Before we get started, read the question carefully. We need to find x squared, not x. Don't call it quits too early!

So, we start here:

$\displaystyle 15x-5\leq54x-18$

Get the x's on one side and the constants on the other.

$\displaystyle 15x-54x\leq -18+5$

$\displaystyle -39x\leq -13$

When we divide by a negative number in an inequality, remember that we need to switch the direction of the sign.

$\displaystyle x\geq \frac{-13}{-39}\geq\frac{1}{3}$

$\displaystyle x\geq\frac{1}{3}$

$\displaystyle x^2\geq\frac{1}{9}$

$\displaystyle 3(2x+6)\leq 21+3x$

We can either divide the other side of the inequality by $\displaystyle 3$ or distribute it. We'll go ahead and distribute it here:

$\displaystyle 6x+18\leq 21+3x$

Now we just solve:

$\displaystyle 6x-3x\leq 21-18$

$\displaystyle 3x \leq 3$

$\displaystyle x \leq 1$

The first thing we must do is solve the given equation for $\displaystyle y$: 

$\displaystyle -12x+2y=8$

$\displaystyle 2y=8+12x$

$\displaystyle y=4+6x$

Since we are looking for values when $\displaystyle y\le46$, we can set up our equation as follows:

$\displaystyle 4+6x\le46$

Solve.

$\displaystyle 6x\le42$

$\displaystyle x\le7$

So, $\displaystyle 7$ is the largest integer of x which makes the statement true.

Our first step will be to solve the given equation for $\displaystyle y$:

$\displaystyle 7y-56x=91$

$\displaystyle 7y=91+56x$

$\displaystyle y=13+8x$

Since we want to know the smallest integer of $\displaystyle x$ for which $\displaystyle y\ge125$, we can set up our equation as

$\displaystyle 13+8x\ge125$

$\displaystyle 8x\ge112$

$\displaystyle x\ge14$

This problem requires first combining like terms that are on opposite sides of the inequality.

First add 3x to both sides of the inequality to get

$\displaystyle 9< 7x+2$.

Then subtract 2 from both sides.

$\displaystyle 7< 7x$

Now, divide 7 from both sides to get x alone.

$\displaystyle 7< x$

which you should switch the order to be

$\displaystyle x>7$.

We begin by moving all of our terms to the left side of the inequality.

$\displaystyle \small x^2+7x+12\leq0$

We then factor.

$\displaystyle \small (x+3)(x+4)\leq0$

That means our left side will equal 0 when $\displaystyle \small x=-4 \;or\;-3$.  However, we also want to know the values when the left side is less than zero.  We can do this using test regions.  We begin by drawing a number line with our two numbers labeled.

We notice that our two numbers divide our line into three regions.  We simply need to try a test value in each region.  We begin with our leftmost region by selecting a number less than $\displaystyle \small -4$.  We then plug that value into the left side of our inequality to see if the result is positive or negative.  Any value (such as $\displaystyle \small -5$) will give us a positive value.

We then repeat this process with the center region by selecting a value between our two numbers.  Any value (such as $\displaystyle \small -3.5$) will result in a negative outcome.

Finally we complete the process with the rightmost region by selecting a value larger than $\displaystyle \small -3$.  Any value (such as $\displaystyle \small -2$) will result in a positive value.

We then label our regions accordingly.

Since we want the result to be less than zero, we want the values between our two numbers.  However, since our left side can be less than or equal to zero, we can also include the two numbers themselves.  We can express this as

$\displaystyle \small -4\leq x\leq-3$

We begin by solving the equation for its zeros. This is done by changing the $\displaystyle \geq$ sign into an $\displaystyle =$ sign. 

$\displaystyle x^2+3x=4\Rightarrow x^2+3x-4=0$

$\displaystyle \\ \Rightarrow(x-1)(x+4)=0\\ \Rightarrow x=-4,1$

Since we know the zeros of the equation, we can then check the areas around the zeros since we naturally have split up the real line into three sections :

$\displaystyle (-\infty,-4),(-4,1),(1,\infty)$

First we check $\displaystyle x=-5$

$\displaystyle (-5)^2+3(-5)-4=25-15-4=6\geq0$

Therefore, the first interval can be included in our answer. Additionally, we know that $\displaystyle x=-4$ satisfies the equation, therefore we can say with certainty that the interval $\displaystyle (-\infty,-4]$ is part of the answer. 

Next we check something in the second interval. Let $\displaystyle x=0$, then

$\displaystyle 0^2+3(0)-4=-4< 0$

Therefore the second interval cannot be included in the answer.

Lastly, we check the third interval. Let $\displaystyle x=2$, then

$\displaystyle 2^2+3(2)-4=4+6-4=6\ge 0$

Which does satisfy the original equation. Therefore the third interval can also be included in the answer. Since we know that $\displaystyle x=1$ satisfies the equation as well, we can include it in the interval as such: $\displaystyle [1,\infty)$

Therefore, 

$\displaystyle x=(-\infty,-4]\cup[1,\infty)$

Method 1:

1) Multiply-out the left side then rewrite the inequality as an equation:

    $\displaystyle x^{2} + x = 12$

2) Now rewrite as a quadratic equation and solve the equation:

    $\displaystyle x^{2} + x - 12 = 0$

    $\displaystyle (x + 4)(x - 3) = 0$

    $\displaystyle x + 4 = 0$     $\displaystyle x - 3 = 0$

    $\displaystyle x = -4$       $\displaystyle x = 3$

3) Next set up intervals using the solutions and test the original inequality to   see where it holds true by using values for $\displaystyle x$ on each interval.

4) The interval between $\displaystyle x = -4$ and $\displaystyle x = 3$ holds true for the original inequality.

5) Solution: $\displaystyle -4 < x < 3$

Method 2:

Using a graphing calculator, find the graph.  The function is below the x-axis (less than $\displaystyle 0$) for the x-values $\displaystyle -4 < x < 3$.  Using interval notation for $\displaystyle x$, $\displaystyle (-4, 3)$.

Method 3:

For the inequality $\displaystyle x(x + 1) < 12$, the variable expression in terms of $\displaystyle x$ is less than $\displaystyle 12$, and an inequality has a range of values that the solution is composed of.  This means that each of the solution values for $\displaystyle x$ are strictly between the two solutions of $\displaystyle x(x + 1) = 12$.  'Between' is for a 'less than' case, 'Outside of' is for a 'greater than' case.

$\displaystyle 3\leq 9 +c$

Subtract 9 from both sides of the inequality.

$\displaystyle 3-9 \leq (9-9) + c$

$\displaystyle -6 \leq c$

Because the variable, in this case $\displaystyle c$ is to the right of the inequality sign, you switch the inequality sign to its opposite.  The $\displaystyle \leq$ becomes a $\displaystyle \geq$ so that the variable is to the left of the inequality sign.

$\displaystyle c\geq -6$

$\displaystyle y + \frac{2}{3} > 7$

Subtract $\displaystyle \frac{2}{3}$ from both sides of the inequality.

$\displaystyle y + \frac{2}{3} - \frac{2}{3} > 7 - \frac{2}{3}$

$\displaystyle y> 6\frac{1}{3}$