We are given two equations:

1. $x^2 + y^2 = 356$ (sum of the squares)
2. $x - y = 6$ (difference)

Here, $x$ is the larger number, and $y$ is the smaller number. We solve as follows:

Step 1: Express $x$ in terms of $y$

From the second equation: $x = y + 6$

Step 2: Substitute $x = y + 6$ into the first equation

$(x^2 + y^2 = 356) \implies ((y + 6)^2 + y^2 = 356)$

Expand $(y + 6)^2$: $(y^2 + 12y + 36) + y^2 = 356$

Simplify: $2y^2 + 12y + 36 = 356$

Subtract 356 from both sides: $2y^2 + 12y - 320 = 0$

Step 3: Simplify the quadratic equation

Divide the entire equation by 2: $y^2 + 6y - 160 = 0$

Step 4: Solve the quadratic equation

Use the quadratic formula: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = 6$, $c = -160$.

Substitute: $y = \frac{-6 \pm \sqrt{6^2 - 4(1)(-160)}}{2(1)}$ $y = \frac{-6 \pm \sqrt{36 + 640}}{2}$ $y = \frac{-6 \pm \sqrt{676}}{2}$ $y = \frac{-6 \pm 26}{2}$

Step 5: Find the two solutions for $y$

$y = \frac{-6 + 26}{2} = \frac{20}{2} = 10$ $y = \frac{-6 - 26}{2} = \frac{-32}{2} = -16$

Step 6: Find $x$

Using $x = y + 6$:

• If $y = 10$, then $x = 10 + 6 = 16$.
• If $y = -16$, then $x = -16 + 6 = -10$.

Final Answer:

• The larger number is $16$.
• The smaller number is $10$.

If you need further details or have any questions, let me know!

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Related Questions:

1. What happens if the difference is negative instead of positive?
2. Can you verify the solution by substituting into the original equations?
3. How would you graph these equations to visualize the solution?
4. What if the numbers must both be positive integers?
5. How does the solution change if $x - y = k$, where $k \neq 6$?

Tip:

Always double-check your solution by substituting the results back into the original equations to ensure accuracy.