Of two numbers, 4 times the smaller one is less than 3 times the larger one by 6. if the sum of the numbers is larger than 6 times their difference by 5, find the number.

Let's denote the two numbers as \( x \) (the smaller number) and \( y \) (the larger number). We are given the following information:

1. **Condition 1**: "4 times the smaller one is less than 3 times the larger one by 6." This translates to:
   \[
   4x = 3y - 6
   \]

2. **Condition 2**: "The sum of the numbers is larger than 6 times their difference by 5." This translates to:
   \[
   x + y = 6(y - x) + 5
   \]

### Step 1: Simplify Condition 2
Rewrite Condition 2 for clarity:
\[
x + y = 6(y - x) + 5
\]
Expanding and rearranging terms, we get:
\[
x + y = 6y - 6x + 5
\]
Bringing terms involving \( x \) and \( y \) to one side, we get:
\[
7x - 5y = -5
\]

So now, we have two equations:
1. \( 4x = 3y - 6 \)
2. \( 7x - 5y = -5 \)

### Step 2: Solve the System of Equations
#### Substitute for \( y \) from the first equation
From the first equation:
\[
4x = 3y - 6
\]
Solving for \( y \), we get:
\[
3y = 4x + 6
\]
\[
y = \frac{4x + 6}{3}
\]

#### Substitute \( y \) in the second equation
Now, substitute \( y = \frac{4x + 6}{3} \) into the second equation:
\[
7x - 5\left(\frac{4x + 6}{3}\right) = -5
\]
Expanding and simplifying:
\[
7x - \frac{20x + 30}{3} = -5
\]
Multiply every term by 3 to clear the fraction:
\[
21x - (20x + 30) = -15
\]
\[
21x - 20x - 30 = -15
\]
\[
x - 30 = -15
\]
\[
x = 15
\]

#### Substitute \( x = 15 \) back to find \( y \)
Now substitute \( x = 15 \) into \( y = \frac{4x + 6}{3} \):
\[
y = \frac{4(15) + 6}{3}
\]
\[
y = \frac{60 + 6}{3}
\]
\[
y = \frac{66}{3}
\]
\[
y = 22
\]

### Solution
The two numbers are:
\[
x = 15 \quad \text{and} \quad y = 22
\]

### Verification
1. **Condition 1**: \( 4x = 3y - 6 \)
   \[
   4 \times 15 = 3 \times 22 - 6
   \]
   \[
   60 = 66 - 6
   \]
   \[
   60 = 60 \quad \text{(True)}
   \]

2. **Condition 2**: \( x + y = 6(y - x) + 5 \)
   \[
   15 + 22 = 6(22 - 15) + 5
   \]
   \[
   37 = 6 \times 7 + 5
   \]
   \[
   37 = 42 + 5
   \]
   \[
   37 = 37 \quad \text{(True)}
   \]

Both conditions are satisfied. Therefore, the solution is correct:

\[
\boxed{15 \text{ and } 22}
\]

Would you like a further breakdown of any part? 

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Here are some related questions to extend this topic:
1. How do we solve systems of linear equations with more than two unknowns?
2. Can you provide an example of solving a system of equations by substitution versus elimination?
3. What are some real-world applications of solving linear equations in two variables?
4. How can inequalities be used alongside equations in system solutions?
5. What is the difference between dependent and independent systems of equations?

**Tip**: When solving systems of equations, check if substitution or elimination is more straightforward based on how the equations are structured.

Of two numbers, 4 times the smaller one is less than 3 times the larger one by 6. If the sum of the numbers is larger than 6 times their difference by 5, find the number.

Learn how to solve a system of linear equations to find two numbers where 4 times the smaller number is less than 3 times the larger by 6, and their sum exceeds 6 times their difference by 5. This problem is suitable for students in Grades 9-10.

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