Problem 27 - Contemporary Precalculus

Question

Use the elimination method to solve the system. $$\begin{aligned}&\frac{1}{x}-\frac{3}{y}=2\\\&\frac{2}{x}+\frac{1}{y}=3\end{aligned}$$

Step-by-Step Solution

Answer

Answer: The solutions to the given system of equations are (0,0) and (5/3, 5/9).

Step 1: Multiply both equations by the correct constants to eliminate terms

In this case, our goal is to eliminate either the x or y terms, so we need to find a common denominator for each variable. Find the least common multiple (LCM) of the x and y terms to make the elimination process easier. The LCM of x and y will be xy. Multiply both equations by xy. $$\begin{aligned} \text{Equation 1:}&\quad xy\left(\frac{1}{x}-\frac{3}{y}\right)=2xy\\\\ \text{Equation 2:}&\quad xy\left(\frac{2}{x}+\frac{1}{y}\right)=3xy \end{aligned}$$

Step 2: Simplify the multiplied equations

Simplify both equations after multiplying with LCM. This will give us: $$\begin{aligned} \text{Equation 1:}&\quad y(1)-x(3)=2xy\\\\ \text{Equation 2:}&\quad y(2)+x(1)=3xy \end{aligned}$$

Step 3: Make the coefficients of either x or y the same in both equations

To eliminate either x or y, we need their coefficients to be the same. In this case, let's multiply equation 1 by 2 so that the coefficients of x in both equations are the same. $$\text{Equation 1:}\quad 2(y-3x)=4xy$$ Now our system looks like this: $$\begin{aligned} \text{Equation 1:}&\quad 2(y-3x)=4xy\\\\ \text{Equation 2:}&\quad 2y+x=3xy \end{aligned}$$

Step 4: Subtract Equation 2 from Equation 1

Now that the coefficients of x are the same in both equations, we can subtract Equation 2 from Equation 1 to eliminate the x variables and solve for y. $$\begin{aligned} \text{Equation 1 - Equation 2:} \ &[2(y-3x)-x] - [2y+x]= [4xy-3xy] \\ &(2y-7x)=-xy \end{aligned}$$ Now, isolate y and solve for it: $$\begin{aligned} y(7x-x) &= 2y\\ y(6x) &= 2y\\ y &= \frac{1}{3}x \end{aligned}$$

Step 5: Substitute the value of y in either Equation 1 or Equation 2 and solve for x

Substitute y in Equation 2: $$\begin{aligned} 2\left(\frac{1}{3}x\right) + x &= 3x\left(\frac{1}{3}x\right) \\ \frac{2}{3}x + x &= x^2 \\ \frac{5}{3}x &= x^2 \end{aligned}$$ Now, solve for x: $$\begin{aligned} x^2-\frac{5}{3}x &= 0\\ x\left(x-\frac{5}{3}\right) &= 0 \end{aligned}$$ So, \(x=0\) or \(x=\frac{5}{3}\).

Step 6: Substitute the values of x back into the equation for y and find the corresponding y values

For \(x=0\): $$y=\frac{1}{3}(0)=0$$ For \(x=\frac{5}{3}\): $$y=\frac{1}{3}\left(\frac{5}{3}\right)=\frac{5}{9}$$

Step 7: Write the solution as ordered pairs

The solutions to the given system of equations are: $$(x,y) = \left\{(0,0), \left(\frac{5}{3}, \frac{5}{9}\right)\right\}$$

Key Concepts

Systems of Equations

When you're dealing with systems of equations, you're essentially working with two or more equations that share the same variables. The solution to these systems is any set of variable values that make both original equations true simultaneously. There are different methods to solve systems, including substitution, elimination, and graphing. Choosing the right method often depends on the structure of the system.

In the given exercise, we have a system where:

• Equation 1: \(\frac{1}{x} - \frac{3}{y} = 2\)
• Equation 2: \(\frac{2}{x} + \frac{1}{y} = 3\)

These equations need to be solved together to find the values of \(x\) and \(y\) that satisfy both equations at once. This requires careful algebraic manipulation to isolate and solve for each variable.

Solving Equations

Solving equations within systems typically involves focusing on one variable first to make the problem more manageable. With the elimination method, the goal is to combine the equations in such a way that one variable is cancelled out, leaving a single equation with one unknown.

In the step-by-step solution, we begin by multiplying both equations by a common denominator, \(xy\), to clear the fractions. This simplifies our equations making it easier to work with:

\[\text{Transformed Equation 1: } y(1) - x(3) = 2xy\] \[\text{Transformed Equation 2: } y(2) + x(1) = 3xy\]

From here, further manipulation is required to make the coefficients of either \(x\) or \(y\) in both equations the same. This allows us to subtract one equation from the other, effectively removing one variable and enabling us to solve for the remaining variable.

Algebraic Manipulation

Algebraic manipulation is a crucial skill when dealing with complex systems of equations. It involves rearranging, combining, and simplifying equations to isolate variables.

For example, in simplifying our system, multiplying one of the equations by a constant (in this case, by 2) helped us align the coefficients of \(x\) to be the same in both equations. After this multiplication, the system was final:

• Transformed Equation 1: \(2(y - 3x) = 4xy\)
• Equation 2: \(2y + x = 3xy\)

Subtracting the second from the first equation eliminated \(x\), leading us to an equation with only \(y\) as the variable.

Through careful manipulation, we were able to find that \(y = \frac{1}{3}x\). Substituting back allowed us to solve for definite values of \(x\) and \(y\), providing solutions to the original problem: \((x,y) = (0,0)\) or \(\left( \frac{5}{3}, \frac{5}{9} \right)\).