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Let's solve the given expressions for \( a = 3 \) and \( b = -2 \):

### a. \( 2a - 5 \)
Substitute \( a = 3 \):

\[
2(3) - 5 = 6 - 5 = 1
\]

So, the result for part **a** is \( 1 \).

### b. \( 3b + 4a \)
Substitute \( b = -2 \) and \( a = 3 \):

\[
3(-2) + 4(3) = -6 + 12 = 6
\]

So, the result for part **b** is \( 6 \).

### Final answers:
- a. \( 2a - 5 = 1 \)
- b. \( 3b + 4a = 6 \)

Would you like more details on any part of this? 

Here are 5 related questions for further practice:
1. What is the value of \( 5a - 2b \) for \( a = 3 \) and \( b = -2 \)?
2. Evaluate \( a^2 + b^2 \) with \( a = 3 \) and \( b = -2 \).
3. Calculate \( 4b - a^2 \) for the same values of \( a \) and \( b \).
4. Find \( 7a + b^3 \) using \( a = 3 \) and \( b = -2 \).
5. Simplify and evaluate \( a(b + a) \) when \( a = 3 \) and \( b = -2 \).

**Tip**: Always double-check your substitutions to ensure accuracy when plugging in values.

Evaluate for a = 3 and b = -2.
a. 2a - 5
b. 3b + 4a

Learn how to evaluate algebraic expressions 2a - 5 and 3b + 4a when a = 3 and b = -2. This problem involves basic substitution and operations, suitable for students in Grades 6-8.

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