finding numbers that work; it's about understanding how different pieces of information, when combined, can give us a complete picture of a situation. In this system of equations, as in many real-life problems, collaboration is key—you often can't solve the puzzle unless you consider all parts as a whole.