Graph the inequality on the numberline and then write it in interval notation.
11z+3(z-1)≤4z-13
Please help me solve and then graph on a numberline along with the interval notation.
11 z + 3 ( z - 1 ) ≤ 4 z -13
11 z + 3 ∙ z + 3 ∙ ( - 1 ) ≤ 4 z -13
11 z + 3 z - 3 ≤ 4 z -13
14 z - 3 ≤ 4 z -13
Subtract 4 z to both sides
14 z - 3 - 4 z ≤ 4 z -13 - 4 z
10 z - 3 ≤ -13
Add 3 to both sides
10 z - 3 + 3 ≤ -13 + 3
10 z ≤ -10
Divide both sides by 10
z ≤ -1
Interval notation:
z ∈ ( - ∞ , 1 ]
My typo.
Sorry.
z ∈ ( - ∞ , - 1 ]
To graph the inequality and write it in interval notation, we need to solve the inequality first.
Let's simplify the inequality step by step:
11z + 3(z - 1) ≤ 4z - 13
First, distribute the 3 into the parentheses:
11z + 3z - 3 ≤ 4z - 13
Combine like terms:
14z - 3 ≤ 4z - 13
Next, isolate the variable by subtracting 4z from both sides:
14z - 4z - 3 ≤ -13
Simplify:
10z - 3 ≤ -13
To isolate the variable, add 3 to both sides:
10z ≤ -10
Finally, divide both sides by 10 to solve for z:
z ≤ -1
Now, let's graph this solution on the number line.
To do this, draw a number line and mark a point at -1. Since the inequality includes "≤" (less than or equal to), we'll fill in the point at -1 with a solid dot to indicate that it is included in the solution.
● (≤)
-∞ --------------------------------------------------------●------→
-1
Next, we'll write the solution in interval notation, which represents the range of values for z. For this inequality, z can be any number less than or equal to -1.
Interval notation: (-∞, -1]