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This lesson focuses on solving equations that contain one or more rational expressions. Since methods for solving this type of equations has already been covered, this lesson will focus on inequalities that contain rational expressions.
### Catch-Up and Review

**Here are a few recommended readings before getting started with this lesson.**

Challenge

Heichi and Dominika like to play basketball. About two months ago, they decided to keep track of how many games they each win. Until now, Dominika has won $24$ out of the $40$ games against Heichi.

a How many games would Dominika have to win in a row to have a $75%$ winning record?

b How many games would Dominika have to win in a row to have a $90%$ winning record?

c Is Dominika able to reach a $100%$ winning record? Explain why or why not.

d Suppose that after reaching a winning record of $90%,$ Dominika had a losing streak. How many games in a row would Dominika have to lose to drop down to a winning record below $50%$ again?

Discussion

A rational equation is an equation that contains at least one rational expression.

When a rational equation has been solved using algebraic methods, it is necessary to ensure that each solution satisfies the original equation because extraneous solutions may appear.

$x1 +2x =2xx+4 $

If a rational equation is expressed as a proportion, it can be solved by using the Cross Products Property. Otherwise, to eliminate the fractions, each side of the equation is multiplied by the least common denominators of the expressions. Rational Equation | Method |
---|---|

$2x+7x =x−12 $ | Cross Products Property |

$x−5x +x+5x =x_{2}−252 $ | LCD |

Method

Rational equations can be solved by using different methods. The main idea is to move the variables out of the denominators. Furthermore, since the domain of rational functions is restricted, some solutions might be extraneous. Therefore, it is necessary to verify the solutions in the original equation. Consider this rational equation as an example.
*expand_more*
*expand_more*
*expand_more*

$x−3-1 =x_{2}−2-x+4 $

The equation is basically a proportion. Therefore, it can be solved by using the Cross Products Property.
1

Use the Cross Products Property

To get the variable terms out of the denominators, the Cross Product Property will be used.

$x−3-1 =x_{2}−2-x+4 ⇓-1⋅(x_{2}−2)=(x−3)(-x+4) $

As shown, the property eliminates the rational expressions in the equation. Now the equation can be solved for $x.$ 2

Solve the Resulting Equation

To solve the resulting equation, use the Distributive Property and combine like terms.
The value that satisfies the equation is $2.$ Next, check if it also satisfies the original equation.

$-1⋅(x_{2}−2)=(x−3)(-x+4)$

Distr

Distribute $-1$

$-x_{2}+2=(x−3)(-x+4)$

$-x_{2}+2=-x_{2}+7x−12$

▼

Solve for $x$

AddEqn

$LHS+x_{2}=RHS+x_{2}$

$2=7x−12$

AddEqn

$LHS+12=RHS+12$

$14=7x$

DivEqn

$LHS/7=RHS/7$

$2=x$

RearrangeEqn

Rearrange equation

$x=2$

3

Check for Extraneous Solutions

The solution found in the previous step might be extraneous. It must be verified in the original equation.
Substiting $x=2$ in the original equation produced a true statement. Therefore, it is a solution to the original equation.

$x−3-1 =x_{2}−2-x+4 $

Substitute

$x=2$

$2−3-1 =?2_{2}−2-2+4 $

$1=?2_{2}−2-2+4 $

▼

Evaluate right-hand side

CalcPow

Calculate power

$1=?4−2-2+4 $

AddSubTerms

Add and subtract terms

$1=?22 $

CalcQuot

Calculate quotient

$1=1✓$

Example

In her chemistry lab, Dominika adds some $60%$ acid solution to $15$ milliliters of a solution with $15%$ acid.

How much of the $60%$ acid solution should she add to create a solution that is $45%$ acid?{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":null,"formTextAfter":"milliliters","answer":{"text":["30"]}}

The percentage of acid in the final solution must equal the total amount of acid divided by the total amount of solution.

Let $x$ be the amount of $60%$ acid solution to be added to $15$ milliliters of a $15%$ acid solution. The amount of each solution in terms of $x$ can be organized in a table.

Original | Added | New | |
---|---|---|---|

Amount of Acid | $0.15(15)$ | $0.6(x)$ | $0.15(15)+0.6(x)$ |

Total Solution | $15$ | $x$ | $15+x$ |

$Percent=Total SolutionAmount of Acid $

The objective is to create a solution that is $45%$ acid, so place that on the left side of the equation. Then substitute the known values into the rest of the formula. $10045 =15+x0.15(15)+0.6(x) $

This proportion is also a rational expression. To solve the equation for $x,$ the Cross Products Property can be used.
$10045 =15+x0.15(15)+0.6(x) $

Multiply

Multiply

$10045 =15+x2.25+0.6x $

CrossMult

Cross multiply

$45(15+x)=100(2.25+0.6x)$

$x=30$

Discussion

To solve a rational equation, the first step is to eliminate all the fractions in the equation. This can be done by multiplying each side of the equation by the least common denominator of the rational expressions. Any time an equation is multiplied by an algebraic expression, an extraneous solution might be introduced, so it will also be necessary to verify the solutions. *expand_more*
*expand_more*
*expand_more*
*expand_more*

$x+25 −x+54 =x_{2}+7x+1016 $

Consider solving the this equation as an example.
1

Find the Least Common Denominator

To find the least common denominator (LCD) of the rational expressions in the equation, their factored form must be found.
Since the LCD is the product of the factors with the highest power appearing in any denominator, the LCD is $(x+2)(x+5).$

$x+25 −x+54 =x_{2}+7x+1016 $

The denominator of the rational expression on the left-hand side cannot be factored further. Factor the denominator of the other expression.
$x_{2}+7x+10$

$(x+2)(x+5)$

Rational Expression | LCD |
---|---|

$x+25 $ | $(x+2)(x+5)$ |

$x+54 $ | |

$(x+2)(x+5)16 $ |

2

Multiply the Equation by the Least Common Denominator

Multiply each side of the equation by the LCD found in the previous step. This will reduce each term and get rid of the denominators.

$x+25 −x+54 =x_{2}+7x+1016 $

MultEqn

$LHS⋅(x+2)(x+5)=RHS⋅(x+2)(x+5)$

$(x+2)(x+5)(x+25 −x+54 )=(x+2)(x+5)(x_{2}+7x+1016 )$

▼

Simplify

Distr

Distribute $(x+2)(x+5)$

$x+25(x+2)(x+5) −x+54(x+2)(x+5) =x_{2}+7x+1016(x+2)(x+5) $

CancelCommonFac

Cancel out common factors

$x+2 5(x+2) (x+5) −x+5 4(x+2)(x+5) =x_{2}+7x+10 16(x+2)(x+5) $

SimpQuot

Simplify quotient

$5(x+5)−4(x+2)=16$

3

Solve the Resulting Equation

Now the resulting equation can be solved by using the Distributive Property and inverse operations.

4

Check for Extraneous Solutions

Finally, check if the solution found in Step $3$ is an extraneous solution or not. To do so, substitute it in the original equation and simplify.
The value made a true statement. Therefore, it is a solution to the equation.

$x+25 −x+54 =x_{2}+7x+1016 $

Substitute

$x=-1$

$-1+25 −-1+54 =?(-1)_{2}+7(-1)+1016 $

$4=?(-1)_{2}+7(-1)+1016 $

▼

Evaluate right-hand side

CalcPow

Calculate power

$4=?1+7(-1)+1016 $

MultPosNeg

$a(-b)=-a⋅b$

$4=?1−7+1016 $

AddSubTerms

Add and subtract terms

$4=?416 $

CalcQuot

Calculate quotient

$4=4✓$

Example

Dominika and Heichi are taking a canoe trip. They are going up the river for $1$ kilometer and then returning to their starting point. The river current flows at $3$ kilometers per hour. The total trip time will be $1$ hour and $30$ minutes.

Assuming that they paddled at a constant rate throughout the trip, find the speed at which Dominika and Heichi are paddling. Round the answer to the two decimal places.{"type":"text","form":{"type":"math","options":{"comparison":"1","nofractofloat":false,"keypad":{"simple":true,"useShortLog":false,"variables":["x"],"constants":["PI"]}},"text":"<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><\/span><\/span>"},"formTextBefore":"About","formTextAfter":"km\/h","answer":{"text":["3.74"]}}

Let $x$ represent the speed, in kilometers per hour, that the canoe would travel with no current. Write two rational expressions for the time it takes to go and return in terms of $x.$

The two students are on the river for $1$ hour and $30$ minutes, or $1.5$ hours. This round trip time represents the time it takes to go upriver and return. Recall the formula that relates distance to time, $d=rt.$ This formula can be maniuplated to represent distance in terms of time.

The sum of the portions is equal to the total round trip time.
To get rid of decimals in the equation, both sides of the equation can be multiplied by $2.$ The equation can then be solved by using the Quadratic Formula.
The solutions for this equation are $x=32±85 .$ Separate them into the positive and negative cases and round the answers to two decimal places.

The second solution does not make sense in the given context as speed cannot be negative. The first solution must be verified in the original equation to determine if it is extraneous or not. If it is a valid solution, substituting $x=3.74$ should equal about $1.5.$
Substituting the solution into the original equation resulted in an identity, so this is a solution to the original equation. Therefore, the speed at which the students can row their canoe is about $3.74$ kilometers per hour.

$d=rt⇔t=rd $

Let $x$ represent the speed, in kilometers per hour, that the canoe would travel with no current. When Dominika and Heichi are traveling with the current, their speed is $x+3km/h,$ and when they travel against the current, their speed is $x−3km/h.$ The trip takes $x+31 $ hours for the outbound portion and $x−31 $ hours for the return portion. Going | Returning | |
---|---|---|

Distance (km) | $1$ | $1$ |

Speed (km/h) | $x+3$ | $x−3$ |

Time (h) | $x+31 $ | $x−31 $ |

$x+31 +x−31 =1.5 $

The least common denominator of the rational expressions is $(x+3)(x−3).$ To eliminate the rational denominators, multiply each side of the equation by the LCD.
$x+31 +x−31 =1.5$

MultEqn

$LHS⋅(x+3)(x−3)=RHS⋅(x+3)(x−3)$

$(x+3)(x−3)(x+31 +x−31 )=(x+3)(x−3)1.5$

▼

Simplify left-hand side

Distr

Distribute $(x+3)(x−3)$

$x+3(x+3)(x−3) +x−3(x+3)(x−3) =(x+3)(x−3)1.5$

CancelCommonFac

Cancel out common factors

$x+3 (x+3) (x−3) +x−3 (x+3)(x−3) =(x+3)(x−3)1.5$

SimpQuot

Simplify quotient

$x−3+x+3=(x+3)(x−3)1.5$

AddTerms

Add terms

$2x=(x+3)(x−3)1.5$

▼

Simplify right-hand side

$2x=1.5x_{2}−13.5$

SubEqn

$LHS−2x=RHS−2x$

$0=1.5x_{2}−2x−13.5$

RearrangeEqn

Rearrange equation

$1.5x_{2}−2x−13.5=0$

$1.5x_{2}−2x−13.5=0$

MultEqn

$LHS⋅2=RHS⋅2$

$3x_{2}−4x−27=0$

▼

Solve using the quadratic formula

UseQuadForm

Use the Quadratic Formula: $a=3,b=-4,c=-27$

$x_{1,2}=2(3)-(-4)±(-4)_{2}−4(3)(-27) $

NegNeg

$-(-a)=a$

$x_{1,2}=2(3)4±(-4)_{2}−4(3)(-27) $

CalcPowProd

Calculate power and product

$x_{1,2}=64±16+324 $

AddTerms

Add terms

$x_{1,2}=64±340 $

SplitIntoFactors

Split into factors

$x_{1,2}=64±4⋅85 $

SqrtProd

$a⋅b =a ⋅b $

$x_{1,2}=64±4 ⋅85 $

CalcRoot

Calculate root

$x_{1,2}=64±2⋅85 $

FactorOut

Factor out $2$

$x_{1,2}=62(2±85 ) $

SimpQuot

Simplify quotient

$x=32±85 $

$x=32±85 $ | |
---|---|

$x_{1}=32+85 $ | $x_{2}=32−85 $ |

$x_{1}=32 +385 $ | $x_{2}=32 −385 $ |

$x_{1}≈3.74$ | $x_{2}≈-2.41$ |

$x+31 +x−31 =?1.5$

Substitute

$x=3.74$

$3.74+31 +3.74−31 =?1.5$

▼

Evaluate left-hand side

AddTerms

Add terms

$6.741 +3.74−31 =?1.5$

SubTerm

Subtract term

$6.741 +0.741 =?1.5$

ExpandFrac

$ba =b⋅0.74a⋅0.74 $

$4.98760.74 +0.741 =?1.5$

ExpandFrac

$ba =b⋅6.74a⋅6.74 $

$4.98760.74 +4.98766.74 =?1.5$

AddFrac

Add fractions

$4.98767.48 =?1.5$

CalcQuot

Calculate quotient

$1.499719…=?1.5$

RoundInt

Round to nearest integer

$1.5=1.5✓$

Example

Heichi is trying to solve the following rational equation.

Is Heichi correct? Explain.
### Answer

### Hint

### Solution

$x_{2}−12 −x−11 =x+11 $

However, he cannot be sure about his solution, so he shares his thoughts with Dominika.
If I clear the denominators I find that the only solution is $x=1,$ but when I substitute $x=1$ into the equation, it does not make any sense. |

Yes, see solution.

Solve the given rational equation. Is the solution in the domain of the rational expressions on both sides?

To check if Heichi's claim is correct or not, the given equation will be solved by using the least common denominator. Notice that $x_{2}−1=(x+1)(x−1),$ so $x_{2}−1$ is the common denominator for the three rational expressions. Multiply the equation by this common denominator and solve for $x.$
Heichi is correct that when the denominators are eliminated, the solution to the resulting equation is $1.$ However, the rest of his statement is also correct because the denominators of the expressions on the left hand side are $0$ when $x=1.$

$x_{2}−12 −x−11 =x+11 $

MultEqn

$LHS⋅(x−1)(x+1)=RHS⋅(x−1)(x+1)$

$(x−1)(x+1)(x_{2}−12 −x−11 )=(x−1)(x+1)(x+11 )$

▼

Solve for $x$

Distr

Distribute $(x−1)(x+1)$

$x_{2}−12(x−1)(x+1) −x−1(x−1)(x+1) =(x−1)(x+1)(x+11 )$

MoveLeftFacToNumOne

$a⋅b1 =ba $

$x_{2}−12(x−1)(x+1) −x−1(x−1)(x+1) =x+1(x−1)(x+1) $

FacDiffSquares

$a_{2}−b_{2}=(a+b)(a−b)$

$(x−1)(x+1)2(x−1)(x+1) −x−1(x−1)(x+1) =x+1(x−1)(x+1) $

CancelCommonFac

Cancel out common factors

$(x−1)(x+1) 2(x−1)(x+1) −x−1 (x−1) (x+1) =x+1 (x−1)(x+1) $

SimpQuot

Simplify quotient

$2−(x+1)=x−1$

Distr

Distribute $-1$

$2−x−1=x−1$

AddEqn

$LHS+1=RHS+1$

$2−x=x$

AddEqn

$LHS+x=RHS+x$

$2=2x$

DivEqn

$LHS/2=RHS/2$

$1=x$

RearrangeEqn

Rearrange equation

$x=1$

$01_{2}−1 2 −$