Problem 6 - Calculus for Biology and Medicine

Question

(a) Determine the solution of $$ \begin{array}{r} -2 x+3 y=5 \\ a x-y=1 \end{array} $$ in terms of \(a\). (b) For which values of \(a\) are there no solutions, exactly one solution, and infinitely many solutions?

Step-by-Step Solution

Answer

The system has no solution when \(a = \frac{2}{3}\). It has exactly one solution for \(a \neq \frac{2}{3}\). It never has infinitely many solutions.

Step 1: Express Variables

We have the system of equations:1. \(-2x + 3y = 5\)2. \(ax - y = 1\)First, express one variable in terms of the other using Equation (2). Solve for \(y\):\[y = ax - 1\]

Step 2: Substitute in First Equation

Substitute \(y = ax - 1\) into Equation (1):\[-2x + 3(ax - 1) = 5\]Expand and simplify:\[-2x + 3ax - 3 = 5\]\[3ax - 2x - 3 = 5\]\[3ax - 2x = 8\]

Step 3: Simplify and Solve for x

Factor out \(x\) from the terms on the left side:\[x(3a - 2) = 8\]Solve for \(x\):\[x = \frac{8}{3a - 2}\]

Step 4: Solve for y

Use the expression \(x = \frac{8}{3a - 2}\) to find \(y\):\[y = ax - 1 = a\left( \frac{8}{3a - 2} \right) - 1\]Simplify:\[y = \frac{8a}{3a - 2} - 1\]

Step 5: Determine Conditions for Solutions

For the system to have a solution, the denominator in the expressions for \(x\) and \(y\) cannot be zero:\[3a - 2 eq 0\]Solve:\[a eq \frac{2}{3}\]

Step 6: Analyze for No Solutions and Infinitely Many Solutions

For no solutions:The equations must be parallel with no intersection, which occurs when the coefficients of \(x\) and \(y\) are proportional but the constant terms are not. Comparing both equations:\[-2/3 eq a/-1 \text{ and } -5/3 eq 1/1\]Since \(-2/3 = a/-1\) implies \(a = \frac{2}{3}\). For \(a = \frac{2}{3}\), both conditions can't hold, hence no further check is needed.For infinitely many solutions:Equations must be equivalent, which cannot be satisfied with different constant terms unless both equations are identical, which only happens for no values of \(a\) unless intentionally made equivalent.

Key Concepts

Systems of Equations

When working with systems of equations, you are dealing with two or more equations that have common variables. The ultimate goal here is to find a set of values for these variables that satisfy all the equations simultaneously. In our original exercise, we are given a system of two equations involving two variables, \(x\) and \(y\):

• \(-2x + 3y = 5\)
• \(ax - y = 1\)

Each equation represents a line in a two-dimensional space, and solving the system means finding the point(s) where these lines intersect. This intersection point is the solution set \((x, y)\), representing the values that work for all equations at once. Depending on the relationship between the lines (e.g., parallel, overlapping), systems can have one solution, no solution, or infinitely many solutions. Recognizing this relationship is key to understanding and solving systems.

Solving Equations

Solving equations is a process where we look for the value(s) of the variable(s) that satisfy the equation. In the context of systems of equations, solving involves finding a common solution for all equations. Let's take a closer look at how we solved the systsem provided:

• We started by rearranging the second equation \(ax - y = 1\) to isolate one variable. We expressed \(y\) in terms of \(x\): \(y = ax - 1\).
• Next, we plugged this expression for \(y\) into the first equation \(-2x + 3y = 5\), leading to an equation involving only \(x\).

By solving this new equation, we could then easily work back to find the value of \(y\). By plugging our value for \(x\) into the rearranged equation \(y = ax - 1\), we smoothly found our \(y\). This method of substitution helps simplify the process of finding variable values step by step.

Substitution Method

The substitution method is a strategic approach to solving systems of equations, especially useful when one of the equations can be quickly reframed to express a variable in terms of the others. Here's how this method plays out in solving the original system:

• First, pick the easier equation and solve for one variable. In our example, we chose \(ax - y = 1\) because it was straightforward to express \(y\) as \(y = ax - 1\).

By substituting this expression into the first equation, we reduced the system of equations to a single-variable equation. The principal advantage of the substitution method is its simplicity in breaking down complex problems into more manageable parts.

• This makes it easier to see potential solutions or issues, like when denominators would be zero, as when \(3a - 2 = 0\) in the example. Such insights are crucial for understanding under which conditions there's no solution or infinitely many solutions, based on the characteristics of linear equations, making it an invaluable technique.