Problem 1 - Beginning and Intermediate Algebra

Question

If you were asked to solve this system by substitution, why would it be easiest to begin by solving for y in the second equation? $$\begin{array}{c}6 x-2 y=-5 \\\3 x+y=4\end{array}$$

Step-by-Step Solution

Answer

It would be easiest to begin by solving for y in the second equation because it has only one y term and fewer coefficients which can simplify the process. By solving for y in \(3x + y = 4\), we get \(y = 4 - 3x\). This expression of y can then be substituted into the first equation and the system can be solved for both x and y values.

Step 1: Solve the second equation for y

To solve the equation \(3x + y = 4\) for y, we will isolate the y term by subtracting 3x from both sides of the equation: \(y = 4 - 3x\) Now we have the value of y in terms of x.

Step 2: Substitute the expression of y into the first equation

We will now substitute the expression for y, \(4 - 3x\), into the first equation in place of y: \(6x - 2(4 - 3x) = -5\)

Step 3: Solve for x

Now, we need to solve the equation for x: \(6x - 8 + 6x = -5\) Combine the x terms and the constant terms: \(12x - 8 = -5\) Add 8 to both sides: \(12x = 3\) Now, divide by 12: \(x = \frac{1}{4}\)

Step 4: Solve for y

Substitute the value of x into the expression for y we found in Step 1: \(y = 4 - 3(\frac{1}{4})\) Now, solve for y: \(y = 4 - \frac{3}{4}\) Add the fractions: \(y = \frac{16}{4} - \frac{3}{4} = \frac{13}{4}\) Thus, the solution of the system of equations is \(x = \frac{1}{4}\) and \(y = \frac{13}{4}\).

Key Concepts

System of Equations

A system of equations consists of two or more equations that share common variables. The goal is to find values for the variables that make all the equations true simultaneously.

This concept is crucial because it allows us to solve problems involving multiple conditions or constraints. For example, consider the following system of equations:

• 6x - 2y = -5
• 3x + y = 4

In this example, both equations have the variables \( x \) and \( y \). We need to find a pair of values that make both equations true at the same time.

Solving systems of equations can be done using several methods, such as substitution, elimination, or graphing. The substitution method is particularly useful when one of the equations is easily solvable for one of the variables, as it simplifies the process substantially.

Isolation of Variables

Isolation of variables refers to manipulating an equation to express one of the variables completely in terms of the others. This is often the first step in the substitution method.

In the context of the provided system of equations, it is often useful to begin by isolating one variable in the simplest equation. In our example, the second equation, \(3x + y = 4\), looks straightforward for isolation. We can solve for \( y \) easily by subtracting \( 3x \) from both sides:

• \( y = 4 - 3x \)

Having \( y \) alone on one side makes it easier to substitute this expression into another equation. The process of isolating \( y \) eliminates the variable from one equation, simplifying our system and making it easier to solve for the other variable.

Solving for y

Solving for \( y \) is a straightforward process where the goal is to find the numerical value of \( y \) for which the equations hold true. In the substitution method, once we have \( y \) isolated, we substitute this expression into the other equation.

For instance, after isolating \( y \) as \( y = 4 - 3x \), we substitute it back into the other equation:

• \(6x - 2(4 - 3x) = -5\)

Simplifying and solving for \( x \), we find \( x = \frac{1}{4} \). Then, we use this \( x \) value in the equation \( y = 4 - 3x \) to find \( y \):

• \( y = 4 - 3\left(\frac{1}{4}\right) = \frac{13}{4} \)

Thus, the solution of the system is \( x = \frac{1}{4} \) and \( y = \frac{13}{4} \).

This demonstrates not only how to isolate variables but also the practical utility of doing so for problem-solving.