Solving Rational Equations

• Cross multiply: We can multiply 4 with$4x-2$, and 8 with $(x+2)$

$(x+2)\times 8=4\times (4x-2)$

• Distributive property:

$8x+16=16x-8$

• Subtract $8x$ from each side:

$16=8x-8$

• Add 8 to each side:

$24=8x$

• Divide each side by 8:

$x=3$

To see if the solution works out, it should be plugged into the equation.

$\frac{4}{3+2}=\frac{8}{4\times 3-2}$

$\frac{4}{5}=\frac{8}{12-2}$

If we divide the right side by 2:

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

This equality shows that the solution works out fine!

Solve $\frac{8}{3x-2}=\frac{2}{x-1}$

Solution:

To solve the equation, cross multiplying should be performed, because it is again in the form of $\frac{a}{b}=\frac{c}{d}$

• Cross multiplying: We should multiply 8 with $(x-1)$, and 2 with $(3x-2)$

$8\times (x-1)=2\times (3x-2)$

• Distributive property:

$8x-8=6x-4$

• Subtract 6x from each side:

$2x-8=-4$

• Add 8 to each side:

$2x=4$

• Divide each side by 2:

$x=2$

Now, let's plug x = 2 into the rational equation to see if it works.

$\frac{8}{3\times 2-2}=\frac{2}{2-1}$

$\frac{8}{6-2}=\frac{2}{1}$

$\frac{8}{4}=2$

As you can see, if we can get the same value on each side when the solution is plugged into the solution, it means it works out as in this case.

What is the solution of $\frac{3}{x}+\frac{5}{4}=\frac{8}{x}$?

Solution:

Steps for solving the rational equation:

In this example, we have denominators as x, 4, and again x. So, the least common denominator can be determined as the multiplication of x and 4, which is 4x. We should use the LCD to multiply with the equation on each side to find the solution.

• Multiply each side by the LCD which is $4x$: $4x(\frac{3}{x}+\frac{5}{4})=4x\left(\frac{8}{x}\right)$

• We should distribute $4x$ to each side and multiply it with the values in the parentheses.

$(4x\times \frac{3}{x})+(4x\times \frac{5}{4})=(4x\times \frac{8}{x})$

$(4\cancel{x}\times \frac{3}{\cancel{x}})+(\cancel{4}x\times \frac{5}{\cancel{4}})=(4\cancel{x}\times \frac{8}{\cancel{x}})$

$(4\times 3)+\left(5x\right)=(4\times 8)$

$12+5x=32$

• Simplify : $12+5x=32$

  To simplify this, we should subtract 12 from each side:

$\cancel{12}+5x-\cancel{12}=32-12$

$5x=20$

• Divide each side by 5 :

$\frac{5x}{5}=\frac{20}{5}$

So the solution for the rational equation is $x=4$, yet we should be sure if it works out. So, again, we should plug this solution into the equation:

$\frac{3}{4}+\frac{5}{4}=\frac{8}{4}$

$\frac{8}{4}=\frac{8}{4}$

Since we have equality again, this solution seems to be working out.

Solve the equation by using LCD: $\frac{1}{2x}+\frac{3}{x+7}=\frac{-1}{x}$

Solution:

• We should distribute the minus sign to the right side:

$\frac{x+7}{2}+3x=-x-7$

• To make it easier to solve the equation, we can multiply each side by 2:

$2\times \left(\frac{x+7}{2}\right)+2\times 3x=2\times (-x-7)$

• We should distribute 2 in the parentheses with multiplication:

$\cancel{2}\times \left(\frac{x+7}{\cancel{2}}\right)+2\times 3x=2\times (-x-7)$

$x+7+6x=-2x-14$

$7x+7=-2x-14$

• We can add $2x$ to each side to simplify the x value:

$7x+7+2x=-2x-14+2x$

$7x+7+2x=\cancel{-2x}-14+\cancel{2x}$

$9x+7=-14$

• We can subtract 7 from each side to simplify again:

$9x+7-7=-14-7$

$9x=-21$

• Divide each side to 9 to leave x value alone:

$\frac{9x}{9}=\frac{-21}{9}$

$x=\frac{-21}{9}=\frac{-7}{3}$

The solution for the rational equation is then $x=\frac{-7}{3}$. Let's see if it works in the equation:

$\frac{1}{2\times \left(\frac{-7}{3}\right)}+\frac{3}{\frac{-7}{3}+7}=\frac{-1}{\frac{-7}{3}}$

$\frac{1}{\frac{-14}{3}}+\frac{3}{\frac{-7}{3}+\frac{21}{3}}=\frac{3}{7}$

$\frac{-3}{14}+\frac{3}{\frac{14}{3}}=\frac{3}{7}$

$\frac{-3}{14}+\frac{9}{14}=\frac{3}{7}$

$\frac{6}{14}=\frac{3}{7}$

We have the equality, so it means that our solution works.

Solve $2-\frac{16}{x-5}=\frac{6}{x}$

Solution:

• Multiply each side by the LCD:

$x\times (x-5)\times (2-\frac{16}{x-5})=x\times (x-5)\times \frac{6}{x}$

• Distribute the LCD to the parentheses:

$(2\times x\times (x-5\left)\right)-(x\times (x-5)\times \frac{16}{x-5})=x\times (x-5)\times \frac{6}{x}$

• We can do simplification:

$(2\times x\times (x-5\left)\right)-(x\times \cancel{(x-5)}\times \frac{16}{\cancel{x-5}})=\cancel{x}\times (x-5)\times \frac{6}{\cancel{x}}$

• Do the multiplications:

$2x^2-10x-16x=6x-30$

• Subtract $6x$ from each side to simplify:

$2x^2-32x=-30$

• Add 30 to each side:

$2x^2-32x+30=0$

• Write in standard form: $2x^2-32x+30=0$

• Divide it by 2:

$\frac{\cancel{2}{x}^2}{\cancel{2}}-\frac{32x}{2}+\frac{30}{2}=0$

$x^2-16x+15=0$

• At this point, we should factor the equation to find the solutions. The important thing is that we have to find two values making up 15, factor also $x^2$ . The relative addition should be giving the middle value of the equation which is $-16x$.

$$\begin{gathered} x^2-16x+15=0 \\ x\iff -15 \\ x\iff -1 \end{gathered}$$

We can divide x² as x times x, and 15 as - 15 and - 1 . When we cross multiply x with - 15 , and the other x with - 1, and add them all together, we should be getting $-16x$ which is satisfied in this way.

As a result, we can write the equation as: $(x-15)\times (x-1)=0$

The values making the parentheses zero are our solutions!

$$\begin{gathered} x-15=0\to x=15 \\ x-1=0\to x=1 \end{gathered}$$

Let's check if these two solutions both work:

• For $x=15$

$2-\frac{16}{15-5}=\frac{6}{15}$

$2-\frac{16}{10}=\frac{6}{15}$

$\frac{2\times 10-16}{10}=\frac{6}{15}$

$\frac{20-16}{10}=\frac{6}{15}$

$\frac{4}{10}=\frac{6}{15}$

Divide the left side by 2, and the right side by 3:

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

So, $x=15$ is one of the solutions.

• For $x=1$ :

$2-\frac{16}{-4}=6$

$2+4=6$

which again works as the solution.

Solve the equation $\frac{x+3}{x-3}+\frac{x}{x-5}=\frac{x+5}{x-5}$

Solution:

• Multiply each side by the LCD which is $(x-3)(x-5)$:

• Distribute the LCD to the parentheses:

• Simplify:

$(\cancel{(x-3)}\times (x-5)\times (\frac{x+3}{\cancel{x-3}}\left)\right)+\left(\right(x-3)\times \cancel{(x-5)}\times (\frac{x}{\cancel{x-5}}\left)\right)=\left(\right(x-3)\times (\cancel{x-5})\times (\frac{x+5}{\cancel{x-5}}\left)\right)$

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

• Do the multiplications:

$(x^2-2x-15)+(x^2-3x)=(x^2+2x-15)$

• Do the additions:

$2x^2-5x-15=x^2+2x-15$

• Cross the values on the right side to the left side:

$2x^2-5x-15-x^2-2x+15=0$

• Simplify:

$x^2-7x=0$

• Since x is common in the equation, we can factor the equation as: $x\times (x-7)=0$

• To find the solutions, we should find the values making the multipliers zero in the equation, which are $x=0$ and $x=7$

When $x=0$ and $x=7$ are both plugged into the equation, it can be seen that they both work out and are both the solutions for the equation:

For $x=0$ :

$-1=-1$

For $x=7$:

$\frac{7+3}{7-3}+\frac{7}{7-5}=\frac{7+5}{7-5}$

$\frac{10}{4}+\frac{7}{2}=\frac{12}{2}$

$\frac{10}{4}+\frac{14}{4}=6$

$\frac{24}{4}=6$

Solve $\frac{4}{x-1}=\frac{8x^2}{x^2-1}-\frac{x}{x+1}$

Solution:

We can solve this type of equation with the LCD method. When we look at the denominators, they are $x-1$, $x^2-1$ which is the multiplication of $x-1$ and $x+1$. So, the least common denominator should be the multiplication of both $x-1$ and $x+1$ , which is $(x-1)\times (x+1)$. Each side of the equation should be multiplied by the LCD to find the solutions.

• Multiply each side by LCD:

• Simplify:

$\left(\cancel{x-1}\right)\times (x+1)\times \frac{4}{\cancel{x-1}}=\left(\cancel{x-1}\right)\times \left(\cancel{x+1}\right)\times \left(\frac{8x^2}{\cancel{x^2-1}}\right)-(x-1)\times \left(\cancel{x+1}\right)\times \left(\frac{x}{\cancel{x+1}}\right)$

$4\times (x+1)=8x^2-x\times (x-1)$

• Distributive property:

$4x+4=8x^2-x^2+x$

• Subtract $4x+4$ from each side:

$4x+4-4x-4=7x^2+x-4x-4$

$7x^2-3x-4=0$

We should factor the equation to find solutions. We can divide 7x² as the multiplication of 7x and x, and - 4 as the multiplication of + 4 and - 1. The goal here is to find the middle value which is - 3x.

$$\begin{gathered} 7x^2-3x-4=0 \\ 7x\iff +4 \\ x\iff -1 \end{gathered}$$

Through cross multiplying and addition like this, we can obtain the middle value.

• Factor: $(7x+4)\times (x-1)=0$

Solutions will be $x=-\frac{4}{7}$ or $x=1$

However, when$x=1$is substituted in the equation, it results in division by zero which gives an undefined result. So, $x=-\frac{4}{7}$remains the only solution that works out. $x=1$ is the extraneous solution!

Solve the equation and check for extraneous solutions: $\frac{18}{x^2-3x}-\frac{6}{x-3}=\frac{5}{x}$

Solution:

We can solve this example with the LCD method. When we look at the denominators, they are: $x^2-3x$ which is the multiplication of $x-3$ and x. So, the least common denominator is $x\times (x-3)$. The solutions can be found by multiplying the equation with the LCD.

• Multiply each side by LCD which is $x\times (x-3)$:

$x\times (x-3)\times (\frac{18}{x^2-3x}-\frac{6}{x-3})=x\times (x-3)\times \left(\frac{5}{x}\right)$

• Distribute the LCD to the parentheses:

$(x\times (x-3)\times (\frac{18}{x^2-3x}\left)\right)-(x\times (x-3)\times (\frac{6}{x-3}\left)\right)=x\times (x-3)\times \left(\frac{5}{x}\right)$

• Simplify:

$(\cancel{x}\times (\cancel{x-3})\times (\frac{18}{\cancel{x^2-3x}}\left)\right)-(x\times (\cancel{x-3})\times (\frac{6}{\cancel{x-3}}\left)\right)=\cancel{x}\times (x-3)\times \left(\frac{5}{\cancel{x}}\right)$

$18-6x=5\times (x-3)$

$18-6x=5x-15$

• Subtract $18-6x$ from each side:

$18-6x-18+6x=5x-15-18+6x$

$0=11x-33$

• Add 33 to each side:

$11x=33$

• Divide each side by 11:

$x=3$

To check for extraneous solutions, the solution found should be plugged into the equation. However, when $x=3$ is plugged in, the equation results as undefined because there is $x-3$ in the denominators. So, there is no solution for the equation!

Steps for solving the rational equation:

• Write the equation:

$6=\frac{8t^2+20}{4t^2-10}$

• To solve the equation, we can cross multiply :

• Distributive property:

$24t^2-60=8t^2+20$

• Subtract $8t^2+20$ from each side:

$16t^2-80=0$

• Add 80 to each side:

$16t^2=80$

• Divide each side by 16:

$\frac{\cancel{16}{t}^2}{\cancel{16}}=\frac{80}{16}$

$t^2=5$

• Take square roots of each side: $\pm 2.24\approx t$

Since -2.24 is not in the domain ($0\le t\le 7$), the only solution remains as +2.24.

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