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 solve the problem, let's work through each part systematically.
### Part A
We need to analyze and compare the two expressions given.
1. **Express the First Expression:**
\[
E_1 = \frac{1}{2}(7x + 48) = \frac{7x}{2} + 24
\]
2. **Express the Second Expression:**
Let's simplify the expression \( E_2 = - (12x - 3) + (x + 5) - \left( \frac{1}{2}x - 3 \right) + 4(x + 5) \).
First, simplify each part individually:
- \( - (12x - 3) = -12x + 3 \)
- \( x + 5 = x + 5 \)
- \( - \left( \frac{1}{2}x - 3 \right) = -\frac{1}{2}x + 3 \)
- \( 4(x + 5) = 4x + 20 \)
Now, combine these results:
\[
E_2 = -12x + 3 + x + 5 - \frac{1}{2}x + 3 + 4x + 20
\]
Combine like terms:
- For \(x\): \( -12x + x - \frac{1}{2}x + 4x = -7.5x \) or \( -\frac{15}{2}x \)
- For constant terms: \( 3 + 5 + 3 + 20 = 31 \)
Therefore,
\[
E_2 = -\frac{15}{2}x + 31
\]
3. **Comparing the Expressions:**
We now have:
- \( E_1 = \frac{7x}{2} + 24 \)
- \( E_2 = -\frac{15}{2}x + 31 \)
To determine the relationship between \( E_1 \) and \( E_2 \), let's find when \( E_1 = E_2 \):
\[
\frac{7x}{2} + 24 = -\frac{15}{2}x + 31
\]
Rearranging gives:
\[
\frac{7x}{2} + \frac{15}{2}x = 31 - 24
\]
\[
\frac{22x}{2} = 7 \implies 11x = 7 \implies x = \frac{7}{11}
\]
Now, testing values around \(\frac{7}{11}\):
- For \( x < \frac{7}{11} \), we can see:
- Let \( x = 0 \):
- \( E_1 = 24 \)
- \( E_2 = 31 \)
- Therefore, \( E_1 < E_2 \).
- For \( x > \frac{7}{11} \):
- Let \( x = 1 \):
- \( E_1 = \frac{7}{2} + 24 = 26.5 \)
- \( E_2 = -\frac{15}{2} + 31 = 23.5 \)
- Therefore, \( E_1 > E_2 \).
Given this analysis, **the correct answer is D**:
- The value of the expression \( E_1 \) is sometimes greater than and sometimes less than the value of the expression \( E_2 \).
### Part B
The explanation above shows how I found the answer. By simplifying both expressions and finding the points of intersection, I was able to conclude that the values of the expressions change based on the value of \( x \).
### Part C
To create a new expression that is always greater than both expressions \( E_1 \) and \( E_2 \), we can simply add a constant that is greater than the maximum values of both expressions at their respective limits.
For example:
\[
E = \frac{1}{2}(7x + 48) + 10 = \frac{7x}{2} + 34
\]
or another option:
\[
E' = -\frac{15}{2}x + 41
\]
Both expressions are just shifted versions of \( E_1 \) and \( E_2 \) with constant adjustments to ensure they are greater than both original expressions for all values of \( x \).
### Part A
We need to analyze and compare the two expressions given.
1. **Express the First Expression:**
\[
E_1 = \frac{1}{2}(7x + 48) = \frac{7x}{2} + 24
\]
2. **Express the Second Expression:**
Let's simplify the expression \( E_2 = - (12x - 3) + (x + 5) - \left( \frac{1}{2}x - 3 \right) + 4(x + 5) \).
First, simplify each part individually:
- \( - (12x - 3) = -12x + 3 \)
- \( x + 5 = x + 5 \)
- \( - \left( \frac{1}{2}x - 3 \right) = -\frac{1}{2}x + 3 \)
- \( 4(x + 5) = 4x + 20 \)
Now, combine these results:
\[
E_2 = -12x + 3 + x + 5 - \frac{1}{2}x + 3 + 4x + 20
\]
Combine like terms:
- For \(x\): \( -12x + x - \frac{1}{2}x + 4x = -7.5x \) or \( -\frac{15}{2}x \)
- For constant terms: \( 3 + 5 + 3 + 20 = 31 \)
Therefore,
\[
E_2 = -\frac{15}{2}x + 31
\]
3. **Comparing the Expressions:**
We now have:
- \( E_1 = \frac{7x}{2} + 24 \)
- \( E_2 = -\frac{15}{2}x + 31 \)
To determine the relationship between \( E_1 \) and \( E_2 \), let's find when \( E_1 = E_2 \):
\[
\frac{7x}{2} + 24 = -\frac{15}{2}x + 31
\]
Rearranging gives:
\[
\frac{7x}{2} + \frac{15}{2}x = 31 - 24
\]
\[
\frac{22x}{2} = 7 \implies 11x = 7 \implies x = \frac{7}{11}
\]
Now, testing values around \(\frac{7}{11}\):
- For \( x < \frac{7}{11} \), we can see:
- Let \( x = 0 \):
- \( E_1 = 24 \)
- \( E_2 = 31 \)
- Therefore, \( E_1 < E_2 \).
- For \( x > \frac{7}{11} \):
- Let \( x = 1 \):
- \( E_1 = \frac{7}{2} + 24 = 26.5 \)
- \( E_2 = -\frac{15}{2} + 31 = 23.5 \)
- Therefore, \( E_1 > E_2 \).
Given this analysis, **the correct answer is D**:
- The value of the expression \( E_1 \) is sometimes greater than and sometimes less than the value of the expression \( E_2 \).
### Part B
The explanation above shows how I found the answer. By simplifying both expressions and finding the points of intersection, I was able to conclude that the values of the expressions change based on the value of \( x \).
### Part C
To create a new expression that is always greater than both expressions \( E_1 \) and \( E_2 \), we can simply add a constant that is greater than the maximum values of both expressions at their respective limits.
For example:
\[
E = \frac{1}{2}(7x + 48) + 10 = \frac{7x}{2} + 34
\]
or another option:
\[
E' = -\frac{15}{2}x + 41
\]
Both expressions are just shifted versions of \( E_1 \) and \( E_2 \) with constant adjustments to ensure they are greater than both original expressions for all values of \( x \).