Problem 21 - Algebra 1

Question

Use linear combinations to solve the system of linear equations. $$\begin{aligned} &2 a+6 z=4\\\ &3 a-7 z=6 \end{aligned}$$

Step-by-Step Solution

Answer

The solution to the system of equations is a = 2, z = 0.

Step 1: Multiplication

Multiply the first equation by 3 and the second one by 2 to make the coefficient of a the same in both equations: \(6a + 18z = 12\) and \(6a - 14z = 12\).

Step 2: Subtraction

Subtract the second equation from the first to eliminate a: \(6a - 6a + 18z - (-14z) = 12 - 12 \rightarrow 32z = 0\)

Step 3: Solve for z

Solve the equation for z by dividing both sides by 32: \(z = 0/32 = 0\)

Step 4: Substitute z into the first equation

Substitute z = 0 into the first equation: \(2a + 6*0 = 4 \rightarrow 2a = 4\)

Step 5: Solve for a

Solve the equation for a by dividing both sides by 2: \(a = 4/2 = 2\)

Key Concepts

Linear Combinations

Linear combinations involve taking multiples of given equations and adding or subtracting them to simplify the problem. In the context of solving linear equations, this method is often used to eliminate one variable by adding or subtracting equations with strategically chosen coefficients. This helps in isolating the other variable. In our example, two linear equations were given:

• Equation 1: \(2a + 6z = 4\)
• Equation 2: \(3a - 7z = 6\)

The goal is to manipulate them such that one variable cancels out. By multiplying the first equation by 3 and the second by 2, their coefficients for \(a\) become equal. Then, subtraction allows the \(a\) variable to be eliminated, simplifying the system to an equation with one unknown.

Solving Systems of Equations

Solving systems of equations involves finding a pair of values that satisfy all equations simultaneously. A system might have one solution, infinitely many, or none at all. Each method—be it graphical, substitution, or elimination—provides a strategic approach to tackle these systems effectively.
In our problem, we used the linear combinations approach as part of the elimination method to solve the system. This ensured we systematically worked towards isolating each variable, eventually leading us to the answer.
Practicing solving systems of equations helps develop mathematical reasoning and problem-solving skills. It’s essential for students to grasp the underlying concepts and explore various solving techniques, enhancing their understanding and application in diverse contexts.

Substitution Method

The substitution method is another technique to solve systems of equations, though not used in our specific problem. However, it's a good to understand how it works for other types of systems.
This method involves:

• Solving one of the equations for one variable.
• Substituting this expression into the other equation.
• Simplifying the new equation to find the value of the second variable.
• Finally, using this value to back-substitute and find the first variable.

The substitution method works well when one equation can be easily solved for one variable. This method provides a clear path to the solution, especially useful in nonlinear systems where linear combinations might not be applicable.

Elimination Method

The elimination method is a favorite for many math enthusiasts when dealing with more straightforward linear equations. It involves combining the equations in a way that cancels out one of the variables, making it easy to solve for the remaining unknown.
In the given solution, we used elimination by aligning the coefficients of \(a\) and subtracting the equations, which allowed \(a\) to vanish from the system. By creating equivalent coefficients and then subtracting or adding the equations, one variable was eliminated. This helped simplify the problem from two variables to one, a crucial step to effectively solve the system.
The elimination method is especially powerful as it applies to all linear systems, unlike substitution which might be more cumbersome in some cases. It's about looking for opportunities to simplify and solve efficiently.