Question

Solve each system of equations. If the system has no solution, state that it is inconsistent. $$ \left\\{\begin{array}{l} \frac{1}{x}+\frac{1}{y}=8 \\ \frac{3}{x}-\frac{5}{y}=0 \end{array}\right. $$

Short answer

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

Step-by-step solution

- Substitute and simplify

Rewrite the second equation in terms of one variable. From \(\frac{3}{x} - \frac{5}{y} = 0\), we can express \(\frac{3}{x} = \frac{5}{y}\). This gives \(\frac{x}{3} = \frac{y}{5}\). From here, cross-multiplying gives \(5x = 3y\), so \(y = \frac{5}{3}x\).

- Substitute into the first equation

Now, substitute \(y = \frac{5}{3}x\) into the first equation: \(\frac{1}{x} + \frac{1}{\frac{5}{3}x} = 8\). Simplify the second term to get \(\frac{1}{x} + \frac{3}{5x} = 8\).

- Combine like terms

Combine the fractions by getting a common denominator: \(\frac{5}{5x} + \frac{3}{5x} = 8\). This gives \(\frac{8}{5x} = 8\).

- Solve for x

Solve for \(x\) by cross-multiplying: \(8 = \frac{8}{5x}\) implies that \(8 \times 5x = 8\). Simplify to get \(40x = 8\), hence \(x = \frac{8}{40} = \frac{1}{5}\).

- Solve for y

Substitute \(x = \frac{1}{5}\) back into \(y = \frac{5}{3}x\): \(y = \frac{5}{3} \times \frac{1}{5} = \frac{1}{3}\).

- Verify the solution

To verify, substitute \(x = \frac{1}{5}\) and \(y = \frac{1}{3}\) back into the original equations and check if they hold true. For the first equation: \(\frac{1}{\frac{1}{5}} + \frac{1}{\frac{1}{3}} = 5 + 3 = 8\), and for the second equation: \(\frac{3}{\frac{1}{5}} - \frac{5}{\frac{1}{3}} = 15 - 15 = 0\). Hence, the solution satisfies both equations.

Key concepts

algebra

Algebra is a branch of mathematics dealing with symbols and the rules for manipulating those symbols. In this problem, we use algebra to solve a system of equations, which involves finding the values of variables that satisfy multiple equations simultaneously. A system of linear equations consists of two or more linear equations in the same variables.
For example, to solve the given system: $$\frac{1}{x}+\frac{1}{y}=8$$ and $$\frac{3}{x}-\frac{5}{y}=0$$, we need to find the values of \(x\) and \(y\) that make both equations true at the same time. The goal is to isolate one variable, replace it in the second equation, and solve it step by step until we find both variables.

substitution method

The substitution method involves solving one of the equations for one variable in terms of the other and then substituting that expression into the other equation. In our problem, we start by rearranging the second equation:
$$ \frac{3}{x} - \frac{5}{y} = 0 $$ to get \(\frac{3}{x} = \frac{5}{y}\). We can express this as: \(\frac{x}{3} = \frac{y}{5}\) and then cross-multiply to find the relationship: \(5x = 3y\).
Once we have \(y = \frac{5}{3}x\), we substitute this back into the first equation to eliminate \(y\) and solve for \(x\). This step-by-step isolation and substitution help simplify and break down the problem into more manageable pieces.

cross-multiplication

Cross-multiplication is a method used to eliminate fractions and solve equations involving ratios. It is particularly handy in solving equations of the form \(\frac{A}{B} = \frac{C}{D}\).
In the given problem:
$$ \frac{3}{x} = \frac{5}{y} $$, cross-multiplication helps us eliminate the fractions. By multiplying both sides by the product of the denominators, we get: \( 3y = 5x \). This equation is easier to work with and lets us solve for one of the variables directly.

fractions simplification

Simplifying fractions is essential when solving equations involving fractions. It makes the equations easier to handle and solve. In this problem, after substituting \(y = \frac{5}{3}x\) into the first equation, we need to simplify the resulting expression:
$$ \frac{1}{x} + \frac{1}{\frac{5}{3}x} = 8 $$
We simplify \(\frac{1}{\frac{5}{3}x}\) to \(\frac{3}{5x}\). Combining these with a common denominator, we get \(\frac{5 + 3}{5x} = 8\), leading to: $$ \frac{8}{5x} = 8 $$
Cross-multiplying finally helps us to find that \(x = \frac{1}{5}\). Simplifying fractions at each step ensures clarity and ease in solving.