Problem 17 - Precalculus: Mathematical for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 3. $$\left\\{\begin{array}{l} \frac{1}{2} x+\frac{1}{3} y=2 \\ \frac{1}{5} x-\frac{2}{3} y=8 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is (10, -9).

Step 1: Clear the fractions

Multiply the first equation by 6 (the least common multiple of 2 and 3) and the second equation by 15 (the least common multiple of 5 and 3) to eliminate fractions.For the first equation: \[6 \times \left(\frac{1}{2}x + \frac{1}{3}y\right) = 6 \times 2 \]Which simplifies to:\[3x + 2y = 12\]For the second equation:\[15 \times \left(\frac{1}{5}x - \frac{2}{3}y\right) = 15 \times 8\]Which simplifies to:\[3x - 10y = 120\]

Step 2: Eliminate one variable

Now let's eliminate one of the variables. Subtract the first equation from the second equation.\[(3x - 10y) - (3x + 2y) = 120 - 12\]This simplifies to:\[-12y = 108\]Divide every term by -12:\[y = -9\]

Step 3: Substitute back to find the other variable

Substitute \(y = -9\) back into one of the original equations to solve for \(x\). We'll use the first equation:\[\frac{1}{2}x + \frac{1}{3}(-9) = 2\]This simplifies to:\[\frac{1}{2}x - 3 = 2\]Add 3 to both sides:\[\frac{1}{2}x = 5\]Multiply both sides by 2:\[x = 10\]

Step 4: Write the solution as an ordered pair

The solution to the system of equations is the ordered pair \((x, y)\). From the calculations, we have \(x = 10\) and \(y = -9\). Therefore, the solution is:\[(10, -9)\]

Key Concepts

Elimination Method

The elimination method is a strategic approach for solving systems of linear equations. It involves combining the equations in such a way that one of the variables is removed, making it easier to solve for the remaining variable. This method is usually preferred when the coefficients of one of the variables are the same or can easily be made the same.

Here's a general overview of how the elimination process works:

• Align the equations one on top of the other for easy comparison.
• Identify a variable to eliminate. It might require adjusting the equations, such as by multiplying, so that adding or subtracting the equations results in the elimination of that variable.
• Add or subtract the equations to wipe out one of the variables.

In our exercise, after clearing fractions (which we'll discuss later), we eliminated `y` by subtracting the first simplified equation from the second. This manipulation allowed us to solely solve for `x` or `y`, simplifying our work significantly.

Fraction Elimination

Linear equations often come with fractions, which can complicate solving them. Fraction elimination is an essential first step in simplifying equations into a manageable form.

The first thing to do is to find the least common multiple (LCM) of the denominators.

• The LCM helps us multiply each term in the equation, thereby eliminating fractions.
• In our problem, the denominators are 2, 3, and 5, so we use 6 and 15 as multiplication factors to clear the fractions from the equations.
• This yields simpler integer-only equations, making calculations cleaner and reducing errors.

By multiplying through to eliminate these fractions, the system of equations becomes more straightforward to work with.

Ordered Pair Solution

Once the equations are solved, the answers are expressed in the form of an ordered pair (x, y). This ordered pair represents the solution to the system of equations, indicating the point where the lines intersect on a graph.

To arrive at the ordered pair, follow these steps:

• Determine the value of one variable using elimination or substitution. In our scenario, we found `y = -9` first.
• Use this value to find the other variable by substituting it back into one of the simplified equations. This led us to `x = 10`.
• Write the solution as an ordered pair in the format `(x, y)`. For our problem, the solution is (10, -9).

This ordered pair solution is crucial as it uniquely satisfies both original equations and confirms the intersection of the lines represented by those equations.