Find all $x$ such that both $2x$ and $3x-7$ are in the interval $(-1,12)$.
To find all $x$ that satisfy both conditions $2x\in(-1,12)$ and $3x-7\in(-1,12)$, we need to find the intersection of these two intervals.
Let's start by considering the condition $2x\in(-1,12)$. This means that $2x$ is greater than -1 and less than 12. We can write this as:
$-1<2x<12$
Next, we'll divide all parts of this inequality by 2 (which is allowed since 2 is positive) to get:
$-\frac{1}{2}<x<6$
Now, let's consider the condition $3x-7\in(-1,12)$. This means that $3x-7$ is greater than -1 and less than 12. We can write this as:
$-1<3x-7<12$
Next, let's add 7 to all parts of this inequality to get:
$6<3x<19$
Finally, we'll divide all parts of this inequality by 3 to get:
$2<x<\frac{19}{3}$
Now, we have two inequalities for $x$: $-\frac{1}{2}<x<6$ and $2<x<\frac{19}{3}$. To find the intersection of these two intervals, we need to find the overlapping range they share.
Looking at the number line, we see that the overlapping range is $2<x<6$.
Therefore, all $x$ that satisfy both conditions are given by $2<x<6$.
To find all $x$ that satisfy both conditions, we need to find the intersection of the two intervals $(-1, 12)$ for $2x$ and $3x-7$.
First, let's find the interval for $2x$:
Since $2x$ is in the interval $(-1, 12)$, we have the inequality:
$$-1 < 2x < 12$$
Dividing the inequality by $2$ (since we want to isolate $x$), we get:
$$-\frac{1}{2} < x < 6$$
So, $x$ is in the interval $\left(-\frac{1}{2}, 6\right)$ for $2x$.
Next, let's find the interval for $3x-7$:
Since $3x-7$ is in the interval $(-1, 12)$, we have the inequality:
$$-1 < 3x-7 < 12$$
Adding $7$ to all sides of the inequality, we get:
$$6 < 3x < 19$$
Dividing the inequality by $3$, we have:
$$2 < x < \frac{19}{3}$$
So, $x$ is in the interval $\left(2, \frac{19}{3}\right)$ for $3x-7$.
To find the values of $x$ that satisfy both conditions, we need to find the intersection of the two intervals $\left(-\frac{1}{2}, 6\right)$ and $\left(2, \frac{19}{3}\right)$.
By comparing the two intervals, we can see that the intersection interval is:
$$\left(2, 6\right)$$
Therefore, all $x$ such that both $2x$ and $3x-7$ are in the interval $(-1, 12)$ is $x \in \left(2, 6\right)$.