Problem 14 - Calculus for Biology and Medicine

Question

In Problems 9-16, reduce the system of equations to upper triangular form and find all the solutions. $$ \begin{aligned} 2 x+3 y &=5 \\ -y &=-2+\frac{2}{3} x \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{11}{4}\), \(y = -\frac{1}{6}\).

Step 1: Simplify Equation 2

Start by simplifying Equation 2, which is originally given as \(-y = -2 + \frac{2}{3}x\). To eliminate the fraction, multiply through by 3 to get:\(-3y = -6 + 2x\). Rewrite this as:\(2x - 3y = 6\).

Step 2: Arrange the Equations in Standard Form

Write both equations from the system in standard form:Equation 1: \(2x + 3y = 5\)Equation 2: \(2x - 3y = 6\).

Step 3: Perform Row Operations to Create Upper Triangular Form

Our goal is to eliminate one of the variables from one of the equations. Subtract Equation 2 from Equation 1 to eliminate \(2x\) from the second equation:\[(2x + 3y) - (2x - 3y) = 5 - 6\].This simplifies to:\[6y = -1\].

Step 4: Solve the Upper Triangular System

Now solve the equation \(6y = -1\) for \(y\):\[y = -\frac{1}{6}\].Substitute \(y = -\frac{1}{6}\) back into Equation 1:\[2x + 3(-\frac{1}{6}) = 5\].Simplify and solve for \(x\):\[2x - \frac{1}{2} = 5\]\[2x = 5 + \frac{1}{2}\]\[2x = \frac{11}{2}\]\[x = \frac{11}{4}\].

Step 5: Verify the Solution

Substitute \(x = \frac{11}{4}\) and \(y = -\frac{1}{6}\) back into the original equations to check:Equation 1: \[2(\frac{11}{4}) + 3(-\frac{1}{6}) = 5\]\[\frac{22}{4} - \frac{1}{2} = 5\]\[\frac{11}{2} - \frac{1}{2} = 5\]\[5 = 5\] (True)Equation 2:Plug into the rearranged equation from Step 1: \(2x - 3y = 6\).\[2(\frac{11}{4}) - 3(-\frac{1}{6}) = 6\]\[\frac{22}{4} + \frac{1}{2} = 6\]\[\frac{11}{2} + \frac{3}{6} = 6\]\[\frac{11}{2} + \frac{1}{2} = 6\]\[6 = 6\] (True)Both equations check out, verifying the solution.

Key Concepts

Systems of Equations

A system of equations consists of two or more equations with a set of variables. The goal is to find the values of these variables that satisfy all equations in the system simultaneously.
To do this, we look for the points of intersection in the graph of the equations, which represents the values where both equations are true.
In linear algebra, particularly, a system of linear equations can be represented by lines in a plane.

• Each equation represents a line.
• The solution is the point where the lines intersect.

For example, given the equations \(2x + 3y = 5\) and \(-y = -2 + \frac{2}{3}x\), we need to find the values of \(x\) and \(y\) that make both equations true at the same time. In our solution process, transforming to standard form and employing row operations helps us tackle such systems systematically.

Upper Triangular Form

Reducing a system of equations to an upper triangular form is a powerful technique.
This form simplifies the process of solving a system of linear equations by focusing on eliminating one variable at a time.
An equation system is in upper triangular form when each equation has fewer variables than the previous one, much like the stairs of an escalator.

• The first equation has the most variables.
• The subsequent equations have progressively fewer variables.

The system, \[ \begin{aligned} 2x + 3y &= 5 \ 2x - 3y &= 6 \end{aligned} \] can be manipulated to get an upper triangular form. This allows us to transform it into a simpler system, where one equation can easily be solved for one variable, leaving a straightforward path to finding the others.

Row Operations

Row operations are an essential step in solving systems of equations, especially when transforming them into upper triangular form.
These operations include:

• Swapping two equations.
• Multiplying an equation by a non-zero constant.
• Adding or subtracting one equation from another.

In this exercise, we used row operations to strategically eliminate variables. For example, we subtract one equation from the other:
\[(2x + 3y) - (2x - 3y) = 5 - 6\]
This effectively removed the \(x\) variable from one of the equations, allowing us to solve for \(y\) directly.

Equation Solutions

Once the upper triangular form is achieved, finding the solution becomes straightforward.
You work from the bottom equation and substitute upwards.

• Solve for one variable.
• Substitute back to find the others.

In our case, the equation \(6y = -1\) led us directly to \(y = -\frac{1}{6}\).
Once \(y\) was known, substituting \(y\) back into the first equation helped us determine \(x\).
This approach guarantees accuracy because each subsequent substitution confirms previous calculations.
The final solutions for our system were \(x = \frac{11}{4}\) and \(y = -\frac{1}{6}\), verified by checking them in the original equations.