Find all $x$ which satisfy both the inequalities $0\ge 54x+144-18x+108$ and $0>12-20x+30x-66$. Express your answer in interval notation, reducing any fractions in your answer.

Question

Find all $x$ which satisfy both the inequalities $0\ge 54x+144-18x+108$ and $0>12-20x+30x-66$. Express your answer in interval notation, reducing any fractions in your answer.

Answer 1

To find the values of $x$ that satisfy both inequalities, we need to solve each inequality separately and then find the intersection of the solution sets.

1) $0\ge 54x+144-18x+108$

Simplifying the inequality, we get:

$0 \ge 36x + 252$

Now, isolate $x$:

$36x \le -252$

Divide both sides of the inequality by 36 (note that dividing by a negative number flips the inequality sign):

$x \le -7$

2) $0 > 12-20x+30x-66$

Simplifying the inequality, we get:

$0 > 30x - 54$

Now, isolate $x$:

$30x > 54$

Divide both sides of the inequality by 30 (note that dividing by a positive number keeps the inequality sign the same):

$x > \frac{9}{5}$

Now, we find the intersection of the solution sets for $x$:

$x \le -7$ and $x > \frac{9}{5}$

This means that the values of $x$ that satisfy both inequalities are in the interval $\boxed{\left(-7, \frac{9}{5}\right]}$.

Answer 2

To solve the first inequality $0\ge 54x+144-18x+108$, we can start by simplifying the expression on the right-hand side:

$54x+144-18x+108$

Combine like terms:

$36x+252$

Now we have the inequality $0\ge 36x+252$. To solve this inequality, we can isolate the variable $x$ by subtracting $252$ from both sides:

$-252\ge 36x$

Now divide both sides by $36$:

$\frac{-252}{36} \ge x$

Simplifying the fraction gives:

$-7 \ge x$

So, the solution to the first inequality is $x\le -7$.

Moving on to the second inequality $0>12-20x+30x-66$, we can simplify the expression on the right-hand side:

$12-20x+30x-66$

Combine like terms:

$10x-54$

Now we have the inequality $0> 10x-54$. To solve this inequality, we can isolate the variable $x$ by adding $54$ to both sides:

$54 > 10x$

Divide both sides by $10$:

$\frac{54}{10} > x$

Reducing the fraction gives:

$\frac{27}{5} > x$

So, the solution to the second inequality is $x< \frac{27}{5}$.

To find the intersection of both solutions, we need to find the values of $x$ that satisfy both inequalities, which means $x$ must satisfy both $x\le -7$ and $x< \frac{27}{5}$.

Since $x$ must be smaller than both $-7$ and $\frac{27}{5}$, the solution is the intersection of these two intervals. Therefore, the solution in interval notation is $(-\infty, \frac{27}{5})$.