Wize High School Grade 11 Math Textbook > Rational Functions

Solving Rational Equations
Rational equations are a type of equation containing at least one rational term, f(x)g(x)\boxed{\displaystyle\frac{f(x)}{g(x)}}g(x)f(x)​​, where f(x) & g(x)f(x)~\&~g(x)f(x) & g(x) are continuous polynomials. 

Rational equations can contain non-permissible values (NPV) that identify the restrictions of the domain. They identify the value of the variable that gives a 0 0~0 in the denominator. 

Wize Tip
The NPV's are equivalent to the vertical asymptote and missing points/holes.

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Example

Solve 2x−1=5x+2\displaystyle\frac{2}{x-1}=\displaystyle\frac{5}{x+2}x−12​=x+25​, stating any NPV's. 

Step 1. 
Identify any non-permissible values.

The NPV's are the vertical asymptotes. 
Therefore, x≠−2,1x\neq{-2,1}x=−2,1 otherwise there is a total of 000 in the denominator. 

Step 2. 
Find the lowest common denominator and multiply each term by it.  

The lowest common denominator is (x−1)(x+2)(x-1)(x+2)(x−1)(x+2). 

(2x−1=5x+2)×(x−1)(x+2)   →    2(x+2)=5(x−1)\Bigg(\displaystyle\frac{2}{x-1}=\displaystyle\frac{5}{x+2}\Bigg)\times(x-1)(x+2)~~~{\color{red}\rightarrow}~~~~2(x+2)=5(x-1)(x−12​=x+25​)×(x−1)(x+2)   →    2(x+2)=5(x−1)

Step 3. 
Solve for ′x′.\green{'x'.}′x′.

2(x+2)=5(x−1)2x+4=5x−53x=9∴ x=3\begin{array}{}
2(x+2)&=&5(x-1)\\\\
2x+4&=&5x-5\\\\
3x&=&9\\\\
\therefore~x&=&3

\end{array}2(x+2)2x+43x∴ x​====​5(x−1)5x−593​

Step 4. 
Verify.

2x−1=5x+223−1=53+222=551=1✓\begin{array}{}
\displaystyle\frac{2}{x-1}&=&\displaystyle\frac{5}{x+2}\\\\
\displaystyle\frac{2}{3-1}&=&\displaystyle\frac{5}{3+2}\\\\
\displaystyle\frac{2}{2}&=&\displaystyle\frac{5}{5}\\\\
1&=&1&\checkmark

\end{array}x−12​3−12​22​1​====​x+25​3+25​55​1​✓​

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Example: Solving Rational Equations
Solve 3xx+5+1x−2=7x2+3x−10\displaystyle\frac{3x}{x+5}+\frac{1}{x-2}=\displaystyle\frac{7}{x^2+3x-10}x+53x​+x−21​=x2+3x−107​, stating any NPV's. 


Step 1. 
Identify any non-permissible values.

The NPV's are the vertical asymptotes. 

(x+5)=0→∴ x=−5(x−2)=0→∴ x=2x2+3x−10=(x+5)(x−2)=0→∴ x=−5,2\begin{array}{rccccccc}
(x+5)&=&0&&\rightarrow&&\therefore~x&=&-5\\\\
(x-2)&=&0&&\rightarrow&&\therefore~x&=&2\\\\
x^2+3x-10=(x+5)(x-2)&=&0&&\rightarrow&&\therefore~x&=&-5, 2

\end{array}(x+5)(x−2)x2+3x−10=(x+5)(x−2)​===​000​​→→→​​∴ x∴ x∴ x​===​−52−5,2​


Therefore, x≠−5,2x\neq{-5,2}x=−5,2 otherwise there is a total of 000 in the denominator. 

Step 2. 
Find the lowest common denominator and multiply each term by it.  

The lowest common denominator is (x+5)(x−2)(x+5)(x-2)(x+5)(x−2). 

(3xx+5+1x−2=7x2+3x−10)×(x+5)(x−2)      →      3x(x−2)+(x+5)=7\Bigg(\displaystyle\frac{3x}{x+5}+\frac{1}{x-2}=\displaystyle\frac{7}{x^2+3x-10}\Bigg)\times(x+5)(x-2)~~~~~~{\color{red}\rightarrow}~~~~~~3x(x-2)+(x+5)=7(x+53x​+x−21​=x2+3x−107​)×(x+5)(x−2)      →      3x(x−2)+(x+5)=7

Step 3. 
Solve for ′x′.\green{'x'.}′x′.

3x(x−2)+(x+5)=73x2−6x+x+5=73x2−5x−2=0(3x2−6x)+(x−2)=0(3x+1)(x−2)=0∴ x=−13,2\begin{array}{rcl}
3x(x-2)+(x+5)&=&7\\\\
3x^2-6x+x+5&=&7\\\\
3x^2-5x-2&=&0\\\\
(3x^2-6x)+(x-2)&=&0\\\\
(3x+1)(x-2)&=&0
\\\\\therefore~x&=&-\displaystyle\frac{1}{3}, 2

\end{array}3x(x−2)+(x+5)3x2−6x+x+53x2−5x−2(3x2−6x)+(x−2)(3x+1)(x−2)∴ x​======​77000−31​,2​

According to the NPV's, x≠2x\neq2x=2. 

Therefore, x=2x=2x=2 is an extraneous solution.

Step 4. 
Verify.

3xx+5+1x−2=7x2+3x−103(−13)(−13)+5+1(−13)−2=7(−13)2+3(−13)−10−314−37=−914−914=−914         ✓\begin{array}{rcl}
\displaystyle\frac{3x}{x+5}+\frac{1}{x-2}&=&\displaystyle\frac{7}{x^2+3x-10}\\\\
\displaystyle\frac{3(-\frac{1}{3})}{(-\frac{1}{3})+5}+\frac{1}{(-\frac{1}{3})-2}&=&\displaystyle\frac{7}{(-\frac{1}{3})^2+3(-\frac{1}{3})-10}
\\\\-\displaystyle\frac{3}{14}-\displaystyle\frac{3}{7}&=&-\displaystyle\frac{9}{14}\\\\
-\displaystyle\frac{9}{14}&=&-\displaystyle\frac{9}{14}~~~~~~~~~\checkmark

\end{array}x+53x​+x−21​(−31​)+53(−31​)​+(−31​)−21​−143​−73​−149​​====​x2+3x−107​(−31​)2+3(−31​)−107​−149​−149​         ✓​  


Therefore, x=−13x=-\displaystyle\frac{1}{3}x=−31​ is the only solution.

Practice: Solving Rational Equations
True or false:

 511 & x=4\displaystyle\frac{5}{11}~\&~x=4115​ & x=4 are solutions to 3=2x+18x−33=\displaystyle\frac{2x+1}{8x-3}3=8x−32x+1​.

Answer

I don't know

Practice: Solving Rational Equations
Given the equation 2x−3−34−x=2x−2x2−x−12\displaystyle\frac{2}{x-3}-\frac{3}{4-x}=\displaystyle\frac{2x-2}{x^2-x-12}x−32​−4−x3​=x2−x−122x−2​, answer the following questions.

Solve for xxx.

x=

(state the larger of the values)

x=

(state the smaller of the values)

I don't know

Extra Practice

Solving Rational Equations

Practice: Solving Rational Equations
Let f(x)=4x2−252x+5f(x)=\dfrac{4x^2-25}{2x+5}f(x)=2x+54x2−25​.  
If f(x)=4x2−2x+1f(x)=4x^2-2x+1f(x)=4x2−2x+1, then what is x?x?x?

Solving Rational Equations

Practice: Solving Rational Equations
Let f(x)=x2−4x−2f(x)=\dfrac{x^2-4}{x-2}f(x)=x−2x2−4​.  
If f(x)=x2−5x−6f(x)=x^2-5x-6f(x)=x2−5x−6, then what is x?x?x?

Solving Rational Equations

Practice: Solving Rational Equations
Given the equation 4xx−5+3x+2=xx2−3x−10\dfrac{4x}{x-5}+\dfrac{3}{x+2}=\dfrac{x}{x^2-3x-10}x−54x​+x+23​=x2−3x−10x​, answer the following questions.

Solving Rational Equations

Practice: Solving Rational Equations
Given the equation 4x+1−1x+3=x+2x2+4x+3\dfrac{4}{x+1}-\dfrac{1}{x+3}=\dfrac{x+2}{x^2+4x+3}x+14​−x+31​=x2+4x+3x+2​, answer the following questions.

Solving Rational Equations

Practice: Solving Rational Equations
True or false:
 8126\dfrac{81}{26}2681​ is a solution to 10=4x−13x+810=\dfrac{4x-1}{3x+8}10=3x+84x−1​.

Solving Rational Equations

Practice: Solving Rational Equations
True or false:
 177\dfrac{17}{7}717​ is a solution to −5=2x+3x−4-5=\dfrac{2x+3}{x-4}−5=x−42x+3​.

Solving Rational Equations

Solve for xxx: 
52x−1−73x+2=96x2+x−2\dfrac{5}{2x-1}-\dfrac{7}{3x+2}=\dfrac{9}{6x^2+x-2}2x−15​−3x+27​=6x2+x−29​

Solving Rational Equations

Solve 2x−3x2−3=5x+1\dfrac{2x-3}{x^2-3}=\dfrac{5}{x+1}x2−32x−3​=x+15​.