Intermediate Single-Variable Algebra

Help Questions

Math › Intermediate Single-Variable Algebra

Questions 1 - 10

Solve: $\frac{2}{3x}+\frac{x}{4}-\frac{1}{x}$

$\frac{1}{4}x-\frac{1}{3x}$

$\frac{1}{4x}-\frac{1}{3}x$

$\frac{1}{12}x$

$\frac{1}{12x}$

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

Explanation

Determine the least common denominator. Each term will need an x-variable, and all three denominators will need a common coefficient.

The least common denominator is .

Convert the fractions.

$\frac{2(4)}{3x(4)}+\frac{x(3x)}{4(3x)}-\frac{1(12)}{x(12)} = \frac{8}{12x}+\frac{3x^2}{12x}-\frac{12}{12x}$

The expression becomes:

$\frac{3x^2-4}{12x} = \frac{x}{4}-\frac{1}{3x}$

The answer is: $\frac{1}{4}x-\frac{1}{3x}$

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.

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

Re-order the terms.

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

Pull out a common factor of negative one on the numerator.

$\frac{-1(2x^2-5x-35)}{x^2+7x} = -\frac{2x^2-5x-35}{x^2+7x}$

The answer is: $-\frac{2x^2-5x-35}{x^2+7x}$

Subtract: $\frac{3}{2-x}-\frac{5}{2x}$

$-\frac{11x-10}{2x^2-4x}$

$-\frac{11x-10}{2x^2-2x}$

$-\frac{x+10}{2x^2-8x}$

$-\frac{4x+10}{2x^2-2x}$

$\frac{x+10}{2x^2-2x}$

Explanation

Multiply the denominators to get the least common denominator. We can then convert both fractions so that the denominators are alike.

$\frac{3}{2-x}-\frac{5}{2x}=\frac{3(2x)}{(2-x)(2x)}-\frac{5(2-x)}{2x(2-x)}$

Simplify both the top and the bottom.

$\frac{6x}{-2x^2+4x} - \frac{10-5x}{-2x^2+4x}$

Combine the numerators as one fraction. Be careful with the second fraction since the entire numerator is a quantity, which means we will need to brace with parentheses.

$\frac{6x-(10-5x)}{-2x^2+4x} =\frac{6x-10+5x}{-2x^2+4x} = \frac{11x-10}{-2x^2+4x}$

Pull out a common factor of negative one in the denominator. This allows us to rewrite the fraction with the negative sign in front of the fraction.

$\frac{11x-10}{-2x^2+4x}= \frac{11x-10}{-(2x^2-4x)}= -\frac{11x-10}{2x^2-4x}$

The answer is: $-\frac{11x-10}{2x^2-4x}$

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

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

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

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

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

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

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

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

Simplify the following polynomial:

$(\frac{2y^{-2}}{-2})^{-1}(\frac{m^2n}{y})^{-2}$

$\frac{-y^4}{m^4n^2}$

$\frac{y^4}{m^4n^2}$

$\frac{-y^3}{m^4n^2}$

$\frac{y^3}{m^4n^2}$

$\frac{y^2}{m^4n^2}$

Explanation

Begin by reversing the numerator and denominator so that the exponents are positive:

$(\frac{2y^{-2}}{-2})^{-1}(\frac{m^2n}{y})^{-2}$

$(\frac{-2}{2y^{-2}})^{1}(\frac{y}{m^2n})^{2}$

$(\frac{-2y^{2}}{2})^{1}(\frac{y}{m^2n})^{2}$

Square the right side of the expression and multiply:

$(\frac{-2y^{2}}{2})(\frac{y^2}{m^4n^2})$

$\frac{-2y^{4}}{2m^4n^2}$

Simplify:

$\frac{-y^{4}}{m^4n^2}$

Expand:

$3x^{2}-25x-18$

$3x^{2}-29x-18$

$3x^{2}-18x-29$

$3x^{2}-18x-25$

None of the other answers

Explanation

To multiple these binomials, you can use the FOIL method to multiply each of the expressions individually.This will give you

or $3x^{2}-25x-18$ .

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.