Problem 60 - Algebra 2

Question

Solve each system by using substitution or elimination. $$ \begin{array}{l}{3 p+6 q=-3} \\ {2 p-3 q=-9}\end{array} $$

Step-by-Step Solution

Answer

\( p = -3 \), \( q = 1 \)

Step 1: Convert equation for substitution

Let's start by solving the first equation for one of the variables to use substitution. We will solve for \( p \): \[ 3p + 6q = -3 \]Dividing every term by 3 to isolate \( p \): \[ p + 2q = -1 \]Therefore, \( p = -1 - 2q \).

Step 2: Substitute into the second equation

Now that we have \( p = -1 - 2q \), substitute this expression into the second equation:\[ 2p - 3q = -9 \]Replace \( p \) with \( -1 - 2q \):\[ 2(-1 - 2q) - 3q = -9 \]

Step 3: Simplify and solve the equation

Simplify the equation from the previous step:\[ 2(-1) + 2(-2q) - 3q = -9 \]\[ -2 - 4q - 3q = -9 \]Combine like terms:\[ -2 - 7q = -9 \]Add 2 to both sides:\[ -7q = -7 \]Divide by -7 to solve for \( q \):\[ q = 1 \]

Step 4: Solve for the other variable

Substitute \( q = 1 \) back into the equation \( p = -1 - 2q \) to find \( p \):\[ p = -1 - 2(1) \]\[ p = -1 - 2 \]\[ p = -3 \]

Step 5: Verify the solution

Substitute \( p = -3 \) and \( q = 1 \) back into the original equations to ensure they are satisfied:For the first equation:\[ 3(-3) + 6(1) = -3 \]\[ -9 + 6 = -3 \]This is true.For the second equation:\[ 2(-3) - 3(1) = -9 \]\[ -6 - 3 = -9 \]This is also true.Both equations are satisfied, confirming that \( p = -3 \) and \( q = 1 \) is the correct solution.

Key Concepts

Substitution Method

The substitution method is a powerful tool to solve a system of equations. It involves isolating one variable in one equation and then replacing that variable in the other equation with the expression obtained. In our exercise, we solved for \( p \) in the equation \( 3p + 6q = -3 \), obtaining \( p = -1 - 2q \). This expression is then substituted into the second equation.
This method is especially useful when one of the equations is easy to solve for one variable, simplifying further calculations. By performing substitution, we reduce the system of two equations to a single equation in one variable, which can be solved using simple algebraic techniques. This systematic approach makes solving complex systems more manageable and straightforward.
For larger systems or when equations are not easily rearranged, other methods like elimination might be more efficient. Use substitution when you can easily isolate a variable and carry it through efficiently.

Elimination Method

The elimination method offers an alternative approach to solve systems of equations by attempting to remove one of the variables. This is done by adding or subtracting equations to eliminate a variable.
Suppose we used elimination in this exercise. We would first multiply the equations so that the coefficients of one of the variables are opposites. Here, multiplying the first equation by 2 and the second by 3 could help make the coefficients of \( p \) opposites:

• First equation: \( 6p + 12q = -6 \)
• Second equation: \( 6p - 9q = -27 \)

Subtracting these, \( 12q - (-9q) \) would eliminate \( p \), leaving a single equation in \( q \).
This method is particularly effective in systems where substitution might complicate matters due to difficult coefficients or when equations are easily aligned for elimination.

Solving Linear Equations

Solving linear equations is a fundamental part of algebra that involves finding the value of the variable that makes the equation true. In our example, after substituting, we got the equation \( -2 - 7q = -9 \).
To solve, simplify using standard algebraic principles:

• Combine like terms to simplify expressions.
• Perform operations such as addition or subtraction to isolate terms containing the variable on one side of the equation.
• Use multiplication or division to solve for the variable completely.

Hence, in our example, adding 2 to both sides and dividing by -7 provided us with \( q = 1 \). This systematic approach ensures a logical pathway to reach the solution, enhancing understanding of fundamental algebraic processes.

Algebraic Manipulation

Algebraic manipulation is an essential skill in solving systems of equations, as it allows for simplification and rearrangement to make solutions clear. Techniques include:

• Distributing multiplication over addition to clear parentheses.
• Combining like terms to simplify expression forms.
• Using inversion operations like adding to cancel out terms or dividing to isolate variables.

In the exercise, we performed several manipulations: substituting \( p = -1 - 2q \) into \( 2p - 3q = -9 \) involved distributing, combining like terms \(-2 - 4q - 3q = -9\), and isolating \( q \). Post finding \( q \), substituting back to find \( p \), demonstrated multi-step algebraic manipulation.
Being proficient with these techniques can streamline solving systems and is invaluable in broader mathematical applications.