Algebra II : Solving Rational Expressions

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #1 : Solving Rational Expressions

Solve for \(\displaystyle x\):

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

Possible Answers:

\(\displaystyle x=\frac{-5}{4}\)

\(\displaystyle x = -10\)

\(\displaystyle x=5\)

\(\displaystyle x = \frac{-10}{3}\)

\(\displaystyle x = 10\)

Correct answer:

\(\displaystyle x = -10\)

Explanation:

To solve this rational equation, start by cross multiplying:

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

Then, distribute the right side:

\(\displaystyle 4x = 8x + 40\)

Finally, subtract \(\displaystyle 4x\) from both sides and bring the \(\displaystyle 40\) over to the left side:

\(\displaystyle -40 = 4x\)

Dividing by \(\displaystyle 4\) gives the answer:

\(\displaystyle x = -10\)

Example Question #132 : Rational Expressions

Solve for \(\displaystyle x\):

\(\displaystyle \frac{x}{x+5} = \frac{2}{x} + 3\)

Possible Answers:

\(\displaystyle \frac{-30\pm \sqrt{157}}{4}\)

\(\displaystyle \frac{-17}{4}, 0\)

\(\displaystyle \frac{-17\pm \sqrt{209}}{4}\)

\(\displaystyle \frac{17\pm \sqrt{209}}{2}\)

\(\displaystyle \frac{23\pm \sqrt{173}}{4}\)

Correct answer:

\(\displaystyle \frac{-17\pm \sqrt{209}}{4}\)

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:

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

\(\displaystyle x^{2} = 2(x+5) + 3(x+5)(x)\)

\(\displaystyle x^{2} = 2x + 10 + 3x^{2} + 15x\)

\(\displaystyle 0 = 2x^{2} +17x +10\)

Then, to solve for \(\displaystyle x\), use the quadratic formula:

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

 

 

 

Example Question #1 : Solving Rational Expressions

Solve for \(\displaystyle x\), given the equation below.

\(\displaystyle \frac{x + 2}{x - 1} = \frac{2x + 4}{x}\)

Possible Answers:

\(\displaystyle -2,\; 2\)

\(\displaystyle -2,\; 4\)

\(\displaystyle 2,\; 4\)

\(\displaystyle 4,\; 4\)

No solutions

Correct answer:

\(\displaystyle -2,\; 2\)

Explanation:

\(\displaystyle \frac{x + 2}{x - 1} = \frac{2x + 4}{x}\)

Begin by cross-multiplying.

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

Distribute the \(\displaystyle x\) on the left side and expand the polynomial on the right.

\(\displaystyle x^2+2x=2x^2+4x-2x-4\)

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

\(\displaystyle x^2-2x=2x^2+2x-4\)

\(\displaystyle 0=x^2-4\)

Now we can isolate and solve for \(\displaystyle x\) by adding \(\displaystyle 4\) to both sides.

\(\displaystyle x^2=4\)

\(\displaystyle x=2, -2\)

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

Solve the rational equation:

\(\displaystyle \small \frac{1}{m+3}+\frac{1}{m-3}=\frac{6}{m^2-9}\)

Possible Answers:

\(\displaystyle \small m=0\)

\(\displaystyle \small m=\frac{1}{3}\)

\(\displaystyle \small m=3\) or \(\displaystyle \small m=-3\)

no solution

\(\displaystyle \small m=6\)

Correct answer:

no solution

Explanation:

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

Example Question #662 : Intermediate Single Variable Algebra

Simplify:

 

\(\displaystyle \frac{2x-3}{3-2x}\)

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle \frac{3}{2}\)

\(\displaystyle \frac{2}{3}\)

\(\displaystyle 0\)

\(\displaystyle -1\)

Correct answer:

\(\displaystyle -1\)

Explanation:

Factor out \(\displaystyle -1\) from the numerator which gives us

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

Hence we get the following

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

which is equal to \(\displaystyle -1\)

Example Question #3 : Solving Rational Expressions

Solve:

 

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

 

Possible Answers:

\(\displaystyle -13\)

\(\displaystyle -7\)

\(\displaystyle 7\)

\(\displaystyle \frac{13}{2}\)

\(\displaystyle 13\)

Correct answer:

\(\displaystyle -13\)

Explanation:

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

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

Now we add or subtract the numerators which gives us:

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

 

Simplifying the above equation gives us the answer which is:

\(\displaystyle -13\)

Example Question #2 : Solving Rational Expressions

\(\displaystyle \frac{y+1}{8y+2} + \frac{3}{y} = \frac{5}{3}\)

Solve for \(\displaystyle y\).

 

Possible Answers:

\(\displaystyle y=-1\)\(\displaystyle y=4\)

\(\displaystyle y=4\)\(\displaystyle y=0\)

\(\displaystyle y=-\frac{5}{2}\)

\(\displaystyle y=2\)\(\displaystyle y=-\frac{9}{37}\)

\(\displaystyle y=6\)\(\displaystyle y=-\frac{7}{15}\)

Correct answer:

\(\displaystyle y=2\)\(\displaystyle y=-\frac{9}{37}\)

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.

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

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

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

To solve, multiply both sides of the equation by \(\displaystyle (y)(8y+2)\), yielding

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

Multiply both sides by 3:

 \(\displaystyle 3y^{2}+75y+18 = 40y^{2}+ 10y\)

Move all terms to the same side:

 \(\displaystyle 37y^{2}-65y-18 = 0\)

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

 \(\displaystyle (37y+9)(y-2)\).

Our solutions are therefore

\(\displaystyle y=2\) 

and

\(\displaystyle y=-\frac{9}{37}\).

Example Question #2 : Solving Rational Expressions

Solve for \(\displaystyle x\):

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

Possible Answers:

\(\displaystyle x = 4\textrm{ or }x = 7\)

\(\displaystyle x = 4\)

\(\displaystyle x = 7\)

\(\displaystyle x = 1 \frac{2}{5} \textrm{ or } x = 4\)

\(\displaystyle x = 1 \frac{2}{5}\)

Correct answer:

\(\displaystyle x = 1 \frac{2}{5} \textrm{ or } x = 4\)

Explanation:

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

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

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

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

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

\(\displaystyle 7x-13= 5 x ^{2 } -20x+15\)

\(\displaystyle 7x -13 -7x + 13= 5 x ^{2 } -20x -7x +15+ 13\)

\(\displaystyle 5 x ^{2 } -27x + 28 = 0\)

Factor this using the \(\displaystyle ac\)-method. We split the middle term using two integers whose sum is \(\displaystyle -27\) and whose product is \(\displaystyle 5 \cdot28 = 140\). These integers are \(\displaystyle -20,-7\):

\(\displaystyle 5 x ^{2 } -20x -7x + 28 = 0\)

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

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

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

Set each factor equal to 0 and solve separately:

\(\displaystyle 5x - 7 = 0\)

\(\displaystyle 5x - 7 + 7 = 0 + 7\)

\(\displaystyle 5x = 7\)

\(\displaystyle 5x \div 5 = 7 \div 5\)

\(\displaystyle x = \frac{7}{5} = 1 \frac{2}{5}\)

 or

\(\displaystyle x - 4 = 0\)

\(\displaystyle x - 4 + 4 = 0+ 4\)

\(\displaystyle x = 4\)

 

 

Example Question #8 : Solving Rational Expressions

Simplify the following expression:

\(\displaystyle \large \frac{x^2-5x+6}{x-2}\)

Possible Answers:

\(\displaystyle \large (x-3)^2\)

\(\displaystyle \large (x-2)^2\)

\(\displaystyle \large x-5\)

\(\displaystyle \large x-3\)

This expression is already simplified.

Correct answer:

\(\displaystyle \large x-3\)

Explanation:

The first step of problems like this is to try to factor the quadratic and see if it shares a factor with the linear polynomial in the denominator. And as it turns out,

 \(\displaystyle \large x^2-5x+6 = (x-2)(x-3)\)

So our rational function is equal to 

\(\displaystyle \large \frac{(x-2)(x-3)}{x-2} = (x-3)\)

which is as simplified as it can get.

Example Question #9 : Solving Rational Expressions

Evaluate the following expression: 

\(\displaystyle \frac{5x+z}{3y}\div\frac{9yz}{2x-4z}\)

Possible Answers:

You cannot divide fractions.

\(\displaystyle \frac{45xyz+9yz^2}{6xy-12yz}\)

\(\displaystyle \frac{27y^2z}{10x^2-18xz-4z^2}\)

\(\displaystyle \frac{10x^2-18xz-4z^2}{27y^2z}\)

\(\displaystyle \frac{6xy-12yz}{45xyz+9yz^2}\)

Correct answer:

\(\displaystyle \frac{10x^2-18xz-4z^2}{27y^2z}\)

Explanation:

When dividing fractions, we multiply by the reciprocal of the second fraction. Therefore, the problem becomes:

\(\displaystyle \frac{5x+z}{3y}\div\frac{9yz}{2x-4z}=\frac{5x+z}{3y}\times\frac{2x-4z}{9yz}\)

\(\displaystyle \frac{(5x+z)(2x-4z)}{(3y)(9yz)}=\frac{10x^2+2xz-20xz-4z^2}{27y^2z}=\frac{10x^2-18xz-4z^2}{27y^2z}\)

Our final unfactored expression is therefore \(\displaystyle \frac{10x^2-18xz-4z^2}{27y^2z}\).

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