Solving Rational Equations

Solving Rational Equations exercise

• Describe how to solve the rational equation.

  Hints

  Here you see an example of the distributive property.

  To solve equations use opposite operations:

  • The opposite operation of addition is subtraction and vice versa.
  • The opposite operation of multiplication is division and vice versa.

  Check your solution by putting it into the starting equation.

  Solution

  The rate is the number of wood pieces divided by the needed time. Let $x$ be the time it takes Jasmine to collect $10$ pieces of wood.

  Thus we get the rate

  $\frac{10~\text{wood pieces}}{x~\text{minutes}}$

  for Jasmine and, with the same measures,

  $\frac{15}{x+10}$

  for Grandpa Lindbergh. Because the rates agree we get the following rational equation:

  $\frac{10}{x}=\frac{15}{x+10}$.

  First we cross multiply to get

  $15x=10(x+10)$.

  Using the distributive property gives us $15x=10x+100$. From now on we use opposite operations to isolate $x$:

  $\begin{array}{rcl} 15x & = & ~10x+100\\ \color{#669900}{-10x} & &\color{#669900}{-10x}\\ 5x & = & ~100\\ \color{#669900}{\div5} & & \color{#669900}{\div5}\\ x & = & ~20 \end{array}$

  Lastly, let's check the solution $\frac{10}{20}=\frac12=\frac{15}{20+10}$ $~~~~~$✓

  Thus we can conclude that Jasmine takes $20$ minutes to collect $10$ pieces of wood, while Grandpa Lindbergh takes $30$ minutes to collect $15$ pieces of wood.

• Calculate how much time it takes General Good and Grandpa Lindbergh to weave $1$ leave.

  Hints

  You can only add or subtract fractions with the same denominator.

  Use FOIL multiplication to check that the quadratic term equals $(2x+9)(x-3)$.

  Put the solutions in the starting equation to check them. They must satisfy this equation.

  Solution

  First let's have a look at the rate for all leaves: together Grandpa and General Good need to weave together $80$ leaves in $3$ hours, it's $180$ minutes. This gives us the rate $\frac{80}{180}$.

  The rate for General Good is $\frac1x$ and for Grandpa $\frac1{x+6}$. Together means „$+$“. So we get $\frac1x+\frac1{x+6}$.

  Now we can state the equation

  $\frac{80}{180}=\frac{1}{x}+\frac{1}{x+6}$.

  To use cross multiplying we first have to sum both fractions on the right side of the equations. For this we need the common denominator, it's $(x)(x+6)$. Now we enlarge the left fraction by $x+6$ and the right one by $x$:

  $\frac{80}{180}=\frac{1(x+6)}{x(x+6)}+\frac{1(x)}{(x+6)(x)}$.

  With the distributive property we get

  $\frac{80}{180}=\frac{x+6}{x^2+6x}+\frac{x}{x^2+6x}$.

  We add fractions with the same denominator by just adding the numerators. Further on we can combine like terms:

  $\frac{80}{180}=\frac{x+6+x}{x^2+6x}=\frac{2x+6}{x^2+6x}$.

  Now we're able to cross multiply:

  $180(2x+6)=80(x^2+6x)$.

  Once again with the distributive property we can multiply out

  $360x+1080=80x^2+480x$.

  Dividing by $40$ this is equivalent to

  $9x+27=2x^2+12x$.

  Using opposite operations and combining like terms lead to a quadratic equation

  $\begin{array}{rclll} 9x+27 & = & ~2x^2+12x\\ \color{#669900}{-9x} & &\color{#669900}{-9x}\\ 27 & = & ~2x^2+12x-9x&|&\text{combining like terms}\\ 27 & = & ~2x^2-3x\\ \color{#669900}{-27} & & \color{#669900}{-27}\\ 0 & = & ~2x^2-3x-27 \end{array}$

  We can factorize this equation as follows $(2x+9)(x-3)=0$.

  Any product is zero if one of the factors is zero. So we have to solve the following equations

  $\begin{array}{rclll} 2x +9& = & ~0\\ \color{#669900}{-9} & &\color{#669900}{-9}\\ 2x & = & ~-9\\ \color{#669900}{\div 2} & & \color{#669900}{\div 2}\\ x & = & ~-\frac92 \end{array}$

  and

  $\begin{array}{rclll} x-3 & = & ~0\\ \color{#669900}{+3} & &\color{#669900}{+3}\\ x & = & ~3 \end{array}$

  You can check your solution, but before you do so, have a look to see if the solutions are meaningful. $x$ represents a time. So $-\frac92$ isn't a meaningful solution at all. The only meaningful solution is $x=3$.

• Determine the rate of gathering berries.

  Hints

  Let $x$ be the time Lisa needs to gather $120$ berries. Then the rate is given by $\frac{120}x$.

  Determine the rate for Tim similarly.

  Keep in mind that the rates must agree.

  Check your solution: the rate for Lisa as well as for Tim must be the same with this solution.

  Solution

  Here we take $x$ for the time Lisa needs to gather $120$ berries. The rate is the number of berries divided by the needed time:

  $\frac{120}{x}$.

  Similarly we establish the rate for Tim. Tim takes $15$ minutes more than Lisa, or $x+15$. During this time he gathers $210$ berries in total. So we get the rate

  $\frac{210}{x+15}$.

  The fact that the rates agree leads to the equation

  $\frac{120}{x}=\frac{210}{x+15}$.

  To get the rate, we have to solve this equation:

  • Cross multiply to $210x=120(x+15)$.
  • Use the distributive property to get $210x=120x+1800$.
  • Solve the equation using opposite operations:

  $\begin{array}{rcl} 210x & = & ~120x+1800\\ \color{#669900}{-120x} & &\color{#669900}{-120x}\\ 90x & = & ~1800\\ \color{#669900}{\div90} & & \color{#669900}{\div90}\\ x & = & ~20 \end{array}$

  So the rate is $\frac{120}{20}=6$ berries per minute.

  We can check the solution as well: $\frac{210}{20+15}=\frac{210}{35}=6$ $~~~~~$✓

• Calculate how much time Jason and Naya can spend maximum to tie $1$ bouquet.

  Hints

  The rate is given by the relation between the number of bouquets and time in minutes.

  So the $2$ hours left for $20$ bouquets is given by the rate: $\frac{20}{120}=\frac16$.

  The equation you have to solve is given by $\frac16=\frac1x+\frac1{x+5}$.

  Enlarge the left fraction by $x+5$ and the right one by $x$.

  You get two solutions but only one is meaningful. This is the maximum time for Naya. Just add $5$ minutes to get the maximum time for Jason.

  Solution

  First we establish the equation to solve:

  • The number of bouquets tied in $2$ hours gives us the rate $\frac{20}{120}=\frac16$.
  • The rate of Naya is given by $\frac1x$.
  • The one of Jason is given by $\frac1{x+5}$, because Jason takes $5$ minutes more than Naya.
  • The total rate is $\frac1x+\frac1{x+5}$.

  So we can state the equation

  $\frac16=\frac{1}{x}+\frac{1}{x+5}$.

  Next we enlarge the left fraction on the right side by $x+5$ and the right one by $x$ to get the common denominator $(x)(x+5)=x^2+5x$:

  $\frac16=\frac{1(x+5)}{x^2+5x}+\frac{1(x)}{x^2+5x}=\frac{2x+5}{x^2+5x}$.

  Now we use cross multiplication:

  $6(2x+5)=x^2+5x$,

  which is, using the distributive property, equivalent to

  $12x+30=x^2+5x$.

  Using opposite operations and combining like terms leads to a quadratic equation

  $\begin{array}{rclll} 12x+30 & = & ~x^2+5x\\ \color{#669900}{-12x} & &\color{#669900}{-12x}\\ 30 & = & ~x^2+5x-12x&|&\text{combining like terms}\\ 30 & = & ~x^2-7x\\ \color{#669900}{-30} & & \color{#669900}{-30}\\ 0 & = & ~x^2-7x-30 \end{array}$

  We can factor this equation as $(x+3)(x-10)=0$.

  Any product is zero if one of the factors is zero. So we have to solve the following equations:

  $\begin{array}{rclll} x +3& = & ~0\\ \color{#669900}{-3} & &\color{#669900}{-3}\\ x & = & ~-3 \end{array}$

  and

  $\begin{array}{rclll} x-10 & = & ~0\\ \color{#669900}{+10} & &\color{#669900}{+10}\\ x & = & ~10 \end{array}$

  $x=-3$ isn't a meaningful solution because $x$ represents a time. So the only solution is $x=10$.

  Thus we can conclude that Naya has at most $10$ minutes and Jason $10+5=15$ minutes to make one bouquet.

• Set up the rational equation.

  Hints

  The rate is the number of wood pieces divided by the needed time.

  Let $x$ be the time Jasmine needs to collect $10$ pieces of wood. So Grandpa Lindbergh needs $10$ minutes more, or $x+10$.

  The rate for Jasmine is $\frac{10}x$.

  Both rates must agree.

  Solution

  To establish en equation corresponding to a word problem we first assign variable(s) to the desired quantity.

  So, we assign $x$ to the time Jasmine needs to collect $10$ pieces of wood. The corresponding rate is then $\frac{10}x$.

  Grandpa Lindbergh takes $10$ minutes more than Jasmine, ior $x+10$, for $15$ pieces of wood. This gives us the rate

  $\frac{15}{x+10}$.

  Those rates have to agree, giving us the equation

  $\frac{10}x=\frac{15}{x+10}$.

• Solve the rational equations.

  Hints

  Remember how to find a solution to a rational equation:

  1. First add or subtract fractions on one or both sides of the equation if it's necessary.
  2. Cross multiply to get a non rational equation.
  3. Solve the resulting equation using opposite operations.
  4. If you get a quadratic equation you have to factor it.

  You can also use the formula to solve quadratic equations:

  $ax^2+bx+c=0$

  $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

  Use the following factorizations:

  • $3x^2-6x-24=(3x-12)(x+2)$
  • $x^2-2x-24=(x+4)(x-6)$

  Three equations have two solutions and only one has one solution.

  Solution

  Here is a general procedure for solving rational equations:

  1. First add or subtract fractions on one or both sides of the equation if it's necessary.
  2. Cross multiply to get a non-rational equation.
  3. Solve the resulting equation using opposite operations.
  4. If you get a quadratic equation you have to factor it.

  $~$

  Let's start with $\frac x{x-2}=\frac1{x+2}$:

  1. Cross multiplying gives us $x-2=x(x+2)=x^2+2x$.
  2. Next we transform this equation, adding $2$ and subtracting $x$ to $x^2+x-2=0$.
  3. Use the formula to solve quadratic equations: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, with $a=1$, $b=1$ and $c=-2$.

  $x=\frac{-1\pm\sqrt{1^2-4(1)(-2)}}{2(1)}=\frac{-1\pm\sqrt9}{2}$.

  So we get two solutions: $x=\frac{-1+3}2=\frac22=1$ and $x=x=\frac{-1-3}2=\frac{-4}2=-2$.

  The next example is given by $\frac6{x}=\frac{3x}{x+4}$.

  1. Cross multiplying gives us $6(x+4)=6x+24=3x^2$.
  2. Next we transform this equation; subtracting $6x$ as well as $24$ gives us $3x^2-6x-24=0$.
  3. You can factorize the quadratic term as follows: $(3x-12)(x+2)$.
  4. Lastly we have to solve two equations:

  $\begin{array}{rclll} 3x -12& = & ~0\\ \color{#669900}{+12} & &\color{#669900}{+12}\\ 3x & = & ~12\\ \color{#669900}{\div 3} & & \color{#669900}{\div 3}\\ x & = & ~4 \end{array}$

  and

  $\begin{array}{rclll} x+2 & = & ~0\\ \color{#669900}{-2} & &\color{#669900}{-2}\\ x & = & ~-2 \end{array}$

  $~$

  Next we have another equation with the variable in the numerator and the denominator on both sides: $\frac{x}{x+4}=\frac{x-4}{x+2}$

  1. Cross multiplying gives us $x(x+2)=(x+4)(x-4)$.
  2. Using the distributive property and FOIL multiplication gives us $x^2+2x=x^2-16$.
  3. Subtracting $x^2$ on both sides leads to $2x=-16$.
  4. Lastly we divide by $2$ to get $x=-8$.

  $~$

  Let's now look at $\frac14=\frac1x+\frac1{x+6}$

  1. Add the fractions on the right side. For this enlarge the left one by $x+6$ and the right one by $x$ to get $\frac{1(x+6)}{x(x+6)}+\frac{1(x)}{(x+6)(x)}=\frac{2x+6}{x^2+6x}$.
  2. Cross multiplying leads to $x^2+6x=4(2x+6)=8x+24$.
  3. Next we transform this equation: first subtract $8x$, combine the like terms, and after subtract $24$ to get $x^2-2x-24=0$.
  4. Factor this equation to $(x+4)(x-6)=0$.
  5. Lastly we have to solve two equations:

  $\begin{array}{rclll} x+4& = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$

  and

  $\begin{array}{rclll} x-6 & = & ~0\\ \color{#669900}{+6} & &\color{#669900}{+6}\\ x & = & ~6 \end{array}$