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

(a) \(x = \frac{8}{3a - 2}\), \(y = a\left(\frac{8}{3a - 2}\right) - 1\). (b) No solutions if \(a = \frac{2}{3}\); one solution if \(a \neq \frac{2}{3}\); never infinitely many solutions.

Step 1: Express y in terms of x using the second equation

From the second equation, rearrange it to express \(y\) in terms of \(x\): \(y = ax - 1\).

Step 2: Substitute y in the first equation

Substitute \(y = ax - 1\) into the first equation: \(-2x + 3(ax - 1) = 5\). Simplify this equation: \(-2x + 3ax - 3 = 5\).

Step 3: Solve the simplified equation for x

Combine like terms to form: \((3a - 2)x - 3 = 5\). Add 3 to both sides to get: \((3a - 2)x = 8\). Solve for \(x\) by dividing both sides by \((3a - 2)\): \(x = \frac{8}{3a - 2}\).

Step 4: Determine the value of y using x

Substitute \(x = \frac{8}{3a - 2}\) back into the expression obtained for \(y\) in Step 1: \(y = a\left(\frac{8}{3a - 2}\right) - 1\).

Step 5: Find values of a for no solutions, one solution, or infinite solutions

For no solutions: \(3a - 2 = 0\) (solve to find \(a = \frac{2}{3}\)), which makes the denominator zero in \(x = \frac{8}{3a - 2}\). For exactly one solution: \(3a - 2 eq 0\), meaning any \(a\) except \(\frac{2}{3}\). For infinitely many solutions: The situation occurs when both equations represent the same line. Equate the coefficients from both equations in slope-intercept form. This condition is never satisfied here unless there is a contradiction that aligns both.

Key Concepts

Systems of Linear Equations

When we talk about systems of linear equations, we refer to a collection of two or more linear equations that share the same set of variables. When you solve them, you aim to find the variable values that satisfy all the equations in the system at once.

In our current exercise, we deal with two equations involving variables \(x\) and \(y\):

• \(-2x + 3y = 5\)
• \(ax - y = 1\)

Each equation represents a line on a 2D plane, and the task is to find the point of intersection (if it exists) of these lines. Finding the solution involves techniques such as substitution or elimination, as seen in the step-by-step solution provided.

Solution Conditions

The solution conditions of a system tell us whether there are one, zero, or many solutions. Understanding these is critical for determining the result of a systems of equations exercise.

• One solution exists if the lines intersect at exactly one point, meaning the lines have different slopes.
• No solution exists if the lines are parallel, meaning they never meet.
• Infinitely many solutions occur if both equations actually describe the same line.

These conditions mandate finding relationships between the coefficients of the terms in the equations. For our exercise, solution conditions depend on the value of \(a\) and whether certain denominators become zero or if equations have equivalent slopes.

Unique Solutions

A unique solution means you find a distinct pair \((x, y)\) that satisfies both equations at once. For this to happen, the lines corresponding to the equations need to have different slopes and intersect at just one point.

In our exercise, a unique solution is found when \(3a - 2 eq 0\). This condition ensures that the first and second equation lines are distinct and cross each other only once on the xy-plane. Hence, solving the simplified form gives us the unique coordinates for \(x\) and \(y\). As stated, this applies to all values of \(a\), except when \(a = \frac{2}{3}\).

Infinite Solutions

When a situation arises with infinite solutions, it implies the equations describe the same line, meaning the two lines lay on top of each other entirely. In such a circumstance, any point on this line is a solution to the system.

In the provided exercise, the system does not naturally result in infinitely many solutions under normal circumstances. However, if one wishes to explore this concept further: for infinite solutions, both equations must be equivalent. This demands checking if the ratios of the coefficients of terms in both equations are constant, a situation not achievable with any \(a\) without altering the original terms and rules.