Problem 26 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}2 x+3 y=3 \\ 4 x-9 y=-4\end{array}\right)\)

Step-by-Step Solution

Answer

The solution to the system is \(\left(\frac{1}{2}, \frac{2}{3}\right)\).

Step 1: Solve one equation for one variable

We will solve the first equation \(2x + 3y = 3\) for \(x\). Begin by isolating \(x\) on one side:\[2x = 3 - 3y\]Then, divide by 2:\[x = \frac{3 - 3y}{2}\]

Step 2: Substitute the expression into the other equation

Take the expression for \(x\) from Step 1 and substitute it into the second equation \(4x - 9y = -4\).Replace \(x\) with \(\frac{3 - 3y}{2}\):\[4\left(\frac{3 - 3y}{2}\right) - 9y = -4\]

Step 3: Simplify and solve for y

Simplify the equation:\[2(3 - 3y) - 9y = -4\]Distribute the 2:\[6 - 6y - 9y = -4\]Combine like terms:\[6 - 15y = -4\]Subtract 6 from both sides:\[-15y = -10\]Divide by -15:\[y = \frac{2}{3}\]

Step 4: Substitute y back to find x

Now that we have \(y = \frac{2}{3}\), substitute it back into the expression for \(x\):\[x = \frac{3 - 3\left(\frac{2}{3}\right)}{2}\]Calculate:\[x = \frac{3 - 2}{2} = \frac{1}{2}\]

Step 5: Write the solution as an ordered pair

The solution to the system is the values of \(x\) and \(y\) as an ordered pair. Therefore, the solution is:\(\left(\frac{1}{2}, \frac{2}{3}\right)\)

Key Concepts

Linear Equations

Linear equations are foundational in algebra. They represent relationships where each term is either a constant or the product of a constant and a single variable. In this exercise, we deal with two linear equations: \(2x + 3y = 3\) and \(4x - 9y = -4\). Linear equations graph as straight lines and their solutions can often be easily visualized as the intersection of these lines on a coordinate plane.

Key characteristics of linear equations include:

• They have one or two variables.
• Each variable has an exponent of 1.
• They do not contain products or powers of variables.

To solve linear equations, we often use methods like substitution and elimination, which are useful in finding where two lines intersect or to solve more complex systems.

Systems of Equations

A system of equations consists of two or more equations that share common variables. In our exercise, we examined the system formed by the equations \(2x + 3y = 3\) and \(4x - 9y = -4\). The goal is to find values for \(x\) and \(y\) that satisfy both equations simultaneously.

For linear systems, the possible solving outcomes are:

• A single solution point (the lines intersect at one unique point).
• No solution (the lines are parallel and never meet).
• Infinitely many solutions (the lines are coincident, i.e., the same line).

The systems of equations can often be solved using methods like substitution, which was used in this exercise. Recognizing these outcomes is essential in determining the nature of the solutions — whether unique, non-existent, or infinite.

Algebraic Solution Methods

Algebraic solution methods like substitution and elimination are effective for solving systems of linear equations. In this exercise, we focused on the substitution method. This method involves solving one of the equations for one variable and substituting this expression into the other equation.

Here’s a clear breakdown of the substitution method:

• Solve one equation for one variable: Start with the equation \(2x + 3y = 3\) and solve for \(x\), yielding \(x = \frac{3 - 3y}{2}\).
• Substitute the expression into the other equation: Plug \(x = \frac{3 - 3y}{2}\) into \(4x - 9y = -4\) to simplify the equation to one variable.
• Solve for the remaining variable: Simplify and resolve to find \(y\), in this case \(y = \frac{2}{3}\).
• Back-substitute to find the other variable: Use the \(y\) value to solve for \(x\), thus obtaining \(x = \frac{1}{2}\).

Algebraic solution methods like substitution provide a structured approach to solving equations that ensures accuracy and consistency across different problems.