Problem 106 - College Algebra with Modeling and Visualization

Question

Solve the system of linear equations $$(a) graphically,\quad(b) numerically,\quad and\quad(c) symbolically.$$ $$\begin{aligned}x-4 y &=15 \\\3 x-2 y &=15\end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 3\) and \(y = -3\).

Step 1: Rearrange Equations for Graphing

Start with the first equation, \(x - 4y = 15\). Rearrange it to the form \(y = mx + b\) for graphing: \(y = \frac{1}{4}x - \frac{15}{4}\). Next, rearrange the second equation, \(3x - 2y = 15\), similarly: \(y = \frac{3}{2}x - \frac{15}{2}\).

Step 2: Graph the Equations

Plot both equations on a coordinate plane. The first line has a slope of \(\frac{1}{4}\) and a y-intercept of \(-\frac{15}{4}\). The second line has a slope of \(\frac{3}{2}\) and a y-intercept of \(-\frac{15}{2}\). Graph these lines to find their intersection point.

Step 3: Find Intersection for Graphical Solution

Observe the graph to find where the two lines intersect. This intersection is the graphical solution to the system.

Step 4: Substitute to Solving Numerically

Substitute one variable in terms of the other using one of the equations. From the first equation, express \(x = 4y + 15\). Substitute \(x = 4y + 15\) into the second equation \(3x - 2y = 15\). This gives: \(3(4y + 15) - 2y = 15\).

Step 5: Solve the Equation Numerically

Expand the substituted equation: \(12y + 45 - 2y = 15\). Combine like terms to get \(10y + 45 = 15\). Solve for \(y\): \(10y = 15 - 45\) which simplifies to \(10y = -30\). Therefore, \(y = -3\).

Step 6: Solve for x Numerically

With \(y = -3\), substitute back into \(x = 4y + 15\). Thus, \(x = 4(-3) + 15 = -12 + 15 = 3\).

Step 7: Verify Solution Symbolically

Substitute \(x = 3\) and \(y = -3\) into the original equations: \(3 - 4(-3) = 15\) and \(3(3) - 2(-3) = 15\). Both equations hold true, confirming the solution is correct.

Key Concepts

Graphical Solution

When solving a system of linear equations graphically, the idea is to represent each equation as a line on the coordinate plane. Both equations need to be rearranged to the form \( y = mx + b \), where \( m \) is the slope and \( b \) is the y-intercept, to make graphing straightforward.
Start by converting the given system of equations:

• For \( x - 4y = 15 \), rearrange to get \( y = \frac{1}{4}x - \frac{15}{4} \).
• For \( 3x - 2y = 15 \), rearrange to get \( y = \frac{3}{2}x - \frac{15}{2} \).

Next, plot these lines on a graph. The first line rises moderately as it moves to the right, while the second line rises much faster due to its steeper slope.
The graphical solution is found at the intersection of these two lines. The point where they cross is the set of values for \( x \) and \( y \) that satisfy both equations simultaneously. This visual method allows for an intuitive understanding of the solution.

Numerical Solution

Numerical solutions involve finding exact values for the variables by substituting and solving the equations mathematically. This method involves expressing one variable in terms of another and substituting it into the second equation.

• From the first equation, express \( x \) as \( x = 4y + 15 \).

Substitute this expression into the second equation \( 3x - 2y = 15 \), resulting in:

• \( 3(4y + 15) - 2y = 15 \).

Simplify the equation by expanding and combining like terms to:

• \( 10y + 45 = 15 \), leading to \( 10y = -30 \), thus \( y = -3 \).

Having found \( y \), substitute back into the expression for \( x \):

• \( x = 4(-3) + 15 = 3 \).

In this way, the numerical solution gives us \( x = 3 \) and \( y = -3 \). This method provides precise values derived through algebraic manipulation.

Symbolic Solution

Symbolic solutions provide a way to verify the correctness of our found solutions by substituting the determined values back into the original equations. It acts as a necessary confirmation step to ensure everything adds up correctly.
Substitute \( x = 3 \) and \( y = -3 \) into the original equations:

• For \( x - 4y = 15 \), substituting gives \( 3 - 4(-3) = 15 \), simplifying to \( 3 + 12 = 15 \).
• For \( 3x - 2y = 15 \), substituting gives \( 3(3) - 2(-3) = 15 \), simplifying to \( 9 + 6 = 15 \).

Both equations hold true, confirming that our found solution of \( x = 3 \) and \( y = -3 \) satisfies the entire system of equations. Symbolically verifying the solution ensures that no missteps occurred during manipulation or calculation.