- Rewrite the inequalities into slope-intercept form: \(y \geq 0.8x - 0.2\) and \(y \leq 2x + 6\).
- Determine the boundary lines \(L_1: y = 0.8x - 0.2\) and \(L_2: y = 2x + 6\).
- Calculate the intersection point by solving the system as a linear equality, resulting in \((-\frac{31}{6}, -\frac{13}{3})\).
- Define the solution region as the set of points \((x, y)\) such that \(\frac{4}{5}x - \frac{1}{5} \leq y \leq 2x + 6\), which is an infinite, wedge-shaped region starting from the intersection point.
Explanation
- Analyzing the System of Inequalities
We are given a system of two linear inequalities involving two variables, \(x\) and \(y\):
\[ \begin{cases} 4x - 5y \leq 1 \ \frac{1}{2}y - x \leq 3 \end{cases} \]
To analyze the solution region, we rewrite each inequality into the slope-intercept form (\(y = mx + b\)). This will allow us to visualize the boundary lines and determine the half-planes they define.
- Transforming the Inequalities
First, we solve the first inequality for \(y\):
\[ 4x - 5y \leq 1 \]
\[ -5y \leq -4x + 1 \]
Dividing by \(-5\) reverses the inequality sign:
\[ y \geq \frac{4}{5}x - \frac{1}{5} \]
Next, we solve the second inequality for \(y\):
\[ \frac{1}{2}y - x \leq 3 \]
\[ \frac{1}{2}y \leq x + 3 \]
Multiplying by \(2\):
\[ y \leq 2x + 6 \]
These two lines, \(L_1: y = \frac{4}{5}x - \frac{1}{5}\) and \(L_2: y = 2x + 6\), act as the boundaries of the solution region.
- Finding the Intersection Point
To find where these boundaries meet, we set the two equations equal to each other:
\[ \frac{4}{5}x - \frac{1}{5} = 2x + 6 \]
Multiplying the entire equation by 5 to clear the fractions:
\[ 4x - 1 = 10x + 30 \]
Rearranging to solve for \(x\):
\[ -6x = 31 \implies x = -\frac{31}{6} \approx -5.17 \]
Now, substitute \(x\) back into the equation for \(L_2\) to find \(y\):
\[ y = 2\left(-\frac{31}{6}\right) + 6 = -\frac{31}{3} + \frac{18}{3} = -\frac{13}{3} \approx -4.33 \]
The two lines intersect at the point \((-\frac{31}{6}, -\frac{13}{3})\).
- Describing the Solution Region
The solution region is the intersection of the two half-planes:
- The region above (or on) the line \(y = \frac{4}{5}x - \frac{1}{5}\).
- The region below (or on) the line \(y = 2x + 6\).
Because the slopes are different (\(m_1 = 0.8\) and \(m_2 = 2\)), the lines intersect. The solution region is an angular, wedge-like section of the plane, bounded by the two intersecting lines at \((-\frac{31}{6}, -\frac{13}{3})\). Any point \((x, y)\) that satisfies both inequalities lies within this infinite, cone-shaped region. The final region consists of all points \((x, y)\) such that \(\frac{4}{5}x - \frac{1}{5} \leq y \leq 2x + 6\).
Examples
In civil engineering or logistics, this type of system of inequalities helps identify "feasible regions" where certain constraints must be satisfied simultaneously. For example, if a company has two manufacturing machines, each with specific operating constraints on the amount of raw materials (x) and power (y) they can consume, these inequalities help determine all possible combinations of material and power that allow both machines to function within their safe operating parameters. By identifying the boundary lines and the intersection point, engineers can visualize the "safe zone" of operation, ensuring that no production limits are exceeded.