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Algebra II : Intermediate Single-Variable Algebra

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Example Questions

Example Question #9 : Factoring Rational Expressions

Simplify to simplest terms.

$\frac{x^2-x-56}{x^2-15x+56}$

Possible Answers:

$\frac{x+7}{x-7}$

Correct answer:

Explanation:

The correct answer is . The numerator and denominator can both be factored to simpler terms:

$\frac{(x-8)(x+7)}{(x-8)(x-7)}$

The (x-8 ) terms will cancel out. Leaving $\frac{x+7}{x-7}$ . While this is an answer choice, it can be simplified further. Factoring out a from the denominator will allow the x-7 terms to cancel out leaving .

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Example Question #11 : Factoring Rational Expressions

Simplify the rational expression by factoring:

$\frac{(x^2+2x+4)(x+2)^2}{(x+2)(x+3)}$

Possible Answers:

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

None of these.

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

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

$\frac{(x^2+2x+4)(x+2)}{(x+3)}$

Correct answer:

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

Explanation:

$\frac{(x^2+2x+4)(x+2)^2}{(x+2)(x+3)}$

To simplify it is best to completely factor all polynomials:

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

Now cancel like terms:

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

Combine like terms:

$\mathbf{\frac{(x+2)^3}{(x+3)}}$

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Example Question #131 : Rational Expressions

Solve for :

$\frac{4}{x+5} = \frac{8}{x}$

Possible Answers:

x=5

$x=\frac{-5}{4}$

x = 10

x = -10

$x = \frac{-10}{3}$

Correct answer:

x = -10

Explanation:

To solve this rational equation, start by cross multiplying:

$\frac{4}{x+5} = \frac{8}{x} \Rightarrow 4x = 8(x+5)$

Then, distribute the right side:

4x = 8x + 40

Finally, subtract from both sides and bring the over to the left side:

-40 = 4x

Dividing by gives the answer:

x = -10

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Example Question #2 : Solving Rational Expressions

Solve for :

$\frac{x}{x+5} = \frac{2}{x} + 3$

Possible Answers:

$\frac{17\pm \sqrt{209}}{2}$

$\frac{-17}{4}, 0$

$\frac{23\pm \sqrt{173}}{4}$

$\frac{-30\pm \sqrt{157}}{4}$

$\frac{-17\pm \sqrt{209}}{4}$

Correct answer:

$\frac{-17\pm \sqrt{209}}{4}$

Explanation:

The first step is to multiply everything by a common denominator. One way to do this is to multiply the entire equation by all three denominators:

$\left ( \frac{x}{x+5} = \frac{2}{x} + 3 \right )*(x+5)(x)*1$

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

$x^{2} = 2x + 10 + 3x^{2} + 15x$

$0 = 2x^{2} +17x +10$

Then, to solve for , use the quadratic formula:

$x = \frac{-17 \pm \sqrt{17^{2}-4(2)(10)} }{2*2} = \frac{-17\pm \sqrt{209}}{4}$

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Example Question #11 : How To Find The Solution To An Equation

Solve for , given the equation below.

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

Possible Answers:

$-2,\; 4$

$2,\; 4$

No solutions

$4,\; 4$

$-2,\; 2$

Correct answer:

$-2,\; 2$

Explanation:

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

Begin by cross-multiplying.

x(x+2) = (x-1)(2x +4)

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

x^2+2x=2x^2+4x-2x-4

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

x^2-2x=2x^2+2x-4

0=x^2-4

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

x^2=4

x=2, -2

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Example Question #1 : How To Find Out When An Equation Has No Solution

Solve the rational equation:

$\small \frac{1}{m+3}+\frac{1}{m-3}=\frac{6}{m^2-9}$

Possible Answers:

$\small m=6$

$\small m=3$ or $\small m=-3$

$\small m=0$

$\small m=\frac{1}{3}$

no solution

Correct answer:

no solution

Explanation:

With rational equations we must first note the domain, which is all real numbers except $\small \small m=-3$ and $\small \small m=3$ . That is, these are the values of $\small m$ that will cause the equation to be undefined. Since the least common denominator of $\small m+3$ , $\small m-3$ , and $\small m^2-9$ is $\small (m+3)(m-3)$ , we can mulitply each term by the LCD to cancel out the denominators and reduce the equation to $\small (m-3)+(m+3)=6$ . Combining like terms, we end up with $\small 2m=6$ . Dividing both sides of the equation by the constant, we obtain an answer of $\small m=3$ . However, this solution is NOT in the domain. Thus, there is NO SOLUTION because $\small m=3$ is an extraneous answer.

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Example Question #3 : Solving Rational Expressions

Simplify:

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

Possible Answers:

$\frac{2}{3}$

$\frac{3}{2}$

Correct answer:

Explanation:

Factor out from the numerator which gives us

$\left -(3-2x \right )$

Hence we get the following

$\frac{\left -(3-2x \right )}{3-2x}$

which is equal to

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Example Question #3 : Solving Rational Expressions

Solve:

$\frac{5}{x+1}-\frac{3x}{\left ( x+1 \right )\left ( x-2 \right )}= \frac{3}{x-2}$

Possible Answers:

$\frac{13}{2}$

Correct answer:

Explanation:

First we convert each of the denominators into an LCD which gives us the following:

$\frac{5\left ( x-2 \right )}{\left ( x+1 \right )\left ( x-2 \right )}- \frac{3x}{\left ( x+1 \right )\left ( x-2 \right )}= \frac{3\left ( x+1 \right )}{\left ( x+1 \right )\left ( x-2 \right )}$

Now we add or subtract the numerators which gives us:

$5\left ( x-2 \right )-3x= 3\left ( x+1 \right )$

Simplifying the above equation gives us the answer which is:

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Example Question #4 : Solving Rational Expressions

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

Solve for .

Possible Answers:

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

y=2 , $y=-\frac{9}{37}$

y=6 , $y=-\frac{7}{15}$

y=-1 , y=4

y=4 , y=0

Correct answer:

y=2 , $y=-\frac{9}{37}$

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 (y)(8y+2) , 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

(37y+9)(y-2) .

Our solutions are therefore

y=2

and

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

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Example Question #2 : Solving Rational Expressions

Solve for :

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

Possible Answers:

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

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

x = 4

x = 7

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

Correct answer:

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

Explanation:

Multiply both sides by (x-1) (x-3) :

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

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

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

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

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

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

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

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 -20,-7 :

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

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

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

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

Set each factor equal to 0 and solve separately:

5x - 7 = 0

5x - 7 + 7 = 0 + 7

5x = 7

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

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

x - 4 = 0

x - 4 + 4 = 0+ 4

x = 4

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Algebra II : Intermediate Single-Variable Algebra

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #9 : Factoring Rational Expressions

Example Question #11 : Factoring Rational Expressions

Example Question #131 : Rational Expressions

Example Question #2 : Solving Rational Expressions

Example Question #11 : How To Find The Solution To An Equation

Example Question #1 : How To Find Out When An Equation Has No Solution

Example Question #3 : Solving Rational Expressions

Example Question #3 : Solving Rational Expressions

Example Question #4 : Solving Rational Expressions

Example Question #2 : Solving Rational Expressions

All Algebra II Resources

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