Problem 12 - Algebra Form and Function

Question

Solve the systems of equations. $$ \left\\{\begin{aligned} r+s &=-3 \\ s-2 r &=6 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Question: Solve the system of linear equations: $$ r + s = -3 \\ s - 2r = 6 $$ Answer: The solution to the system of equations is r = -3 and s = 0.

Step 1: Isolate a variable in one of the equations

We will isolate the variable, s, in the first equation: $$ r + s = -3 \implies s = -r - 3 $$

Step 2: Substitute the expression for s in the second equation

Now substitute the expression obtained in step 1, \(s = -r - 3\), into the second equation to eliminate the variable s: $$ (-r - 3) - 2r = 6 $$

Step 3: Solve for the remaining variable

Now we have an equation with just one variable, r, which we will solve for: $$ -3r - 3 = 6 \implies -3r = 9 \implies r = -3 $$

Step 4: Substitute the value of r into the expression for s

Now that we have the value of r, we can substitute it back into the expression for s from step 1: $$ s = -(-3) - 3 = 3 - 3 = 0 $$

Step 5: Check the solutions in both original equations

Finally, we will check that r = -3 and s = 0 satisfy both original equations: Equation 1: $$ (-3) + 0 = -3 $$ Equation 2: $$ 0 - 2(-3) = 6 $$ Both solutions check out, so the final answer is r = -3 and s = 0.

Key Concepts

Substitution Method

The substitution method is a powerful approach to solving systems of equations. It involves replacing one variable with an expression obtained from another equation. This simplifies the system to a single equation with one variable.
Let's break down how to apply the substitution method:

• Step 1: Solve one equation for one variable.
  In our exercise, we solved the first equation for \(s\): \(r + s = -3\) resulting in \(s = -r - 3\).
• Step 2: Substitute the expression into the other equation.
  Taking \(s = -r - 3\) and substituting it into the second equation \(s - 2r = 6\), we get \((-r - 3) - 2r = 6\).
• Step 3: Simplify and solve the resulting equation
  This gives us one equation in terms of one variable, which we can then solve with more straightforward algebraic methods.

The substitution method is especially useful when one of the equations is easy to solve for one variable. It minimizes the number of variables, making the calculation simpler and clearer.

Solving Linear Equations

Solving linear equations is at the heart of handling systems of equations. Once a variable is isolated through substitution, solving linear equations becomes your next task.
Here's what you need to know:

• Combine like terms: Start by simplifying both sides of the equation. For example, in \(-3r - 3 = 6\), we bring together all terms involving \(r\), resulting in \(-3r = 9\).
• Isolate the variable: Next, solve for the variable by performing algebraic operations. In this case, divide both sides by \(-3\) to find \(r = -3\).

Solving linear equations often involves simple operations like addition, subtraction, multiplication, or division to isolate the variable. The aim is to simplify the equation step by step until the value of the variable is revealed.

Algebraic Manipulation

Algebraic manipulation is a skillful balancing act that helps reshape equations to facilitate solving. When solving systems of equations, it's all about moving parts around to find unknowns.
Some key algebraic manipulations include:

• Rearranging equations: Start by getting all terms involving a specific variable on one side. This might require moving terms across the equal sign, like transforming \(r + s = -3\) to \(s = -r - 3\).
• Simplifying expressions: Break down complex expressions into simpler forms by combining like terms or factoring when necessary. In our example, combining terms in \(-r - 3 - 2r = 6\) simplifies the equation.
• Balancing the equation: Perform identical operations on both sides to maintain the balance. This is crucial when isolating variables or solving equations.

Mastering algebraic manipulation can greatly enhance your problem-solving ability, making complex systems of equations more approachable by peeling back layers of mathematical complexity.