\(\displaystyle 12< 3n-5< 27\)

\(\displaystyle 17< 3n< 32\)

\(\displaystyle \frac{17}{3}< n< \frac{32}{3}\)

The lowest value n can take is 6.

Our first step is to simplify the expression. We need to remember our order of operations or PEMDAS.

First distribute the 0.4 to the binomial.

\(\displaystyle 0.4(50-x)\times10+0.6x\times5 < 0\)

\(\displaystyle (20-0.4x)\times 10 +0.6x \times 5< 0\)

Now distribute the 10 to the binomial.

\(\displaystyle 200-4x+0.6x \times 5< 0\)

Now multiply 0.6 by 5

\(\displaystyle 200-4x+3x< 0\)

\(\displaystyle 200-x< 0\)

\(\displaystyle -x< -200\)

Remember to flip the sign of the inequation when multiplying or dividing by a negative number.

\(\displaystyle x>200\)

220 will make the expression negative.

To solve this problem all we need to do is solve for \(\displaystyle x\): 

\(\displaystyle 3x+12\leq48\)

\(\displaystyle 3x\leq36\)

\(\displaystyle x\leq12\)

Solve for \(\displaystyle m\) in both inequalities:

(1)

\(\displaystyle 9-3m\leq5\)

Subtract 9 from each side of the inequality:

\(\displaystyle -3m\leq-4\)

Then, divide by \(\displaystyle -3\). Remember to switch the direction of the sign from "less than or equal to" to "greater than or equal to."

\(\displaystyle m\geq\frac{4}{3}\)

(2)

\(\displaystyle 2m+7\leq13-m\)

Add \(\displaystyle m\) to both sides of the inequality:

\(\displaystyle 3m+7\leq13\)

Subtract 7 from each side of the inequality:

\(\displaystyle 3m\leq6\)

Divide each side of the inequality by 3:

\(\displaystyle m\leq2\)

From solving both inequalities, we find \(\displaystyle m\) such that:

\(\displaystyle \frac{4}{3}\leq m\leq 2\)

\(\displaystyle \frac{4}{3}\approx 1.33\)

Only \(\displaystyle \frac{3}{2}\)  is in that interval \(\displaystyle (\frac{3}{2}=1.5)\).

\(\displaystyle \frac{5}{4}\) and \(\displaystyle \frac{2}{3}\) are less than 1.33, so they are not in the interval.

\(\displaystyle \frac{7}{3}\) and \(\displaystyle \frac{8}{3}\) are more than 2, so they are not in the interval.

The inequality that is presented in the problem is:

\(\displaystyle 8-7x\leq11x+4\)

Start by moving your variables to one side of the inequality and all other numbers to the other side:

\(\displaystyle -7x-11x\leq4-8\)

\(\displaystyle -18x\leq-4\)

Divide both sides of the equation by \(\displaystyle -18\). Remember to flip the direction of the inequality's sign since you are dividing by a negative number!

\(\displaystyle x\geq\frac{-4}{-18}\)

Reduce:

\(\displaystyle x\geq\frac{2}{9}\)

The only answer choice with a value greater than \(\displaystyle \frac{2}{9}\) is \(\displaystyle \frac{4}{9}\).

First let's solve each of the inequalities:

 \(\displaystyle 6m-7>11\)

\(\displaystyle 6m>18\)

\(\displaystyle m>3\)

\(\displaystyle n< 2n-9\)

\(\displaystyle -n< -9\)

Don't forget to flip the direction of the sign when dividing by a negative number:

\(\displaystyle n>9\)

\(\displaystyle m+n>12\) is the correct answer. \(\displaystyle m\) is an integer greater than 3 and \(\displaystyle n\) is greater than 9. Therefore, the sum of \(\displaystyle m\) and \(\displaystyle n\) is greater than 12.

\(\displaystyle m+n< -12\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are two positive integers as \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The sum of two positive integers cannot be a negative number.

\(\displaystyle \frac{m}{n}< 0\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are two positive integers as \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The division of two positive numbers is positive and therefore cannot be less than 0.

\(\displaystyle m*n< 3\) is not true. \(\displaystyle m\) is greater than 3 and \(\displaystyle n\) is greater than 9. The product of \(\displaystyle m\) and \(\displaystyle n\) cannot be less than 3.

\(\displaystyle -m*n>0\) is not true. \(\displaystyle m\) and \(\displaystyle n\) are positive. Therefore, the product of \(\displaystyle -m\) and \(\displaystyle n\) is negative and cannot be greater than 0.

First, we solve both inequalities:

\(\displaystyle 3-x>8-2x\)

\(\displaystyle 3-x+2x>8\)

\(\displaystyle x>5\)

\(\displaystyle y-2>9\)

\(\displaystyle y>11\)

If \(\displaystyle m\) satisfies both inequalities, then \(\displaystyle m\) is greater than 5 AND \(\displaystyle m\) is greater than 11. Therefore \(\displaystyle m\) is greater than 11.

\(\displaystyle m>11\) is the correct answer.

We start by simplifying our inequality like any other equation:

\(\displaystyle 7x-4< 13x+26\)

\(\displaystyle -6x< 30\)

Now we must remember that when we divide by a negative, the inequality is flipped, so we obtain:

\(\displaystyle x>-5\)

Solving inequalities is very similar to solving equations, but we need to remember an important rule:

If we multiply or divide by a negative number, we must switch the direction of the inequality. So a "greater than" sign will become a "less than" sign and vice versa.

We are given

\(\displaystyle 4x-7\geq13x+11\)

Start by moving the \(\displaystyle 7\) and the \(\displaystyle 13x\) over:

\(\displaystyle 4x-7({\color{Blue} -13x})({\color{Red} +7})\geq13x+11({\color{Red} +7})({\color{Blue} -13x})\)

Simplify to get the following:

\(\displaystyle -9x\geq18\)

Then, we will divide both sides of the equation by \(\displaystyle -9\). Remember to switch the direction of the inequality sign!

\(\displaystyle \frac{-9x}{-9}\geq\frac{18}{-9}\)

\(\displaystyle \small x\leq\frac{18}{-9}\)

So,

\(\displaystyle \small \small x\leq-2\)

The question is equivalent to asking when \(\displaystyle x > y\) and \(\displaystyle x > z\) are both true statements. 

We can solve for \(\displaystyle y\) in the first equation:

\(\displaystyle x = 2y- 6\)

\(\displaystyle 2y = x + 6\)

\(\displaystyle y = \frac{1}{2}(x+6)\)

\(\displaystyle y = \frac{1}{2}x+3\)

Now use substitution to solve the inequality

\(\displaystyle x > y\)

\(\displaystyle x > \frac{1}{2}x+3\)

\(\displaystyle x - \frac{1}{2}x > \frac{1}{2}x+3 - \frac{1}{2}x\)

\(\displaystyle \frac{1}{2}x > 3\)

\(\displaystyle 2 \cdot \frac{1}{2}x >2 \cdot 3\)

\(\displaystyle x > 6\)

Therefore, \(\displaystyle x > y\) if and only if \(\displaystyle x > 6\).

Similarly, we can solve for \(\displaystyle z\) in the first equation:

\(\displaystyle x = 10 - 5z\)

\(\displaystyle x + 5z - x = 10 - 5z + 5z - x\)

\(\displaystyle 5z = 10-x\)

\(\displaystyle z = \frac{1}{5}(10-x)\)

\(\displaystyle z = 2 -\frac{1}{5}x\)

Now use substitution to solve the inequality

\(\displaystyle x > z\)

\(\displaystyle x >2 -\frac{1}{5}x\)

\(\displaystyle x+ \frac{1}{5}x >2 -\frac{1}{5}x + \frac{1}{5}x\)

\(\displaystyle \frac{6}{5}x >2\)

\(\displaystyle \frac{5}{6} \cdot \frac{6}{5}x >\frac{5}{6} \cdot 2\)

\(\displaystyle x > \frac{5}{3} = 1\frac{2}{3}\)

Therefore, \(\displaystyle x > z\) if and only if \(\displaystyle x > 1\frac{2}{3}\).

We combine these results, and conclude that both \(\displaystyle x > y\) and \(\displaystyle x > z\) hold - that is, \(\displaystyle x\) is the greatest of the three - if and only if \(\displaystyle x > 6\).