Problem 32 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{rl}6 x-y & =3 \\ 5 x+3 y & =-9\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \((0, -3)\).

Step 1: Express One Variable in Terms of the Other

From the first equation, solve for one of the variables. Let's solve for \( y \):\[ 6x - y = 3 \]Rearranging gives:\[ y = 6x - 3 \]

Step 2: Substitute Expression in Second Equation

Substitute \( y = 6x - 3 \) into the second equation:\[ 5x + 3(6x - 3) = -9 \]Simplify and solve for \( x \):\[ 5x + 18x - 9 = -9 \]

Step 3: Solve for x

Combine like terms in the equation:\[ 23x - 9 = -9 \]Add 9 to both sides:\[ 23x = 0 \]Divide both sides by 23:\[ x = 0 \]

Step 4: Substitute Back to Find y

Now that we know \( x = 0 \), substitute back into \( y = 6x - 3 \):\[ y = 6(0) - 3 \]This simplifies to:\[ y = -3 \]

Step 5: Conclusion

Thus, the solution to the system of equations is:\[ (x, y) = (0, -3) \]

Key Concepts

System of Equations

A system of equations is a collection of two or more equations set together involving the same set of variables. In algebra, these are often equations containing the variables, such as \( x \) and \( y \). The example problem contains the following system:

• \( 6x - y = 3 \)
• \( 5x + 3y = -9 \)

The goal when dealing with such a system is to find values for \( x \) and \( y \) such that both equations are true simultaneously. This means we look for the intersection point of the lines when these equations are graphed. Solving a system of equations is a fundamental skill in algebra and helps in understanding relationships between different quantities.

Solving Systems

When solving systems of equations, several methods can be used, such as:

• Graphing
• Substitution
• Elimination

In the given exercise, the substitution method is applied. This method is handy when one of the equations can be easily solved for one variable. The solution process involves: - Solving one equation for one variable - Substituting this expression into the other equation - Solving for the remaining variable After finding the value of the first variable, it's substituted back to find the value of the second variable. This method is perfect for linear equations like our exercise, where equations can be manipulated to reveal values easily.

Algebraic Techniques

Algebraic techniques are the tools and strategies used to rearrange and solve equations. In the substitution method, the key algebraic technique utilized is isolating one variable. This involves:- Rearranging an equation to express it with one variable isolated on one side. In our example, \( y \) was isolated in the first equation: \( y = 6x - 3 \).- Substituting this expression into the other equation to solve for the other variable. Remember to:

• Perform the same operation on both sides of an equation to maintain equality.
• Simplify expressions fully by combining like terms.
• Pay attention to signs to avoid mistakes in arithmetic operations.

Linear Equations

Linear equations are algebraic equations where each variable is raised to the power of one. They form straight lines when graphed on a coordinate plane. An essential characteristic of linear equations is their simplicity, thus allowing techniques like substitution and elimination to be effective in solving them. Examples from the exercise are:

• \( 6x - y = 3 \)
• \( 5x + 3y = -9 \)

Each of these is a linear equation because the highest exponent of the variables is one.Linear equations can model real-world scenarios and are foundational in understanding more complex algebraic concepts. Their graph provides a visual representation of solutions, with the intersection point representing the solution set in the context of linear systems.