Solve the Fractional System: 1/3x - 4y = 5 and x + 6y = 9

Q: Why does substitution work better than elimination here?

The second equation already has $ x $ by itself (with coefficient 1), making it perfect for substitution. Elimination would require more steps to align coefficients properly.

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

Substitute both values into both original equations:Equation 1: $ \frac{1}{3}(11) - 4(-\frac{1}{3}) = \frac{11}{3} + \frac{4}{3} = 5 ✓ $Equation 2: $ 11 + 6(-\frac{1}{3}) = 11 - 2 = 9 ✓ $

Q: What if I get confused with the negative fraction?

Take your time with signs! When you have $ y = -\frac{1}{3} $, remember that $ 6y = 6 \times (-\frac{1}{3}) = -2 $. Double-check each arithmetic step.

Q: Can I solve this system using elimination instead?

Yes! You could eliminate $ x $ by multiplying the first equation by 3 to get $ x - 12y = 15 $, then subtracting from $ x + 6y = 9 $. But substitution is more direct here.

Linear Systems with Fractional Coefficients

Solve the above set of equations and choose the correct answer.

$\begin{cases} \frac{1}{3}x-4y=5 \\ x+6y=9 \end{cases}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve the system of equations

00:03 Multiply by 3 to eliminate the fraction

00:11 Now this is the system of equations

00:20 Subtract between the equations

00:31 Collect like terms

00:41 Isolate Y

00:56 Substitute the value of Y to find the value of X

01:12 Isolate X

01:26 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

Solve the above set of equations and choose the correct answer.

$\begin{cases} \frac{1}{3}x-4y=5 \\ x+6y=9 \end{cases}$

Step-by-step solution

To solve this system of equations, we are going to use the substitution method:

Given the equations:

$\begin{cases} \frac{1}{3}x - 4y = 5 \quad \text{(Equation 1)} \\ x + 6y = 9 \quad \text{(Equation 2)} \end{cases}$

First, we solve Equation 2 for $x$ :

$x = 9 - 6y$

Substitute this expression for $x$ into Equation 1:

$\frac{1}{3}(9 - 6y) - 4y = 5$

Multiply through by 3 to eliminate fractions:

$9 - 6y - 12y = 15$

Combine like terms:

$9 - 18y = 15$

Subtract 9 from both sides:

$-18y = 6$

Divide both sides by -18:

$y = -\frac{1}{3}$

Substitute $y = -\frac{1}{3}$ back into the expression for $x$ from Equation 2:

$x = 9 - 6(-\frac{1}{3})$

$x = 9 + 2$

$x = 11$

Thus, the solution to the system of equations is:

$x = 11, y = -\frac{1}{3}$ .

Final Answer

$x=11,y=-\frac{1}{3}$

Key Points to Remember

Essential concepts to master this topic

Method: Use substitution or elimination to solve two-equation systems
Technique: Clear fractions by multiplying: $\frac{1}{3}x$ becomes $x$ when multiplied by 3
Check: Substitute $x = 11, y = -\frac{1}{3}$ into both original equations ✓

Common Mistakes

Avoid these frequent errors

Not multiplying all terms when clearing fractions
Don't multiply just $\frac{1}{3}x$ by 3 and forget the other terms = wrong equation! This breaks the balance and leads to incorrect solutions. Always multiply every single term on both sides by the same number.

Practice Quiz

Test your knowledge with interactive questions

Solve the following equations:

$\begin{cases} 2x+y=9 \\ x=5 \end{cases}$

FAQ

Everything you need to know about this question

Should I clear the fraction first or use substitution directly?

You can do either! In this problem, substitution works well without clearing fractions first. But if fractions make you uncomfortable, multiply the first equation by 3 to get $x - 12y = 15$ before substituting.

Why does substitution work better than elimination here?

The second equation already has $x$ by itself (with coefficient 1), making it perfect for substitution. Elimination would require more steps to align coefficients properly.

How do I check my answer when there are fractions involved?

Substitute both values into both original equations:

Equation 1: $\frac{1}{3}(11) - 4(-\frac{1}{3}) = \frac{11}{3} + \frac{4}{3} = 5 ✓$
Equation 2: $11 + 6(-\frac{1}{3}) = 11 - 2 = 9 ✓$

What if I get confused with the negative fraction?

Take your time with signs! When you have $y = -\frac{1}{3}$ , remember that $6y = 6 \times (-\frac{1}{3}) = -2$ . Double-check each arithmetic step.

Can I solve this system using elimination instead?

Yes! You could eliminate $x$ by multiplying the first equation by 3 to get $x - 12y = 15$ , then subtracting from $x + 6y = 9$ . But substitution is more direct here.

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