Technical Explanation: Break Down Each Fraction in the System of Equations

Q: Why don't we clear the fractions first like in single equations?

With systems involving $ \frac{1}{x} $ and $ \frac{1}{y} $, clearing fractions creates quadratic terms that are much harder to solve. The elimination method works directly with these fractional terms!

Q: How do I know which equations to add or subtract?

Look for terms that will cancel out. Here, $ +\frac{1}{y} $ and $ -\frac{1}{y} $ have opposite signs, so adding the equations eliminates $ \frac{1}{y} $.

Q: What if both variables have the same sign in both equations?

Then you'd subtract one equation from the other to eliminate a variable. The key is making one variable disappear completely.

Q: Can I solve this using substitution instead?

Yes, but it's more complicated! You'd solve for $ \frac{1}{y} $ from one equation and substitute into the other. Elimination is usually faster for this type of system.

Q: How do I check my answer when there are fractions?

Substitute your values back into both original equations. For $ x = 1, y = 1 $: First equation gives $ \frac{3}{1} + \frac{1}{1} = 4 ✓ $, second gives $ \frac{5}{1} - \frac{1}{1} = 4 ✓ $

Systems of Equations with Fractional Terms

$\frac{3}{x}+\frac{1}{y}=4$

$\frac{5}{x}-\frac{1}{y}=4$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:04 Let's combine the equations

00:14 Let's group terms

00:21 Let's isolate X

00:33 This is the value of X

00:37 Now let's substitute X to find the value of Y

00:51 Let's isolate Y

01:05 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$\frac{3}{x}+\frac{1}{y}=4$

$\frac{5}{x}-\frac{1}{y}=4$

Step-by-step solution

We begin by examining the given system of equations:

$\frac{3}{x} + \frac{1}{y} = 4$ --- (1)

$\frac{5}{x} - \frac{1}{y} = 4$ --- (2)

Let's eliminate $\frac{1}{y}$ by adding equations (1) and (2):

$\left(\frac{3}{x} + \frac{1}{y}\right) + \left(\frac{5}{x} - \frac{1}{y}\right) = 4 + 4$

$\frac{3}{x} + \frac{5}{x} = 8$

$\frac{8}{x} = 8$

Solving for $x$ , we have:

$x = 1$

Now, substitute $x = 1$ back into equation (1):

$\frac{3}{1} + \frac{1}{y} = 4$

$3 + \frac{1}{y} = 4$

$\frac{1}{y} = 1$

Solving for $y$ , we obtain:

$y = 1$

Thus, the solution to the system of equations is $x = 1$ and $y = 1$ .

The correct answer choice is: $x=1,y=1$ .

Final Answer

$x=1,y=1$

Key Points to Remember

Essential concepts to master this topic

Elimination Method: Add or subtract equations to eliminate one variable
Example Technique: Adding $\frac{3}{x} + \frac{1}{y} = 4$ and $\frac{5}{x} - \frac{1}{y} = 4$ eliminates $\frac{1}{y}$
Verification: Substitute $x = 1, y = 1$ back: $\frac{3}{1} + \frac{1}{1} = 3 + 1 = 4$ ✓

Common Mistakes

Avoid these frequent errors

Trying to clear fractions before using elimination
Don't multiply each equation by xy to eliminate denominators first = creates complicated quadratic terms! This makes the system much harder to solve. Always use elimination or substitution directly with the fractional terms as they are.

Practice Quiz

Test your knowledge with interactive questions

$\begin{cases} x+y=8 \\ x-y=6 \end{cases}$

FAQ

Everything you need to know about this question

Why don't we clear the fractions first like in single equations?

With systems involving $\frac{1}{x}$ and $\frac{1}{y}$ , clearing fractions creates quadratic terms that are much harder to solve. The elimination method works directly with these fractional terms!

How do I know which equations to add or subtract?

Look for terms that will cancel out. Here, $+\frac{1}{y}$ and $-\frac{1}{y}$ have opposite signs, so adding the equations eliminates $\frac{1}{y}$ .

What if both variables have the same sign in both equations?

Then you'd subtract one equation from the other to eliminate a variable. The key is making one variable disappear completely.

Can I solve this using substitution instead?

Yes, but it's more complicated! You'd solve for $\frac{1}{y}$ from one equation and substitute into the other. Elimination is usually faster for this type of system.

How do I check my answer when there are fractions?

Substitute your values back into both original equations. For $x = 1, y = 1$ : First equation gives $\frac{3}{1} + \frac{1}{1} = 4 ✓$ , second gives $\frac{5}{1} - \frac{1}{1} = 4 ✓$

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