Algebra II - Intermediate Single-Variable Algebra

$\frac{2-x}{x^2}+\frac{4}{x}$

$\frac{3x+2}{x^2}$

$-\frac{3x+2}{x^2}$

$\frac{3x-2}{x^2}$

$\frac{3x+2}{x}$

$\frac{3}{x^2}$

Explanation

To combine these rational expressions, first find the common denominator. In this case, it is . Then, offset the second equation so that you get the correct denominator: $\frac{4}{x}=\frac{4x}{x^2}$ . Then, combine the numerators: . Put your numerator over the denominator for your answer: $\frac{3x+2}{x^2}$ .

Add: $\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x}$

$\frac{49}{12x}$

$\frac{49}{12}x$

$\frac{49}{24x}$

$\frac{14}{3x}$

$\frac{14}{3}x$

Explanation

Determine the least common denominator in order to add the numerator.

Each denominator shares an term. The least common denominator is since it is divisible by each coefficient of the denominator.

Convert the fractions.

$\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x} = \frac{3(6)}{2x(6)}+\frac{4(4)}{3x(4)}+\frac{5(3)}{4x(3)}$

Simplify the top and bottom.

$\frac{18}{12x}+\frac{16}{12x}+\frac{15}{12x}=\frac{49}{12x}$

The answer is: $\frac{49}{12x}$

Which of the following is the same after completing the square?

$(x+\frac{1}{6})^2 =\frac{73}{36}$

$(x-\frac{1}{6})^2 =\frac{73}{36}$

$(x+\frac{1}{6})^2 =\frac{1}{2}$

$(x+\frac{1}{3})^2 =\frac{7}{2}$

$(x-\frac{1}{3})^2 =\frac{7}{2}$

Explanation

Divide by three on both sides.

$\frac{3x^2+x-6 }{3}=\frac{0}{3}$

$x^2+\frac{1}{3}x-2 = 0$

Add two on both sides.

$x^2+\frac{1}{3}x-2 +2= 0+2$

$x^2+\frac{1}{3}x = 2$

To complete the square, we will need to divide the one-third coefficient by two, which is similar to multiplying by one half, square the quantity, and add the two values on both sides.

$x^2+\frac{1}{3}x+ (\frac{1}{3}\cdot \frac{1}{2})^2 = 2+ (\frac{1}{3}\cdot \frac{1}{2})^2$

Simplify both sides.

$x^2+\frac{1}{3}x+ \frac{1}{36} = 2+ \frac{1}{36}$

Factor the left side, and combine the terms on the right.

$(x+\frac{1}{6})^2 = \frac{72}{36}+ \frac{1}{36}$

The answer is: $(x+\frac{1}{6})^2 =\frac{73}{36}$

Simplify: $\frac{x^2-x}{x^3-x} \cdot \frac{x+1}{x-1}$

$\frac{1}{x-1}$

$\frac{x}{x-1}$

$\frac{x}{x+1}$

$\frac{1}{x^2-1}$

Explanation

Rewrite the left fraction using common factors.

$\frac{x^2-x}{x^3-x} \cdot \frac{x+1}{x-1} = \frac{x(x-1)}{x(x^2-1)} \cdot \frac{x+1}{x-1}$

Cancel out common terms.

$\frac{x+1}{x^2-1}$

Factorize the bottom term.

$\frac{x+1}{x^2-1} =\frac{x+1}{(x+1)(x-1)} = \frac{1}{x-1}$

The answer is: $\frac{1}{x-1}$

Solve for .

$\frac{x}{4}=\frac{3}{4}$

Explanation

To solve for the variable , isolate the variable on one side of the equation with all other constants on the other side. To accomplish this perform the opposite operation to manipulate the equation.

First cross multiply.

$\frac{x}{4}=\frac{3}{4}$

Next, divide by four on both sides.

$\frac{4x}{4}=\frac{12}{4}=\frac{4\cdot 3}{4}=3$

Add: $\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x}$

$\frac{49}{12x}$

$\frac{49}{12}x$

$\frac{49}{24x}$

$\frac{14}{3x}$

$\frac{14}{3}x$

Explanation

Determine the least common denominator in order to add the numerator.

Each denominator shares an term. The least common denominator is since it is divisible by each coefficient of the denominator.

Convert the fractions.

$\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x} = \frac{3(6)}{2x(6)}+\frac{4(4)}{3x(4)}+\frac{5(3)}{4x(3)}$

Simplify the top and bottom.

$\frac{18}{12x}+\frac{16}{12x}+\frac{15}{12x}=\frac{49}{12x}$

The answer is: $\frac{49}{12x}$

Factor .

Cannot be factored any further.

Explanation

This is a difference of squares. The difference of squares formula is _a_2 – _b_2 = (a + b)(a – b).

In this problem, a = 6_x_ and b = 7_y_:

36_x_2 – 49_y_2 = (6_x_ + 7_y_)(6_x_ – 7_y_)

Solve for by completing the square.

$x=0.24\text{ and }x=-8.24$

$x=0.39\text{ and }x=-4.39$

$x=1.24\text{ and }x=-10.24$

$x=0.97\text{ and }x=-8.97$

Explanation

Start by adding to both sides so that the terms with the are together on the left side of the equation.

Now, look at the coefficient of the -term. To complete the square, divide this coefficient by , then square the result. Add this term to both sides of the equation.

Rewrite the left side of the equation in the squared form.

Take the square root of both sides.

$x+4=\pm \sqrt{18}$

Now solve for .

$x=-4+\sqrt{18}=0.24$

$x=-4-\sqrt{18}=-8.24$

Round to two places after the decimal.

Add: $\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x}$

$\frac{49}{12x}$

$\frac{49}{12}x$

$\frac{49}{24x}$

$\frac{14}{3x}$

$\frac{14}{3}x$

Explanation

Determine the least common denominator in order to add the numerator.

Each denominator shares an term. The least common denominator is since it is divisible by each coefficient of the denominator.

Convert the fractions.

$\frac{3}{2x}+\frac{4}{3x}+\frac{5}{4x} = \frac{3(6)}{2x(6)}+\frac{4(4)}{3x(4)}+\frac{5(3)}{4x(3)}$

Simplify the top and bottom.

$\frac{18}{12x}+\frac{16}{12x}+\frac{15}{12x}=\frac{49}{12x}$

The answer is: $\frac{49}{12x}$

Simplify: $\frac{5}{x}-\frac{2x}{x+7}$

$-\frac{2x^2-5x-35}{x^2+7x}$

$-\frac{2x^2+5x-35}{x^2+7x}$

$-\frac{2x^2-5x-35}{x+7}$

$\frac{2x^2-5x-35}{x+7}$

$\frac{2x-5}{x-7}$

Explanation

In order to add the numerators, we will need the least common denominator.

Multiply the denominators together.

Convert both fractions by multiplying both the top and bottom by what was multiplied to get the denominator. Rewrite the fractions and combine as one single fraction.

$\frac{5(x+7)}{x(x+7)}-\frac{2x(x)}{(x)(x+7)} = \frac{5x+35-2x^2}{x^2+7x}$