Math Problem Statement

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.

Solution

Let's denote the two numbers as xx (the smaller number) and yy (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=3y64x = 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(yx)+5x + y = 6(y - x) + 5

Step 1: Simplify Condition 2

Rewrite Condition 2 for clarity: x+y=6(yx)+5x + y = 6(y - x) + 5 Expanding and rearranging terms, we get: x+y=6y6x+5x + y = 6y - 6x + 5 Bringing terms involving xx and yy to one side, we get: 7x5y=57x - 5y = -5

So now, we have two equations:

  1. 4x=3y64x = 3y - 6
  2. 7x5y=57x - 5y = -5

Step 2: Solve the System of Equations

Substitute for yy from the first equation

From the first equation: 4x=3y64x = 3y - 6 Solving for yy, we get: 3y=4x+63y = 4x + 6 y=4x+63y = \frac{4x + 6}{3}

Substitute yy in the second equation

Now, substitute y=4x+63y = \frac{4x + 6}{3} into the second equation: 7x5(4x+63)=57x - 5\left(\frac{4x + 6}{3}\right) = -5 Expanding and simplifying: 7x20x+303=57x - \frac{20x + 30}{3} = -5 Multiply every term by 3 to clear the fraction: 21x(20x+30)=1521x - (20x + 30) = -15 21x20x30=1521x - 20x - 30 = -15 x30=15x - 30 = -15 x=15x = 15

Substitute x=15x = 15 back to find yy

Now substitute x=15x = 15 into y=4x+63y = \frac{4x + 6}{3}: y=4(15)+63y = \frac{4(15) + 6}{3} y=60+63y = \frac{60 + 6}{3} y=663y = \frac{66}{3} y=22y = 22

Solution

The two numbers are: x=15andy=22x = 15 \quad \text{and} \quad y = 22

Verification

  1. Condition 1: 4x=3y64x = 3y - 6 4×15=3×2264 \times 15 = 3 \times 22 - 6 60=66660 = 66 - 6 60=60(True)60 = 60 \quad \text{(True)}

  2. Condition 2: x+y=6(yx)+5x + y = 6(y - x) + 5 15+22=6(2215)+515 + 22 = 6(22 - 15) + 5 37=6×7+537 = 6 \times 7 + 5 37=42+537 = 42 + 5 37=37(True)37 = 37 \quad \text{(True)}

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

15 and 22\boxed{15 \text{ and } 22}

Would you like a further breakdown of any part?


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.

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Math Problem Analysis

Mathematical Concepts

Algebra
Systems of Linear Equations

Formulas

4x = 3y - 6
x + y = 6(y - x) + 5

Theorems

Linear System Solution Techniques

Suitable Grade Level

Grades 9-10