Intermediate Single-Variable Algebra

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Math › Intermediate Single-Variable Algebra

Questions 1 - 10

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:

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

Simplify the following polynomial:

$ab^{-2}(a^2b+a^{-1}b^3-a^3b^{-1})$

$\frac{a^3}{b}+b-\frac{a^4}{b^3}$

$\frac{a^2}{b}+b-\frac{a^4}{b^3}$

$\frac{a^3}{b}+b^2-\frac{a^4}{b^3}$

$\frac{a^3}{b}+a-\frac{a^4}{b^3}$

$\frac{a^3}{b}+b-\frac{a^3}{b^4}$

Explanation

To simplify the polynomial, begin by multiplying the first binomial by every term within the parentheses:

$ab^{-2}(a^2b+a^{-1}b^3-a^3b^{-1})$

$aa^2b^{-2}b+aa^{-1}b^{-2}b^3-aa^3b^{-2}b^{-1}$

Now, combine like terms:

$a^3b^{-1}+b^1-a^4b^{-3}$

Convert the polynomial into fraction form:

$\frac{a^3}{b}+b-\frac{a^4}{b^3}$

$\frac{y+1}{8y+2} + \frac{3}{y} = \frac{5}{3}$

Solve for .

, $y=-\frac{9}{37}$

, $y=-\frac{7}{15}$

$y=-\frac{5}{2}$

Explanation

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

$\frac{y+1}{8y+2}\cdot \frac{y}{y}$ becomes $\frac{y^{2}+y}{(y)(8y+2)}$ .

$\frac{3}{y}\cdot \frac{8y+2}{8y+2}$ becomes $\frac{24y+6}{(y)(8y+2)}$ .

Now add the two fractions: $\frac{y^{2}+25y+6}{(y)(8y+2)}$

To solve, multiply both sides of the equation by , yielding

$y^{2}+25y+6 = \frac{(40y^{2}+10y)}{3}$ .

Multiply both sides by 3:

$3y^{2}+75y+18 = 40y^{2}+ 10y$

Move all terms to the same side:

$37y^{2}-65y-18 = 0$

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

Our solutions are therefore

and

$y=-\frac{9}{37}$ .

If , , and , what is ?

Explanation

To find , we must start inwards and work our way outwards, i.e. starting with :

We can now use this value to find as follows:

Our final answer is therefore

$\frac{y+1}{8y+2} + \frac{3}{y} = \frac{5}{3}$

Solve for .

, $y=-\frac{9}{37}$

, $y=-\frac{7}{15}$

$y=-\frac{5}{2}$

Explanation

The two fractions on the left side of the equation need a common denominator. We can easily do find one by multiplying both the top and bottom of each fraction by the denominator of the other.

$\frac{y+1}{8y+2}\cdot \frac{y}{y}$ becomes $\frac{y^{2}+y}{(y)(8y+2)}$ .

$\frac{3}{y}\cdot \frac{8y+2}{8y+2}$ becomes $\frac{24y+6}{(y)(8y+2)}$ .

Now add the two fractions: $\frac{y^{2}+25y+6}{(y)(8y+2)}$

To solve, multiply both sides of the equation by , yielding

$y^{2}+25y+6 = \frac{(40y^{2}+10y)}{3}$ .

Multiply both sides by 3:

$3y^{2}+75y+18 = 40y^{2}+ 10y$

Move all terms to the same side:

$37y^{2}-65y-18 = 0$

This looks like a complicated equation to factor, but luckily, the only factors of 37 are 37 and 1, so we are left with

Our solutions are therefore

and

$y=-\frac{9}{37}$ .

Solve for :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$x = 1 \frac{2}{5} \textrm{ or } x = 4$

$x = 1 \frac{2}{5}$

$x = 4\textrm{ or }x = 7$

Explanation

Multiply both sides by :

$\frac{3}{x-1} + \frac{4}{x- 3 } = 5$

$\frac{3}{x-1} \cdot (x-1) (x-3)+ \frac{4}{x- 3 } \cdot (x-1) (x-3) = 5\cdot (x-1) (x-3)$

$3 \cdot (x-3)+4\cdot (x-1) = 5\cdot (x-1) (x-3)$

$3x-9 + 4x-4 = 5 (x ^{2 } -4x+3)$

$7x-13= 5 x ^{2 } -20x+15$

$7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13$

$5 x ^{2 } -27x + 28 = 0$

Factor this using the -method. We split the middle term using two integers whose sum is and whose product is $5 \cdot28 = 140$ . These integers are :

$5 x ^{2 } -20x -7x + 28 = 0$

$\left (5 x ^{2 } -20x \right ) - \left (7x - 2 8\right ) = 0$

$5x \left (x-4 \right ) - 7 \left (x-4 \right ) = 0$

$\left (5x - 7 \right )\left (x-4 \right ) = 0$

Set each factor equal to 0 and solve separately:

$5x \div 5 = 7 \div 5$

$x = \frac{7}{5} = 1 \frac{2}{5}$

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: $\frac{49}{12x}$

Solve for , given the equation below.

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

No solutions

Explanation

$\frac{x + 2}{x - 1} = \frac{2x + 4}{x}$

Begin by cross-multiplying.

Distribute the on the left side and expand the polynomial on the right.

Combine like terms and rearrange to set the equation equal to zero.

Now we can isolate and solve for by adding to both sides.

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.