To solve this equation, repeatedly do the same thing to both sides of the equation until the variable is isolated on one side of the < sign and the answer on the other.

3x + 7 < \( \frac{x}{1} \)
1 x (3x + 7) < x
(1 x 3x) + (1 x 7) < x
3x + 7 < x
3x + 7 - x < 0
3x - x < -7
2x < -7
x < \( \frac{-7}{2} \)
x < -3\(\frac{1}{2}\)

To factor a quadratic expression, apply the FOIL method (First, Outside, Inside, Last) in reverse. First, find the two Last terms that will multiply to produce 5 as well and sum (Inside, Outside) to equal 6. For this problem, those two numbers are 1 and 5. Then, plug these into a set of binomials using the square root of the First variable (y²):

y² + 6y + 5
y² + (1 + 5)y + (1 x 5)
(y + 1)(y + 5)

Parallel lines are lines that share the same slope (steepness) and therefore never intersect. A transversal occurs when a set of parallel lines are crossed by another line. All of the angles formed by a transversal are called interior angles and angles in the same position on different parallel lines equal each other (a° = w°, b° = x°, c° = z°, d° = y°) and are called corresponding angles. Alternate interior angles are equal (a° = z°, b° = y°, c° = w°, d° = x°) and all acute angles (a° = c° = w° = z°) and all obtuse angles (b° = d° = x° = y°) equal each other. Same-side interior angles are supplementary and add up to 180° (e.g. a° + d° = 180°, d° + c° = 180°).

To find the diagonal we need to know the length of one of the square's sides. We know the area and the area of a square is the length of one side squared:

a = s²

so the length of one side of the square is:

s = \( \sqrt{a} \) = \( \sqrt{16} \) = 4

The Pythagorean theorem defines the square of the hypotenuse (diagonal) of a triangle with a right angle as the sum of the squares of the other two sides:

c² = a² + b²
c² = 4² + 4²
c² = 32
c = \( \sqrt{32} \) = \( \sqrt{16 x 2} \) = \( \sqrt{16} \) \( \sqrt{2} \)
c = 4\( \sqrt{2} \)

A monomial contains one term, a binomial contains two terms, and a polynomial contains more than two terms.