Problem 31 - Precalculus Mathematics for Calculus

Question

Find all solutions of the system of equations. $$\left\\{\begin{array}{c} \frac{2}{x}-\frac{3}{y}=1 \\ -\frac{4}{x}+\frac{7}{y}=1 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{5}\) and \(y = \frac{1}{3}\).

Step 1: Write the equations in terms of fractions

The given system of equations is:\[\frac{2}{x} - \frac{3}{y} = 1\] and \[-\frac{4}{x} + \frac{7}{y} = 1\]. We want to express these equations as functions of \( x \) and \( y \).

Step 2: Rewrite the fractions using substitutions

Let's use the substitutions \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). This transforms the equations into:\[2u - 3v = 1\] and \[-4u + 7v = 1\].

Step 3: Solve the system of equations for \( u \) and \( v \)

We solve \[2u - 3v = 1\] and \[-4u + 7v = 1\]. Start by multiplying the first equation by 2, and the second one as it is, we get:\[\begin{align*} 4u - 6v &= 2\ -4u + 7v &= 1\end{align*}\]. When we add these equations together, the \( u \) terms cancel, yielding:\[v = 3\].

Step 4: Substitute back to find \( u \)

Substitute \( v = 3 \) in the first transformed equation: \[2u - 3(3) = 1\], simplifying gives \[2u - 9 = 1\]. Solving for \( u \), we add 9 to both sides: \[2u = 10\], then divide by 2: \[u = 5\].

Step 5: Convert \( u \) and \( v \) back to \( x \) and \( y \)

Recall the substitutions \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Solving these, we get \( \frac{1}{x} = 5 \), thus \( x = \frac{1}{5} \), and \( \frac{1}{y} = 3 \), thus \( y = \frac{1}{3} \).

Step 6: Verify the solution

Substitute \((x, y) = \left(\frac{1}{5}, \frac{1}{3}\right)\) back into the original equations to check. For the first equation: \[\frac{2}{1/5} - \frac{3}{1/3} = 10 - 9 = 1\]. For the second equation: \[-\frac{4}{1/5} + \frac{7}{1/3} = -20 + 21 = 1\]. Both equations are satisfied, confirming our solution is correct.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations, making it easier to find solutions. It involves substituting expressions to simplify complex equations into more manageable forms.

In our example problem, we're dealing with fractions in the form of \( \frac{2}{x} - \frac{3}{y} = 1 \) and \( -\frac{4}{x} + \frac{7}{y} = 1 \). To apply the substitution method effectively, we introduce new variables. Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \).

This step converts the original system into a pair of simpler linear equations: \( 2u - 3v = 1 \) and \( -4u + 7v = 1 \). This transformation makes the problem easier to handle without the complexity of fractions.

Once rewritten, solving for one variable in terms of the other becomes straightforward. The substitution method highlights how changing perspectives can simplify equation systems. It's especially useful when dealing with fractions or complex expressions.

Linear Equations

Linear equations are equations of the first order and involve two variables. They are represented in the general form \( ax + by = c \).

In solving the system \( 2u - 3v = 1 \) and \( -4u + 7v = 1 \), we deal with linear equations. These equations are called 'linear' because they represent straight lines when graphed on a coordinate plane.

To solve such a system, you can use various methods like substitution, elimination, or graphical methods. In our case, eliminating one of the variables by strategically adding or subtracting the two equations is possible once they're simplified.

Here, multiplying the first equation by 2 results in the systematic elimination of \( u \) when both equations are added. This yields the value of \( v \) directly, and consequently, the value of \( u \) is easily calculated.

Understanding how to manipulate and solve linear equations is crucial, as they are foundational in mathematics and appear frequently in practical applications.

Solution Verification

After solving a system of equations, it's crucial to verify that the solution is correct. This step ensures that no mistakes were made during calculations and that the proposed solution satisfies both original equations.

Given our solution for \( u = 5 \) and \( v = 3 \), we converted these back to the original variables \( x \) and \( y \): \( x = \frac{1}{5} \) and \( y = \frac{1}{3} \).

Verification involves substituting these values back into the original equations. For the first equation \( \frac{2}{1/5} - \frac{3}{1/3} \), we simplify and check if the left side equals the right side, producing \( 10 - 9 = 1 \). Similarly, for the second equation \( -\frac{4}{1/5} + \frac{7}{1/3}, \) the computation yields \( -20 + 21 = 1 \).

Both equations are satisfied, confirming the correctness of our solution. This step underscores the importance of verification to ensure accuracy and increase confidence in the solved results.