Find all real numbers $x$ such that $3x - 7 \le 5x +9$. Give your answer as an interval.
ok, going by Liams's interpretation ....
3x-7 ≤ 5x+9
-2x ≤ 16
x ≥ -8
Thank you!
To find all real numbers $x$ that satisfy the inequality $3x - 7 \le 5x + 9$, we need to solve it step by step. Here's how:
Step 1: Subtract $3x$ from both sides of the inequality, which gives $-7 \le 2x + 9$.
Step 2: Subtract $9$ from both sides of the inequality, which gives $-16 \le 2x$.
Step 3: Divide both sides of the inequality by $2$ (since we want to isolate $x$), which gives $-8 \le x$.
So, the solution to the inequality is $x \ge -8$. In interval notation, this can be written as $[-8, \infty)$, where $[-8$ means inclusive of $-8$ and $\infty)$ means approaching infinity (no upper bound).
@Cherry I think it would be easier to understand if you put it like this: Find all real numbers x such that 3x-7< or = to 5x+9.
Just use standard keyboard letters to type your math expression.
I can't make you your post.