Problem Analysis:
The task involves solving a series of equations to find the value of the parameter \( x \). The equations are provided in a table format, and we need to determine which method (substitution or elimination) is most appropriate for each equation. Additionally, there are some word problems related to quantities and costs.
Let's solve each part step by step.
---
Part 1: Solve the Equations
####
Equation 1:
\[ 3x - 7 = 4x + 5 \]
-
Method: Elimination
-
Solution:
\[
3x - 7 = 4x + 5
\]
Subtract \( 3x \) from both sides:
\[
-7 = x + 5
\]
Subtract 5 from both sides:
\[
x = -12
\]
####
Equation 2:
\[ 6x + 8 = 6x + 8 \]
-
Method: Substitution
-
Solution:
This equation is an identity because both sides are exactly the same. Therefore, it holds true for all values of \( x \).
####
Equation 3:
\[ \frac{3}{4}x + \frac{1}{2} = \frac{1}{4}x + \frac{3}{2} \]
-
Method: Elimination
-
Solution:
First, eliminate the fractions by multiplying through by 4 (the least common denominator):
\[
4 \left( \frac{3}{4}x + \frac{1}{2} \right) = 4 \left( \frac{1}{4}x + \frac{3}{2} \right)
\]
Simplify:
\[
3x + 2 = x + 6
\]
Subtract \( x \) from both sides:
\[
2x + 2 = 6
\]
Subtract 2 from both sides:
\[
2x = 4
\]
Divide by 2:
\[
x = 2
\]
####
Equation 4:
\[ 0 = 0 \]
-
Method: Substitution
-
Solution:
This is a trivial identity, meaning it is always true regardless of the value of \( x \).
####
Equation 5:
\[ 4x - 3 = 4(x - 1) + 1 \]
-
Method: Elimination
-
Solution:
Expand the right-hand side:
\[
4x - 3 = 4x - 4 + 1
\]
Simplify the right-hand side:
\[
4x - 3 = 4x - 3
\]
This is an identity, so it holds true for all values of \( x \).
####
Equation 6:
\[ 6x + 8 = 6x + 9 \]
-
Method: Substitution
-
Solution:
Subtract \( 6x \) from both sides:
\[
8 = 9
\]
This is a contradiction, so there is no solution.
---
Part 2: Quantity and Cost Problems
####
Problem 1:
The quantity in Column A is given as \( x \), and the quantity in Column B is also given as \( x \). Determine if the quantities are equal.
-
Solution:
Since both columns have the same expression \( x \), the quantities are equal.
####
Problem 2:
The quantity in Column A is given as \( x \), and the quantity in Column B is given as \( x + 1 \). Determine if the quantities are equal.
-
Solution:
Since \( x \neq x + 1 \) for any real number \( x \), the quantities are not equal.
---
Part 3: Graphing and Function Evaluation
####
Function Definition:
\[ f(x) = 3x - 5 \]
####
Step 1: Find the value of the function when \( x = 0 \)
-
Solution:
Substitute \( x = 0 \) into the function:
\[
f(0) = 3(0) - 5 = -5
\]
####
Step 2: Determine when \( f(x) = 0 \)
-
Solution:
Set the function equal to 0 and solve for \( x \):
\[
3x - 5 = 0
\]
Add 5 to both sides:
\[
3x = 5
\]
Divide by 3:
\[
x = \frac{5}{3}
\]
---
Final Answer
\[
\boxed{-12, \text{all } x, 2, \text{all } x, \text{all } x, \text{no solution}, \text{equal}, \text{not equal}, -5, \frac{5}{3}}
\]
Parent Tip: Review the logic above to help your child master the concept of 10th grade algebra 2 worksheet.
