Solving Simultaneous Equations with Fractions: Substitution Method Challenge

Q: Why do we need to clear fractions first?

Clearing fractions makes the equations much easier to work with! Without fractions, you can focus on solving the system instead of managing complex denominators and arithmetic.

Q: What if I get different LCDs for each equation?

That's normal! Each equation can have its own LCD. Multiply the first equation by its LCD, then multiply the second equation by its LCD. Work with each equation separately.

Q: Why did the second equation give us x directly?

After clearing fractions, the y terms canceled out: $ 2x - 4y + 4x + 4y = 150 $ became $ 6x = 150 $. This lucky cancellation made substitution very easy!

Q: How can I check if my final answer is correct?

Substitute $ x = 25 $ and $ y = -153 $ back into both original equations (with fractions). If both sides equal 12 and 15 respectively, your solution is correct!

Simultaneous Equations with Complex Fractions

Solve the following equations for x and y using the substitution method.

$\begin{cases} \frac{-x+3y}{6}+\frac{5x-y}{3}=12 \\ \frac{2x-4y}{10}-\frac{-2x-2y}{5}=15 \end{cases}$

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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 6 to eliminate fractions

00:16 Open parentheses properly, multiply by each factor

00:25 Collect like terms

00:29 Isolate Y

00:33 This is Y expressed in terms of X

00:40 Multiply by 10 to eliminate fractions

00:51 Open parentheses properly, multiply by each factor

01:01 Collect like terms

01:11 This is the value of X

01:17 Substitute the value of X to find Y

01:27 And this is the solution to the problem

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

Solve the following equations for x and y using the substitution method.

$\begin{cases} \frac{-x+3y}{6}+\frac{5x-y}{3}=12 \\ \frac{2x-4y}{10}-\frac{-2x-2y}{5}=15 \end{cases}$

Step-by-step solution

To solve the system of equations using the substitution method, follow these detailed steps:

We start with the given system:

\begin{cases} \frac{-x+3y}{6}+\frac{5x-y}{3} = 12 \\ \frac{2x-4y}{10}-\frac{-2x-2y}{5} = 15 \end{cases}

Step 1: Clear Fractions

For the first equation, multiply through by 6 (the least common multiple of 6 and 3):

(-x + 3y) + 2(5x - y) = 72

-x + 3y + 10x - 2y = 72

9x + y = 72 \quad \text{(Equation 1)}

For the second equation, multiply through by 10 (the LCM of 10 and 5):

\frac{2x-4y}{10} \times 10 - \frac{-2x-2y}{5} \times 10 = 150

(2x - 4y) + (-4x - 4y) = 150

2x - 4y + 4x + 4y = 150

6x = 150

x = 25 \quad \text{(Equation 2)}

Step 2: Substitute & Solve

Since we have $x = 25$ from Equation 2, substitute $x = 25$ into Equation 1:

9(25) + y = 72

225 + y = 72

y = 72 - 225

y = -153

Therefore, the solution to the system is $\boxed{x = 25, y = -153}$ .

Final Answer

$x=25,y=-153$

Key Points to Remember

Essential concepts to master this topic

Clearing Fractions: Multiply each equation by LCD to eliminate all denominators
Substitution Method: From $6x = 150$ , get $x = 25$ then substitute
Verification: Check $x = 25, y = -153$ in both original equations ✓

Common Mistakes

Avoid these frequent errors

Not multiplying every term by the LCD
Don't multiply only the fractions by the LCD and leave other terms unchanged = incorrect equation! This creates unbalanced equations with wrong coefficients. Always multiply every single term in the equation by the LCD, including constants.

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

Why do we need to clear fractions first?

Clearing fractions makes the equations much easier to work with! Without fractions, you can focus on solving the system instead of managing complex denominators and arithmetic.

How do I find the LCD for each equation?

Look at all denominators in each equation separately. For equation 1: denominators are 6 and 3, so LCD = 6. For equation 2: denominators are 10 and 5, so LCD = 10.

What if I get different LCDs for each equation?

That's normal! Each equation can have its own LCD. Multiply the first equation by its LCD, then multiply the second equation by its LCD. Work with each equation separately.

Why did the second equation give us x directly?

After clearing fractions, the y terms canceled out: $2x - 4y + 4x + 4y = 150$ became $6x = 150$ . This lucky cancellation made substitution very easy!

How can I check if my final answer is correct?

Substitute $x = 25$ and $y = -153$ back into both original equations (with fractions). If both sides equal 12 and 15 respectively, your solution is correct!

What if my answer has large numbers like -153?

Large numbers are fine! Systems with fractions often produce surprising results. The key is careful arithmetic and always double-checking your work by substitution.

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