Problem 28 - Algebra and Trigonometry

Question

\(21-48=\) Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example \(6 .\) $$ \left\\{\begin{array}{l}{4 x-3 y=28} \\ {9 x-y=-6}\end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \((-2, -12)\).

Step 1: Begin with the System of Equations

The system of equations is \( \begin{align*} 4x - 3y &= 28 \ 9x - y &= -6 \end{align*} \). Our goal is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve for y in the Second Equation

Take the second equation, \(9x - y = -6\), and solve for \(y\). Add \(y\) to both sides and then add \(6\) to each side to isolate \(y\). The rewritten equation is \( y = 9x + 6 \).

Step 3: Substitute y in the First Equation

Substitute \(y = 9x + 6\) into the first equation, \(4x - 3y = 28\), giving \(4x - 3(9x + 6) = 28\). Simplify by distributing \(-3\), resulting in \(4x - 27x - 18 = 28\).

Step 4: Simplify the Equation

Combine like terms from \(4x - 27x - 18 = 28\) to get \(-23x - 18 = 28\). Add \(18\) to both sides to obtain \(-23x = 46\).

Step 5: Solve for x

Divide both sides of \(-23x = 46\) by \(-23\) to find \(x\). Thus, \(x = -2\).

Step 6: Substitute x Back to Solve for y

Using \(x = -2\) in the equation \(y = 9x + 6\), substitute to find \(y = 9(-2) + 6 = -18 + 6 = -12\).

Step 7: Write the Solution as an Ordered Pair

The solution to the system of equations is \((x, y) = (-2, -12)\).

Key Concepts

Solving Systems of Linear Equations

A system of linear equations consists of two or more linear equations with the same set of variables. The main goal is to find the values of these variables that satisfy all equations in the system simultaneously. In our example, we have the system: \[ \begin{align*} 4x - 3y &= 28 \ 9x - y &= -6 \end{align*} \] To solve such a system, you can use several methods, including graphing, substitution, or elimination. Each method has its own advantages, but they all aim to find the point, often expressed as an ordered pair, where all equations intersect. When dealing with systems, checking the solution by substituting the values back into the original equations ensures accuracy. It is crucial to determine if the system has a single solution, no solution, or infinitely many solutions.

Substitution Method

The substitution method is a straightforward approach to solving systems of linear equations. This technique involves solving one of the equations for one variable and then substituting that expression into the other equation.- **Step 1**: Choose one of the equations and solve for one variable in terms of the other. - **Step 2**: Substitute this expression into the other equation.- **Step 3**: Solve for the remaining variable.- **Step 4**: Substitute back to find the other variable.In our exercise, we first solved the second equation for \( y \): \( y = 9x + 6 \). Then, we substituted this into the first equation, transforming it to a single-variable equation that could be simplified and solved. This method works well when one of the equations is easily rearranged.

Ordered Pairs

Ordered pairs are a way to represent the solution to a system of equations. They are written in the form \((x, y)\), where \(x\) and \(y\) are the coordinates that satisfy both equations in the system. For a set of linear equations, this ordered pair represents the point where the lines intersect on a graph. In our solved exercise, the ordered pair \((-2, -12)\) is the solution meaning that when \(x = -2\), \(y = -12\) satisfies both original equations. Understanding ordered pairs is key in visualizing and verifying the solution graphically.

Algebraic Solutions

Algebraic solutions refer to finding the values of variables using algebraic manipulations rather than graphically. These solutions are precise and rely on applying arithmetic operations, like addition, subtraction, multiplication, and division. The step-by-step process shown in the solution is an example of using algebra to solve a set of linear equations. It involves:

• Isolating one variable using one equation.
• Substituting it into another equation to solve for the second variable.
• Using the result to find the first variable.

Algebraic solutions help solve complex systems without needing a visual graph. They are essential tools, especially when the graphical representation of equations is impractical.