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The problem you're working on states:

- The sum of two numbers is 22.
- Three times one number increased by five is 35.

### a) What are the equations you would use?
Let the two numbers be \(x\) and \(y\). From the problem, we can write two equations:

1. The sum of the two numbers:  
   \[
   x + y = 22
   \]
2. Three times one number (let's say \(x\)) increased by five equals 35:  
   \[
   3x + 5 = 35
   \]

These are the two equations.

### b) What are the two numbers?
We solve the system of equations.

From the second equation \(3x + 5 = 35\), subtract 5 from both sides:
\[
3x = 30
\]
Now divide by 3:
\[
x = 10
\]

Substitute \(x = 10\) into the first equation \(x + y = 22\):
\[
10 + y = 22
\]
Solve for \(y\):
\[
y = 12
\]

Thus, the two numbers are \(x = 10\) and \(y = 12\).

---

Would you like further details on the steps or have any other questions? Here are five related questions you could explore:

1. How do you solve systems of equations using substitution?
2. Can the second equation be solved if the form were slightly different, e.g., \(2x + 5 = 35\)?
3. How do you solve this system using elimination instead of substitution?
4. How would the equations change if the sum of the numbers was different?
5. What if both equations involved multiplication of both \(x\) and \(y\)?

**Tip:** Always double-check your solutions by plugging them back into the original equations to verify their correctness.

The sum of two numbers is 22. Three times one number increased by five is 35.

Learn how to solve a system of equations where the sum of two numbers is 22 and three times one number increased by five equals 35. This problem covers key algebraic concepts and involves solving using the substitution method. Suitable for students in Grades 7-8.

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