Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.
(5 + 2b)+(2b + 3/2)
(-14 + 3/2b)-(1 + 8/2b)
(-10 + b)+(7b -5)
(7/2b – 3)-(8 + 6b)
8b – 15
4b + 13/2
-5/2b – 11
-15 – 5/2b
The Correct Answer and Explanation is :
To match the pairs of equivalent expressions, we need to simplify each expression and see which simplified version corresponds to the given options. Let’s break down each expression step by step.
Expression 1: ((5 + 2b) + (2b + \frac{3}{2}))
- Combine like terms:
[
5 + 2b + 2b + \frac{3}{2} = 5 + 4b + \frac{3}{2}
] - Convert (5) to a fraction:
[
5 = \frac{10}{2} \quad \Rightarrow \quad \frac{10}{2} + \frac{3}{2} = \frac{13}{2}
] - Thus, this simplifies to:
[
4b + \frac{13}{2}
]
Expression 2: ((-14 + \frac{3}{2}b) – (1 + \frac{8}{2}b))
- Distributing the negative sign:
[
-14 + \frac{3}{2}b – 1 – \frac{8}{2}b = -15 + \frac{3}{2}b – 4b
] - Convert (\frac{8}{2}b) to (4b):
[
-15 + \left(\frac{3}{2} – 4\right)b = -15 – \frac{5}{2}b
]
Expression 3: ((-10 + b) + (7b – 5))
- Combine like terms:
[
-10 + b + 7b – 5 = -15 + 8b
]
Expression 4: ((\frac{7}{2}b – 3) – (8 + 6b))
- Distributing the negative sign:
[
\frac{7}{2}b – 3 – 8 – 6b = \frac{7}{2}b – 11 – 6b
] - Convert (6b) to (\frac{12}{2}b):
[
\frac{7}{2}b – 11 – \frac{12}{2}b = -\frac{5}{2}b – 11
]
Summary of Simplified Expressions:
- Expression 1: (4b + \frac{13}{2})
- Expression 2: (-15 – \frac{5}{2}b)
- Expression 3: (8b – 15)
- Expression 4: (-\frac{5}{2}b – 11)
Matching with Options:
- (4b + \frac{13}{2}) matches with the simplified Expression 1.
- (-15 – \frac{5}{2}b) matches with Expression 2.
- (8b – 15) matches with Expression 3.
- (-\frac{5}{2}b – 11) matches with Expression 4.
Final Pairs:
- ((5 + 2b)+(2b + 3/2)) → (4b + \frac{13}{2})
- ((-14 + 3/2b)-(1 + 8/2b)) → (-15 – \frac{5}{2}b)
- ((-10 + b)+(7b -5)) → (8b – 15)
- ((7/2b – 3)-(8 + 6b)) → (-\frac{5}{2}b – 11)
This process of simplification helps to identify equivalent expressions and confirm that each expression indeed matches one of the options provided.