Question

Compare and Contrast: Two equations are listed below. Solve each equation and compare the solutions. Choose the statement that is true about both solutions.
Equation 1 Equation 2
|5x + 6| = 41 |2x + 13| = 28

Answer 1

Answer:

The absolute vale of x in equation 2 is greater than the absolute value of x in equation 1

Step-by-step explanation:

Equation 1

|5x + 6| = 41.......................finding the absolute value of x

5x+6=41

5x=41-6

5x=35........................divide by 5 both sides

x=35/5= 7

|x|=7

Equation 2

|2x + 13| = 28........................find the absolute value of x

2x+13=28

2x=28-13

2x=15........................................divide by 2 both sides

x=15/2 =7.5

|x|=7.5

Conclusion

The absolute vale of x in equation 2 is greater than the absolute value of x in equation 1

Answer 2

Answer:

Solution of inequality 1:

$x \in [\frac{-47}{5}, 7]$

Solution of Inequality 2:

$x \in [\frac{-41}{2}, \frac{15}{2}]$

Step-by-step explanation:

We are given two inequalities:

Inequality 1

$\mid 5x + 6 \mid = 41\\-41 \leq 5x + 6 \leq 41\\-47 \leq 5x \leq 35\\\frac{-47}{5} \leq x \leq 7\\x \in [\frac{-47}{5}, 7]$

Inequality 2

$\mid 2x + 13 \mid = 28\\-28 \leq 2x +13 \leq 28\\-41 \leq 2x \leq 15\\\frac{-41}{2} \leq x \leq \frac{15}{2}\\x \in [\frac{-41}{2}, \frac{15}{2}]$

Solution of inequality 1:

$x \in [\frac{-47}{5}, 7]$

Solution of Inequality 2:

$x \in [\frac{-41}{2}, \frac{15}{2}]$