Solve for X: Finding the Value in (2/5)x + (3/4)x = 1

Q: Why can't I just add the fractions 2/5 + 3/4 directly?

You can only add fractions when they have the same denominator. Think of it like adding 2 apples and 3 oranges - you need to convert them to the same unit first!

Q: What does it mean to multiply by the reciprocal?

The reciprocal of $ \frac{23}{20} $ is $ \frac{20}{23} $ (flip the fraction). When you multiply a number by its reciprocal, you get 1, which helps isolate the variable.

Q: Can I solve this by clearing fractions first?

Yes! Multiply the entire equation by 20 (the LCD): $ 20 \cdot \frac{2}{5}x + 20 \cdot \frac{3}{4}x = 20 \cdot 1 $ gives you $ 8x + 15x = 20 $. Same answer!

Linear Equations with Fractional Coefficients

Solve for X:

$\frac{2}{5}x+\frac{3}{4}x=1$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solve

00:04 We want to isolate the unknown X

00:08 We'll multiply by the common denominator to eliminate fractions

00:28 We'll reduce what we can

00:44 We'll solve each multiplication separately

00:54 We'll collect terms

01:00 We'll isolate the unknown X

01:19 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 for X:

$\frac{2}{5}x+\frac{3}{4}x=1$

Step-by-step solution

To solve the equation $\frac{2}{5}x + \frac{3}{4}x = 1$ , we will follow these steps:

Step 1: Find a common denominator to combine the fractions on the left-hand side.
Step 2: Simplify the equation.
Step 3: Isolate the variable $x$ .

Now, let's work through each step:
Step 1: The denominators are 5 and 4. The least common denominator (LCD) is 20.
Convert each term: $\frac{2}{5} = \frac{2 \times 4}{5 \times 4} = \frac{8}{20}$ and $\frac{3}{4} = \frac{3 \times 5}{4 \times 5} = \frac{15}{20}$ .

Step 2: Combine the fractions: $\frac{8}{20}x + \frac{15}{20}x = \frac{23}{20}x$ .
The equation now is $\frac{23}{20}x = 1$ .

Step 3: Solve for $x$ by multiplying both sides by the reciprocal of $\frac{23}{20}$ , which is $\frac{20}{23}$ .
Thus, $x = 1 \times \frac{20}{23} = \frac{20}{23}$ .

Therefore, the solution to the equation is $\boxed{\frac{20}{23}}$ .

Final Answer

$\frac{20}{23}$

Key Points to Remember

Essential concepts to master this topic

Common Denominator: Find LCD to combine fractions with same variable
Technique: Convert $\frac{2}{5} = \frac{8}{20}$ and $\frac{3}{4} = \frac{15}{20}$ to add coefficients
Check: Substitute $x = \frac{20}{23}$ back: $\frac{2}{5} \cdot \frac{20}{23} + \frac{3}{4} \cdot \frac{20}{23} = 1$ ✓

Common Mistakes

Avoid these frequent errors

Adding fractions without finding common denominator first
Don't just add $\frac{2}{5} + \frac{3}{4} = \frac{5}{9}$ by adding numerators and denominators = completely wrong result! This violates fraction addition rules. Always find the LCD (20 here) and convert both fractions before combining.

Practice Quiz

Test your knowledge with interactive questions

$$ x+7=14 $$

$x=\text{?}$

FAQ

Everything you need to know about this question

Why can't I just add the fractions 2/5 + 3/4 directly?

You can only add fractions when they have the same denominator. Think of it like adding 2 apples and 3 oranges - you need to convert them to the same unit first!

How do I find the LCD of 5 and 4?

List multiples of each number: 5 (5, 10, 15, 20, 25...) and 4 (4, 8, 12, 16, 20, 24...). The smallest number that appears in both lists is your LCD.

What does it mean to multiply by the reciprocal?

The reciprocal of $\frac{23}{20}$ is $\frac{20}{23}$ (flip the fraction). When you multiply a number by its reciprocal, you get 1, which helps isolate the variable.

Can I solve this by clearing fractions first?

Yes! Multiply the entire equation by 20 (the LCD): $20 \cdot \frac{2}{5}x + 20 \cdot \frac{3}{4}x = 20 \cdot 1$ gives you $8x + 15x = 20$ . Same answer!

Why is my answer a fraction instead of a whole number?

Many equations have fractional solutions - that's completely normal! The important thing is that $\frac{20}{23}$ makes the original equation true when substituted back.

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