Question
(a)
Martin is considering the expressions
and
. He wants to know if one expression is greater than the other for all values of
.
Part A
Which statement about the relationship between the expressions is true?
A
The value of the expression 12(7x+48)\frac{1}{2}\left(7x+48\right)
2
1
(7x+48) is always equal to the value of the expression −(12x−3)+(x+5)-\left(\frac{1}{2}x-3\right)+\left(x+5\right)−(
2
1
x−3)+(x+5).
B
The value of the expression 12(7x+48)\frac{1}{2}\left(7x+48\right)
2
1
(7x+48) is always less than the value of the expression −(12x−3)+4(x+5)-\left(\frac{1}{2}x-3\right)+4\left(x+5\right)−(
2
1
x−3)+4(x+5).
C
The value of the expression 12(7x+48)\frac{1}{2}\left(7x+48\right)
2
1
(7x+48) is always greater than the value of the expression −(12x−3)+4(x+5)-\left(\frac{1}{2}x-3\right)+4\left(x+5\right)−(
2
1
x−3)+4(x+5).
D
The value of the expression 12(7x+48)\frac{1}{2}\left(7x+48\right)
2
1
(7x+48)is sometimes greater than and sometimes less than the value of the expression −(12x−3)+4(x+5)-\left(\frac{1}{2}x-3\right)+4\left(x+5\right)−(
2
1
x−3)+4(x+5).
(b)
Part B
Show or explain how you found your answer to Part A.
Enter your work or your explanation in the box provided.
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(c)
Part C
Write a new expression that always has a greater value than both of these expressions.
Enter your expressions in the box provided.
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Answers
GPT-4o mini
To determine the relationship between the expressions \( \frac{1}{2}(7x+48) \) and \( -(12x-3)+(x+5)-\left(\frac{1}{2}x-3\right)+4(x+5) \), we need to simplify each expression and compare them.
### Part A
1. **Simplifying the first expression:**
\[
E_1 = \frac{1}{2}(7x+48) = \frac{7x}{2} + 24
\]
2. **Simplifying the second expression:**
\[
E_2 = -(12x-3)+(x+5)-\left(\frac{1}{2}x-3\right)+4(x+5)
\]
Expand each term:
\[
= -12x + 3 + x + 5 - \frac{1}{2}x + 3 + 4x + 20
\]
Combine like terms:
\[
= (-12x + x - \frac{1}{2}x + 4x) + (3 + 5 + 3 + 20)
\]
\[
= (-12 + 1 - 0.5 + 4)x + 31
\]
\[
= -7.5x + 31
\]
So, the simplified expressions are:
- \( E_1 = \frac{7}{2}x + 24 \)
- \( E_2 = -7.5x + 31 \)
3. **Finding the relationship between \( E_1 \) and \( E_2 \):**
To determine if one expression is always greater than, less than, or equal to the other, we can set them equal to each other and analyze the implications.
\[
\frac{7}{2}x + 24 = -7.5x + 31
\]
Rearranging gives:
\[
\frac{7}{2}x + 7.5x = 31 - 24
\]
\[
\frac{7}{2}x + \frac{15}{2}x = 7
\]
\[
\frac{22}{2}x = 7
\]
\[
11x = 7 \implies x = \frac{7}{11}
\]
4. **Evaluate expressions at a value above and below \( x = \frac{7}{11} \):**
- For \( x = 0 \):
\[
E_1 = 24, \quad E_2 = 31 \quad \text{(so } E_1 < E_2\text{)}
\]
- For \( x = 1 \):
\[
E_1 = \frac{7}{2} + 24 = 3.5 + 24 = 27.5
\]
\[
E_2 = -7.5 + 31 = 23.5 \quad \text{(so } E_1 > E_2\text{)}
\]
Since the comparison shows that \( E_1 \) is less than \( E_2 \) when \( x < \frac{7}{11} \) and greater than \( E_2 \) when \( x > \frac{7}{11} \), it indicates:
### Final Answer for Part A
**D:** The value of the expression \( \frac{1}{2}(7x+48) \) is sometimes greater than and sometimes less than the value of the expression \( -(12x-3)+(x+5)-\left(\frac{1}{2}x-3\right)+4(x+5) \).
### Part B
To explain:
1. I simplified both expressions to their linear forms.
2. I found the intersection point by setting them equal.
3. I evaluated at specific points to determine the relationship—showing that one is greater for some values of \( x \) and less for others.
### Part C
To write an expression that is always greater than both \( E_1 \) and \( E_2 \):
A simple expression that would always be greater than both (regardless of \( x \)) could be:
\[
E_{new} = 50
\]
This constant number is greater than both simplified expressions for any value of \( x \).
### Part A
1. **Simplifying the first expression:**
\[
E_1 = \frac{1}{2}(7x+48) = \frac{7x}{2} + 24
\]
2. **Simplifying the second expression:**
\[
E_2 = -(12x-3)+(x+5)-\left(\frac{1}{2}x-3\right)+4(x+5)
\]
Expand each term:
\[
= -12x + 3 + x + 5 - \frac{1}{2}x + 3 + 4x + 20
\]
Combine like terms:
\[
= (-12x + x - \frac{1}{2}x + 4x) + (3 + 5 + 3 + 20)
\]
\[
= (-12 + 1 - 0.5 + 4)x + 31
\]
\[
= -7.5x + 31
\]
So, the simplified expressions are:
- \( E_1 = \frac{7}{2}x + 24 \)
- \( E_2 = -7.5x + 31 \)
3. **Finding the relationship between \( E_1 \) and \( E_2 \):**
To determine if one expression is always greater than, less than, or equal to the other, we can set them equal to each other and analyze the implications.
\[
\frac{7}{2}x + 24 = -7.5x + 31
\]
Rearranging gives:
\[
\frac{7}{2}x + 7.5x = 31 - 24
\]
\[
\frac{7}{2}x + \frac{15}{2}x = 7
\]
\[
\frac{22}{2}x = 7
\]
\[
11x = 7 \implies x = \frac{7}{11}
\]
4. **Evaluate expressions at a value above and below \( x = \frac{7}{11} \):**
- For \( x = 0 \):
\[
E_1 = 24, \quad E_2 = 31 \quad \text{(so } E_1 < E_2\text{)}
\]
- For \( x = 1 \):
\[
E_1 = \frac{7}{2} + 24 = 3.5 + 24 = 27.5
\]
\[
E_2 = -7.5 + 31 = 23.5 \quad \text{(so } E_1 > E_2\text{)}
\]
Since the comparison shows that \( E_1 \) is less than \( E_2 \) when \( x < \frac{7}{11} \) and greater than \( E_2 \) when \( x > \frac{7}{11} \), it indicates:
### Final Answer for Part A
**D:** The value of the expression \( \frac{1}{2}(7x+48) \) is sometimes greater than and sometimes less than the value of the expression \( -(12x-3)+(x+5)-\left(\frac{1}{2}x-3\right)+4(x+5) \).
### Part B
To explain:
1. I simplified both expressions to their linear forms.
2. I found the intersection point by setting them equal.
3. I evaluated at specific points to determine the relationship—showing that one is greater for some values of \( x \) and less for others.
### Part C
To write an expression that is always greater than both \( E_1 \) and \( E_2 \):
A simple expression that would always be greater than both (regardless of \( x \)) could be:
\[
E_{new} = 50
\]
This constant number is greater than both simplified expressions for any value of \( x \).