Problem 31

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

Verified

Answer

The solutions are $ x = \frac{13}{2} $ and $ y = -\frac{13}{3} $.

1Step 1: Express First Equation in Terms of One Variable

Let's express the first equation, $ \frac{2}{x} - \frac{3}{y} = 1 $, in terms of one variable. First, we express $ \frac{2}{x} $ as $ a $ and $ \frac{3}{y} $ as $ b $. Therefore the equation becomes $ a - b = 1 $. Hence, $ a = 1 + b $, which translates back to $ \frac{2}{x} = 1 + \frac{3}{y} $.

2Step 2: Express Second Equation in Terms of One Variable

For the second equation, $ -\frac{4}{x} + \frac{7}{y} = 1 $, we express $ -\frac{4}{x} $ as $ c $ and $ \frac{7}{y} $ as $ d $. This gives $ c + d = 1 $ so that $ c = 1 - d $ and translates back to $ -\frac{4}{x} = 1 - \frac{7}{y} $.

3Step 3: Eliminate Variables by Multiplying and Subtracting Equations

We already have $ \frac{2}{x} = 1 + \frac{3}{y} $ and $ -\frac{4}{x} = 1 - \frac{7}{y} $. Multiply the first equation by 2: $ 2 \cdot \left( \frac{2}{x} = 1 + \frac{3}{y} \right) $. This leads to $ \frac{4}{x} = 2 + \frac{6}{y} $. Now subtract this from the second equation: $ -\frac{4}{x} = 1 - \frac{7}{y} $. Add them to eliminate $ \frac{4}{x} $, which results in: $ 0 = 3 + \frac{13}{y} $.

4Step 4: Solve for $ y $

Rearrange $ 0 = 3 + \frac{13}{y} $ to solve for $ y $. Subtract 3 from both sides to obtain $ -3 = \frac{13}{y} $. Multiply both sides by $ y $ to get $ -3y = 13 $. Divide by -3 to solve for $ y $, leading to $ y = -\frac{13}{3} $.

5Step 5: Solve for $ x $ Using $ y $

Substitute $ y = -\frac{13}{3} $ back into $ \frac{2}{x} = 1 + \frac{3}{y} $. This becomes $ \frac{2}{x} = 1 - \frac{9}{13} $, simplifying to $ \frac{2}{x} = \frac{4}{13} $. Solving for $ x $, we multiply by $ x $ and then divide by 2, resulting in $ x = \frac{26}{4} = \frac{13}{2} $.

6Step 6: Verify Solution

Verify that $ x = \frac{13}{2} $ and $ y = -\frac{13}{3} $ satisfy both original equations. Substitute into the first and second equations to check: $ \frac{2}{x} - \frac{3}{y} = \frac{2}{\left(\frac{13}{2}\right)} - \frac{3}{\left(-\frac{13}{3}\right)} = 1 $ and $ -\frac{4}{x} + \frac{7}{y} = -\frac{4}{\left(\frac{13}{2}\right)} + \frac{7}{\left(-\frac{13}{3}\right)} = 1 $. Both validate as true, confirming our solutions.

Key Concepts

Equation SolvingSubstitution MethodVariable EliminationRational Equations

Equation Solving

Equations are a mathematical way to represent problems and think of solutions. An equation has an equal sign, showing two expressions are the same. Solving equations means finding values for unknowns that make the equation true. There are many methods to solve equations, including substitution, elimination, and matrices, to name a few. In a system of equations, we're looking at multiple equations at the same time, usually with more than one variable to find. The solution is the set of values that satisfy all equations in the system. Regular practice with different types of equations helps in mastering these concepts and makes you a better problem solver.

Substitution Method

The substitution method is a technique for solving systems of equations, especially handy when dealing with algebraic equations. We express one variable in terms of the other using one equation and substitute this into the other. Let’s digest this with an example. If you have equations like:

Equation 1: $ y = 2x + 3 $
Equation 2: $ x + y = 9 $

You solve for one variable, let's take $ y $ from Equation 1, then use this expression to replace $ y $ in Equation 2. This results in a single equation with one variable:\[ x + (2x + 3) = 9 \]This is easier to solve for $ x $, and after finding $ x $, one substitutes back to find $ y $. Substitution is powerful as it reduces the system into something manageable.

Variable Elimination

Variable elimination, also known as the elimination method, allows for removing one variable to easily solve a system of equations. This is achieved by adding or subtracting equations to cancel out one of the variables.

Rational Equations

Rational equations feature fractions or ratios involving polynomials in their expressions. Solving them involves finding a common denominator or multiplying through by the least common multiple to eliminate fractions, streamlining the equation solving process. Take, for example, the equation:\[ \frac{2}{x} - \frac{3}{y} = 1 \]By identifying and multiplying through by a common denominator of all fractions, you can convert the equation to an easier form without fractions. Care is necessary to avoid undefined situations in fractions, such as division by zero. Introducing variables like $ a = \frac{2}{x} $ and $ b = \frac{3}{y} $ can help to simplify the manipulation and understanding initially.With problems involving rational equations, look out for the values that make denominators zero, as they must be excluded from the set of solutions. Rational equations teach us to handle fractions and open the path to understanding further algebraic structures.

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