who is correct and why?
chloe jonas
3<2x-5<7 3<2x-5<7
3<2x<12 8<2x<7
3/2<x<6 4<x<7/2
2) solve each inequality for x. assume a is constant and a>0
a. -3<ax+1< or equal to 5
b. -1/ax+6<1 or 2-ax>8
3) create an example of a compound inequality containing or that has infinitely many solutions.
chole 3<2x-5<7
3<2x<12
3/2<x<6
jonas
3<2x-5<7
8<2x<7
4<x<7/2
3<2x-5<7.
3+5<2x<7+5
8<2x<12
4<X<6.
Note: Add 5 to BOTH sides of the compound inequality.
2a. -3<ax+1<=5.
-3-1<ax<=5-1
-4<ax<=4
-4/a<X<=4/a.
To determine who is correct in the given statements, we need to solve each inequality and compare the solutions.
Statement 1:
Chloe: 3 < 2x - 5 < 7
To solve this inequality, add 5 to all parts of the inequality:
8 < 2x < 12
Now, divide all parts of the inequality by 2:
4 < x < 6
Jonas: 3 < 2x - 5 < 7
Following the same steps as above:
8 < 2x < 12
Divide by 2:
4 < x < 6
Both Chloe and Jonas are correct in this case since they both arrived at the same solution.
Statement 2:
a. -3 < ax + 1 ≤ 5
Subtract 1 from all parts:
-4 < ax ≤ 4
Divide by a (remembering that a > 0):
-4/a < x ≤ 4/a
Now, Chloe and Jonas have different solutions since Chloe's solution is -4/a < x ≤ 4/a, while Jonas' solution is -4 < x ≤ 4.
b. -1/(ax) + 6 < 1 or 2 - ax > 8
Let's solve these inequalities separately:
-1/(ax) + 6 < 1
Subtract 6 from all parts:
-1/(ax) < -5
Multiply by -ax (since a > 0):
1 > 5ax
Divide by 5 (remembering a > 0):
1/5 > ax
Thus, Chloe's solution here is 1/5 > ax.
2 - ax > 8
Subtract 2 from all parts:
-ax > 6
Multiply by -1 (note that a > 0, so no changes in direction):
ax < -6
Here, Jonas' solution is ax < -6.
Both Chloe and Jonas are correct in this case as their solutions are for different inequalities.
Statement 3:
An example of a compound inequality containing "or" and having infinitely many solutions could be:
x > 5 or x < -5.
This compound inequality has infinitely many solutions because any value of x that is greater than 5 or less than -5 would satisfy the inequality.