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Solve each system. $$\begin{aligned} &\frac{2}{x}+\frac{3}{2 y}=\frac{11}{4}\\\ &\frac{5}{2 x}-\frac{2}{y}=\frac{3}{2} \end{aligned}$$

Short Answer

Expert verified
The solution is \( x = 1 \) and \( y = 2 \).

Step by step solution

01

Clear the fractions

Multiply both equations by the least common multiple of the denominators to clear the fractions. For the first equation, multiply by 4: \[ 4 \times \frac{2}{x} + 4 \times \frac{3}{2y} = 4 \times \frac{11}{4} \]which simplifies to: \[ \frac{8}{x} + \frac{6}{y} = 11 \].For the second equation, multiply by 2: \[ 2 \times \frac{5}{2x} - 2 \times \frac{2}{y} = 2 \times \frac{3}{2} \]which simplifies to: \[ \frac{5}{x} - \frac{4}{y} = 3 \].
02

Substitute

Let's introduce new variables to simplify calculations. Let \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). The equations become:\[ 8u + 6v = 11 \] and \[ 5u - 4v = 3 \].
03

Solve for one variable

Solve the second equation for \( u \):\[ 5u - 4v = 3 \quad \rightarrow \ \ 5u = 3 + 4v \quad \rightarrow \ \ u = \frac{3 + 4v}{5} \].
04

Substitute back into the first equation

Substitute \( u = \frac{3 + 4v}{5} \) into the first equation:\[ 8 \times \frac{3 + 4v}{5} + 6v = 11 \quad \rightarrow \ \ \frac{24 + 32v}{5} + 6v = 11 \quad \rightarrow \ \ 24 + 32v + 30v = 55 \quad \rightarrow \ \ 24 + 62v = 55 \quad \rightarrow \ \ 62v = 31 \quad \rightarrow \ \ v = \frac{31}{62} = \frac{1}{2} \].
05

Solve for the other variable

Substitute \( v = \frac{1}{2} \) back into \( u = \frac{3 + 4v}{5} \):\[ u = \frac{3 + 4 \times \frac{1}{2}}{5} \quad \rightarrow \ \ u = \frac{3 + 2}{5} \quad \rightarrow \ \ u = \frac{5}{5} = 1 \].
06

Resubstitute into original variables

Remember that \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \). Substitute back:\[ \frac{1}{x} = 1 \quad \rightarrow \ \ x = 1 \] and \[ \frac{1}{y} = \frac{1}{2} \quad \rightarrow \ \ y = 2 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

substitution method
The substitution method is an effective way to solve systems of equations by replacing one variable with an expression involving another variable. This simplifies the system, making it easier to solve. For example, if you have two equations, you can solve one equation for a variable and then substitute that expression into the other equation.
This way, you're working with a single variable, which often simplifies the solving process.

  1. Step 1: Solve one of the equations for one of the variables.
  2. Step 2: Substitute this expression into the other equation.
  3. Step 3: Solve for the remaining variable.
  4. Step 4: Substitute back to find the other variable.
This method is especially useful when dealing with linear equations or when one variable is easy to isolate.
fraction elimination
Fraction elimination involves getting rid of the fractions in an equation, making it simpler to work with. This is typically done by finding the least common multiple (LCM) of the denominators and then multiplying every term in the equation by the LCM.
Consider the equation \(\frac{2}{x} + \frac{3}{2y} = \frac{11}{4}\). The LCM of the denominators (x, 2y, and 4) helps to clear the fractions:
For example:
\[4 \times \frac{2}{x} + 4 \times \frac{3}{2y} = 4 \times \frac{11}{4}\] Simplifying yields: \(\frac{8}{x} + \frac{6}{y} = 11\).
After clearing the fractions, the system is simpler and more straightforward to solve.
Follow these steps:
  • Identify the denominators in the equations.
  • Find the LCM of the denominators.
  • Multiply each term by the LCM.
  • Simplify the resulting equations.
This technique makes equations less cumbersome and easier to manipulate.
linear equations
Linear equations represent relationships where each term is either a constant or the product of a constant and a single variable. The general form is \(ax + by + c = 0\), where \(a, b,\) and \(c\) are constants.
Solving systems of linear equations means finding values for the variables that satisfy all given equations simultaneously.
Key steps include:
  • Setting equations in standard form \(ax + by = c\).
  • Using methods like substitution or elimination to solve for variables.
  • Checking solutions by substituting back into the original equations.
When dealing with a system of linear equations, it's beneficial to simplify the equations by clearing any fractions and introducing new variables if necessary, as these steps simplify the calculation process.
Common solutions methods like the substitution method can then be used more effectively when the system consists of linear equations.
introduction of new variables
Introducing new variables can simplify complex equations, making them easier to solve. In systems of equations, new variables can replace complicated terms or expressions.
For instance, in the given system: \( \frac{2}{x} + \frac{3}{2y} = \frac{11}{4} \) and \(\frac{5}{2x} - \frac{2}{y} = \frac{3}{2} \),
introducing variables like \( u = \frac{1}{x} \) and \( v = \frac{1}{y} \), transforms the system into simpler linear equations: \( 8u + 6v = 11 \) and \( 5u - 4v = 3 \).
Steps typically include:
  • Identifying complicated terms or fractions to be replaced.
  • Introducing new variables for these terms.
  • Transforming the original equations using these new variables.
  • Solving the simpler system.
After solving for the new variables, reverse the substitution to find the original variables. This approach can be highly effective in streamlining and solving systems of equations.

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Most popular questions from this chapter

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