Problem 44 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} y=-2 x-165 \\ x=\frac{2}{3} y+5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system is \( x = -45 \) and \( y = -75 \).

Step 1: Understand the System Given

We are provided with two equations in a system: \[ 1. \ y = -2x - 165 \] \[ 2. \ x = \frac{2}{3}y + 5 \] Our goal is to find the intersection point, if it exists, that satisfies both equations.

Step 2: Substitute y from Equation (1) into Equation (2)

Since we know the value of \( y \) from the first equation, \( y = -2x - 165 \), we can substitute this into the second equation:\[ x = \frac{2}{3}(-2x - 165) + 5 \]

Step 3: Simplify the Substitution

Let's simplify the equation by distributing \( \frac{2}{3} \):\[ x = \frac{2}{3}(-2x) + \frac{2}{3}(-165) + 5 \]\[ x = -\frac{4}{3}x - 110 + 5 \]

Step 4: Solve for x

Combine and rearrange terms:\[ x + \frac{4}{3}x = -110 + 5 \]\[ \frac{3}{3}x + \frac{4}{3}x = -105 \]\[ \frac{7}{3}x = -105 \] Multiply both sides by \( \frac{3}{7} \) to solve for \( x \):\[ x = -105 \times \frac{3}{7} \]\[ x = -45 \]

Step 5: Find y using x's value

Substitute \( x = -45 \) back into the first equation to find \( y \):\[ y = -2(-45) - 165 \]\[ y = 90 - 165 \]\[ y = -75 \]

Step 6: Verify the Solution

Use both \( x = -45 \) and \( y = -75 \) in the original equations to ensure they hold:1. \( y = -2x - 165 \rightarrow -75 = -2(-45) - 165 \rightarrow -75 = 90 - 165 \rightarrow -75 = -75 \)2. \( x = \frac{2}{3}y + 5 \rightarrow -45 = \frac{2}{3}(-75) + 5 \rightarrow -45 = -50 + 5 \rightarrow -45 = -45 \)Both are satisfied.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of linear equations. It involves solving one of the equations for one variable and then plugging that expression into the other equation. This allows us to find the value of one variable, which we can then substitute back to find the other variable.
In our exercise, we began with two equations:

• \( y = -2x - 165 \)
• \( x = \frac{2}{3}y + 5 \)

We started by solving the first equation for \( y \). Then, we substituted that expression into the second equation. Now, our second equation had only one variable, \( x \), making it easier to solve. Once we found \( x \), we substituted it back into the expression for \( y \) to find its corresponding value. It’s a straightforward method that streamlines the process and makes it manageable.

Dependent Equations

Dependent equations occur in a system of equations where both equations describe the same line. This means that instead of intersecting at a single point, every point on one line is also a point on the other. When this happens, the system does not have a unique solution; instead, it has infinitely many solutions.

In our given exercise, the equations represent two different lines in the plane, since after substitution and solving, we arrived at distinct values for \( x \) and \( y \). Thus, the dependent situation does not apply here, but knowing this concept helps in quickly identifying similar lines in other problems.

Consistent System

A consistent system is one where there is at least one set of values for the variables that satisfies all equations simultaneously. In simple terms, the lines represented by the equations intersect in at least one point in the coordinate plane.

In our solved exercise, the result was \( x = -45 \) and \( y = -75 \). When we substitute these values back into both original equations, they satisfy each equation, confirming that this system is consistent. If the lines intersected at every point (were identical), it would also be consistent but specifically dependent. Our lines intersect at precisely one point, signifying a single unique solution.