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Solve Rational Equations

Solve rational equations

Solve Rational Equations

We have already solved linear equations that contained fractions. We found the LCD of all the fractions in the equation and then multiplied both sides of the equation by the LCD to “clear” the fractions.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then, we will have an equation that does not contain rational expressions and thus is much easier for us to solve. But because the original equation may have a variable in a denominator, we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution to a rational equation.

We note any possible extraneous solutions, c, by writing \(x\ne c\) next to the equation.

How to Solve a Rational Equation

Try it.

Solve: \(\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.\)

Solution

The steps of this method are shown.

We always start by noting the values that would cause any denominators to be zero.

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Use Rational Functions

Working with functions that are defined by rational expressions often lead to rational equations. Again, we use the same techniques to solve them.

Example

Try it.

For rational function, \(f(x)=\frac{2x-6}{{x}^{2}-8x+15},\) ⓐ find the domain of the function, ⓑ solve \(f(x)=1,\) and ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

	\({x}^{2}-8x+15=0\)
Factor the trinomial.	\((x-3)(x-5)=0\)
Use the Zero Product Property.	\(x-3=0\ x-5=0\)
Solve.	\(x=3\ x=5\)
	The domain is all real numbers except \(x\ne 3,x\ne 5\text{.}\)

ⓑ


Substitute in the rational expression.
Factor the denominator.
Multiply both sides by the LCD, \((x-3)(x-5).\)
Simplify.
Solve.
Factor.
Use the Zero Product Property.
Solve.

However, \(x=3\) is outside the domain of this function, so we discard that root as extraneous.

Solve a Rational Equation for a Specific Variable

When we solved linear equations, we learned how to solve a formula for a specific variable. Many formulas used in business, science, economics, and other fields use rational equations to model the relation between two or more variables. We will now see how to solve a rational equation for a specific variable.

When we developed the point-slope formula from our slope formula, we cleared the fractions by multiplying by the LCD.

\(\begin{array}{llllll} & & & m & = & \frac{y-{y}_{1}}{x-{x}_{1}} \\ \text{Multiply both sides of the equation by}\ x-{x}_{1}. & & & m(x-{x}_{1}) & = & (\frac{y-{y}_{1}}{x-{x}_{1}})(x-{x}_{1}) \\ \text{Simplify.} & & & m(x-{x}_{1}) & = & y-{y}_{1} \\ \text{Rewrite the equation with the}\ y\ \text{terms on the left.} & & & y-{y}_{1} & = & m(x-{x}_{1})\end{array}\)

In the next example, we will use the same technique with the formula for slope that we used to get the point-slope form of an equation of a line through a point in Chapter 3. We will add one more step to solve for y.

Example

Try it.

Solve:\(m=\frac{y-2}{x-3}\) for \(y.\)

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equation by the LCD, \(x-3.\)
Simplify.
Isolate the term with y.

Remember to multiply both sides by the LCD in the next example.

Example

Try it.

Solve: \(\frac{1}{c}+\frac{1}{m}=1\) for c.

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, cm.
Distribute.
Simplify.
Collect the terms with c to the right.
Factor the expression on the right.
To isolate c, divide both sides by \(m-1.\)
Simplify by removing common factors.
Notice that even though we excluded \(c=0,m=0\) from the original equation, we must also now state that \(m\ne 1.\)

Key Concepts

How to solve equations with rational expressions.
1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check:
  - If any values found in Step 1 are algebraic solutions, discard them.
  - Check any remaining solutions in the original equation.

Solve Rational Equations

Solve Rational Equations

In the following exercises, solve each rational equation.

Try it.

\(\frac{1}{a}+\frac{2}{5}=\frac{1}{2}\)

Solution

\(a=10\)

Try it.

\(\frac{6}{3}-\frac{2}{d}=\frac{4}{9}\)

Try it.

\(\frac{4}{5}+\frac{1}{4}=\frac{2}{v}\)

Solution

\(v=\frac{40}{21}\)

Try it.

\(\frac{3}{8}+\frac{2}{y}=\frac{1}{4}\)

Try it.

\(1-\frac{2}{m}=\frac{8}{{m}^{2}}\)

Solution

\(m=-2,m=4\)

Try it.

\(1+\frac{4}{n}=\frac{21}{{n}^{2}}\)

Try it.

\(1+\frac{9}{p}=\frac{-20}{{p}^{2}}\)

Solution

\(p=-5,p=-4\)

Try it.

\(1-\frac{7}{q}=\frac{-6}{{q}^{2}}\)

Try it.

\(\frac{5}{3v-2}=\frac{7}{4v}\)

Solution

\(v=14\)

Try it.

\(\frac{8}{2w+1}=\frac{3}{w}\)

Try it.

\(\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{{x}^{2}-16}\)

Solution

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

Try it.

\(\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{{y}^{2}-81}\)

Try it.

\(\frac{8}{z-10}-\frac{7}{z+10}=\frac{5}{{z}^{2}-100}\)

Solution

\(z=-145\)

Try it.

\(\frac{9}{a+11}-\frac{6}{a-11}=\frac{6}{{a}^{2}-121}\)

Try it.

\(\frac{-10}{q-2}-\frac{7}{q+4}=1\)

Solution

\(q=-18,q=-1\)

Try it.

\(\frac{2}{s+7}-\frac{3}{s-3}=1\)

Try it.

\(\frac{v-10}{{v}^{2}-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{w+8}{{w}^{2}-11w+28}=\frac{5}{w-7}+\frac{2}{w-4}\)

Try it.

\(\frac{x-10}{{x}^{2}+8x+12}=\frac{3}{x+2}+\frac{4}{x+6}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{y-5}{{y}^{2}-4y-5}=\frac{1}{y+1}+\frac{1}{y-5}\)

Try it.

\(\frac{b+3}{3b}+\frac{b}{24}=\frac{1}{b}\)

Solution

\(b=-8\)

Try it.

\(\frac{c+3}{12c}+\frac{c}{36}=\frac{1}{4c}\)

Try it.

\(\frac{d}{d+3}=\frac{18}{{d}^{2}-9}+4\)

Solution

\(d=2\)

Try it.

\(\frac{m}{m+5}=\frac{50}{{m}^{2}-25}+6\)

Try it.

\(\frac{n}{n+2}-3=\frac{8}{{n}^{2}-4}\)

Solution

\(n=1\)

Try it.

\(\frac{p}{p+7}-8=\frac{98}{{p}^{2}-49}\)

Try it.

\(\frac{q}{3q-9}-\frac{3}{4q+12}=\frac{7{q}^{2}+6q+63}{24{q}^{2}-216}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{r}{3r-15}-\frac{1}{4r+20}=\frac{3{r}^{2}+17r+40}{12{r}^{2}-300}\)

Try it.

\(\frac{s}{2s+6}-\frac{2}{5s+5}=\frac{5{s}^{2}-3s-7}{10{s}^{2}+40s+30}\)

Solution

\(s=\frac{5}{4}\)

Try it.

\(\frac{t}{6t-12}-\frac{5}{2t+10}=\frac{{t}^{2}-23t+70}{12{t}^{2}+36t-120}\)

Try it.

\(\frac{2}{{x}^{2}+2x-8}-\frac{1}{{x}^{2}+9x+20}=\frac{4}{{x}^{2}+3x-10}\)

Solution

\(x=-\frac{4}{3}\)

Try it.

\(\frac{5}{{x}^{2}+4x+3}+\frac{2}{{x}^{2}+x-6}=\frac{3}{{x}^{2}-x-2}\)

Try it.

\(\frac{3}{{x}^{2}-5x-6}+\frac{3}{{x}^{2}-7x+6}=\frac{6}{{x}^{2}-1}\)

Solution

no solution

Try it.

\(\frac{2}{{x}^{2}+2x-3}+\frac{3}{{x}^{2}+4x+3}=\frac{6}{{x}^{2}-1}\)

Solve Rational Equations that Involve Functions

Try it.

For rational function, \(f(x)=\frac{x-2}{{x}^{2}+6x+8},\) ⓐ find the domain of the function ⓑ solve \(f(x)=5\) ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except \(x\ne \text{-}2\) and \(x\ne \text{-}4.\) ⓑ \(x=-3,x=-\frac{14}{5}\) ⓒ \((-3,5),(-\frac{14}{5},5)\)

Try it.

For rational function, \(f(x)=\frac{x+1}{{x}^{2}-2x-3},\) ⓐ find the domain of the function ⓑ solve \(f(x)=1\) ⓒ find the points on the graph at this function value.

Try it.

For rational function, \(f(x)=\frac{2-x}{{x}^{2}-7x+10},\) ⓐ find the domain of the function ⓑ solve \(f(x)=2\) ⓒ find the points on the graph at this function value.

Solution

ⓐ The domain is all real numbers except \(x\ne 2\) and \(x\ne 5.\) ⓑ \(x=\frac{9}{2},\) ⓒ \((\frac{9}{2},2)\)

Try it.

For rational function, \(f(x)=\frac{5-x}{{x}^{2}+5x+6},\)
ⓐ find the domain of the function
ⓑ solve \(f(x)=3\)
ⓒ the points on the graph at this function value.

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

Try it.

\(\frac{C}{r}=2\pi\) for \(r.\)

Solution

\(r=\frac{C}{2\pi }\)

Try it.

\(\frac{I}{r}=P\) for \(r.\)

Try it.

\(\frac{v+3}{w-1}=\frac{1}{2}\) for \(w.\)

Solution

\(w=2v+7\)

Try it.

\(\frac{x+5}{2-y}=\frac{4}{3}\) for \(y.\)

Try it.

\(a=\frac{b+3}{c-2}\) for \(c.\)

Solution

\(c=\frac{b+3+2a}{a}\)

Try it.

\(m=\frac{n}{2-n}\) for \(n.\)

Try it.

\(\frac{1}{p}+\frac{2}{q}=4\) for \(p.\)

Solution

\(p=\frac{q}{4q-2}\)

Try it.

\(\frac{3}{s}+\frac{1}{t}=2\) for \(s.\)

Try it.

\(\frac{2}{v}+\frac{1}{5}=\frac{3}{w}\) for \(w.\)

Solution

\(w=\frac{15v}{10+v}\)

Try it.

\(\frac{6}{x}+\frac{2}{3}=\frac{1}{y}\) for \(y.\)

Try it.

\(\frac{m+3}{n-2}=\frac{4}{5}\) for \(n.\)

Solution

\(n=\frac{5m+23}{4}\)

Try it.

\(r=\frac{s}{3-t}\) for \(t.\)

Try it.

\(\frac{E}{c}={m}^{2}\) for \(c.\)

Solution

\(c=\frac{E}{{m}^{2}}\)

Try it.

\(\frac{R}{T}=W\) for \(T.\)

Try it.

\(\frac{3}{x}-\frac{5}{y}=\frac{1}{4}\) for \(y.\)

Solution

\(y=\frac{20x}{12-x}\)

Try it.

\(c=\frac{2}{a}\ +\frac{b}{5}\) for \(a.\)

Condensed — the full section is in OpenStax Intermediate Algebra 2e.

Solve Rational Equations

Here is an example we did when we worked with linear equations:


We multiplied both sides by the LCD.
Then we distributed.
We simplified—and then we had an equation with no fractions.
Finally, we solved that equation.

We will use the same strategy to solve rational equations. We will multiply both sides of the equation by the LCD. Then we will have an equation that does not contain rational expressions and thus is much easier for us to solve.

But because the original equation may have a variable in a denominator we must be careful that we don’t end up with a solution that would make a denominator equal to zero.

So before we begin solving a rational equation, we examine it first to find the values that would make any denominators zero. That way, when we solve a rational equation we will know if there are any algebraic solutions we must discard.

An algebraic solution to a rational equation that would cause any of the rational expressions to be undefined is called an extraneous solution.

We note any possible extraneous solutions, c, by writing \(x\ne c\) next to the equation.

How to Solve Equations with Rational Expressions

Try it.

Solve: \(\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.\)

Solution

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Solve a Rational Equation for a Specific Variable

We’ll start with a formula relating distance, rate, and time. We have used it many times before, but not usually in this form.

Example

Try it.

Solve: \(\frac{D}{T}=R\ \text{for}\ T.\)

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, T.
Simplify.
Divide both sides by R to isolate T.
Simplify.

uses the formula for slope that we used to get the point-slope form of an equation of a line.

Example

Try it.

Solve: \(m=\frac{x-2}{y-3}\ \text{for}\ y.\)

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, \(y-3\).
Simplify.
Isolate the term with y.
Divide both sides by m to isolate y.
Simplify.

Be sure to follow all the steps in . It may look like a very simple formula, but we cannot solve it instantly for either denominator.

Example

Try it.

Solve \(\frac{1}{c}+\frac{1}{m}=1\ \text{for}\ c.\)

Solution


Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, \(cm\).
Distribute.
Simplify.
Collect the terms with c to the right.
Factor the expression on the right.
To isolate c, divide both sides by \(m-1\).
Simplify by removing common factors.

Notice that even though we excluded \(c=0\ \text{and}\ m=0\) from the original equation, we must also now state that \(m\ne 1\).

Key Concepts

Strategy to Solve Equations with Rational Expressions
1. Note any value of the variable that would make any denominator zero.
2. Find the least common denominator of all denominators in the equation.
3. Clear the fractions by multiplying both sides of the equation by the LCD.
4. Solve the resulting equation.
5. Check.
- If any values found in Step 1 are algebraic solutions, discard them.
- Check any remaining solutions in the original equation.

Solve Rational Equations

Solve Rational Equations

In the following exercises, solve.

Try it.

\(\frac{1}{a}+\frac{2}{5}=\frac{1}{2}\)

Solution

\(10\)

Try it.

\(\frac{5}{6}+\frac{3}{b}=\frac{1}{3}\)

Try it.

\(\frac{5}{2}-\frac{1}{c}=\frac{3}{4}\)

Solution

\(\frac{4}{7}\)

Try it.

\(\frac{6}{3}-\frac{2}{d}=\frac{4}{9}\)

Try it.

\(\frac{4}{5}+\frac{1}{4}=\frac{2}{v}\)

Solution

\(\frac{40}{21}\)

Try it.

\(\frac{3}{7}+\frac{2}{3}=\frac{1}{w}\)

Try it.

\(\frac{7}{9}+\frac{1}{x}=\frac{2}{3}\)

Solution

\(-9\)

Try it.

\(\frac{3}{8}+\frac{2}{y}=\frac{1}{4}\)

Try it.

\(1-\frac{2}{m}=\frac{8}{{m}^{2}}\)

Solution

\(-2,4\)

Try it.

\(1+\frac{4}{n}=\frac{21}{{n}^{2}}\)

Try it.

\(1+\frac{9}{p}=\frac{-20}{{p}^{2}}\)

Solution

\(-5,-4\)

Try it.

\(1-\frac{7}{q}=\frac{-6}{{q}^{2}}\)

Try it.

\(\frac{1}{r+3}=\frac{4}{2r}\)

Solution

\(-6\)

Try it.

\(\frac{3}{t-6}=\frac{1}{t}\)

Try it.

\(\frac{5}{3v-2}=\frac{7}{4v}\)

Solution

\(14\)

Try it.

\(\frac{8}{2w+1}=\frac{3}{w}\)

Try it.

\(\frac{3}{x+4}+\frac{7}{x-4}=\frac{8}{{x}^{2}-16}\)

Solution

\(\frac{-4}{5}\)

Try it.

\(\frac{5}{y-9}+\frac{1}{y+9}=\frac{18}{{y}^{2}-81}\)

Try it.

\(\frac{8}{z-10}+\frac{7}{z+10}=\frac{5}{{z}^{2}-100}\)

Solution

\(-\frac{1}{3}\)

Try it.

\(\frac{9}{a+11}+\frac{6}{a-11}=\frac{7}{{a}^{2}-121}\)

Try it.

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

Solution

\(-2,\ -1\)

Try it.

\(\frac{3}{r+10}-\frac{4}{r-4}=1\)

Try it.

\(\frac{1}{t+7}-\frac{5}{t-5}=1\)

Solution

\(-5,-1\)

Try it.

\(\frac{2}{s+7}-\frac{3}{s-3}=1\)

Try it.

\(\frac{v-10}{{v}^{2}-5v+4}=\frac{3}{v-1}-\frac{6}{v-4}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{w+8}{{w}^{2}-11w+28}=\frac{5}{w-7}+\frac{2}{w-4}\)

Try it.

\(\frac{x-10}{{x}^{2}+8x+12}=\frac{3}{x+2}+\frac{4}{x+6}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{y-3}{{y}^{2}-4y-5}=\frac{1}{y+1}+\frac{8}{y-5}\)

Try it.

\(\frac{z}{16}+\frac{z+2}{4z}=\frac{1}{2z}\)

Solution

\(-4\)

Try it.

\(\frac{a}{9}+\frac{a+3}{3a}=\frac{1}{a}\)

Try it.

\(\frac{b+3}{3b}+\frac{b}{24}=\frac{1}{b}\)

Solution

\(-8\)

Try it.

\(\frac{c+3}{12c}+\frac{c}{36}=\frac{1}{4c}\)

Try it.

\(\frac{d}{d+3}=\frac{18}{{d}^{2}-9}+4\)

Solution

\(2\)

Try it.

\(\frac{m}{m+5}=\frac{50}{{m}^{2}-25}+6\)

Try it.

\(\frac{n}{n+2}=\frac{8}{{n}^{2}-4}+3\)

Solution

\(1\)

Try it.

\(\frac{p}{p+7}=\frac{98}{{p}^{2}-49}+8\)

Try it.

\(\frac{q}{3q-9}-\frac{3}{4q+12}\)
\(\ =\frac{7{q}^{2}+6q+63}{24{q}^{2}-216}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{r}{3r-15}-\frac{1}{4r+20}\)
\(\ =\frac{3{r}^{2}+17r+40}{12{r}^{2}-300}\)

Try it.

\(\frac{s}{2s+6}-\frac{2}{5s+5}\)
\(\ =\frac{5{s}^{2}-s-18}{10{s}^{2}+40s+30}\)

Solution

\(\text{no solution}\)

Try it.

\(\frac{t}{6t-12}-\frac{5}{2t+10}\)
\(\ =\frac{{t}^{2}-23t+70}{12{t}^{2}+36t-120}\)

Solve a Rational Equation for a Specific Variable

In the following exercises, solve.

Try it.

\(\frac{C}{r}=2\pi \ \text{for}\ r\)

Solution

\(r=\frac{C}{2\pi }\)

Try it.

\(\frac{I}{r}=P\ \text{for}\ r\)

Try it.

\(\frac{V}{h}=lw\ \text{for}\ h\)

Solution

\(h=\frac{v}{lw}\)

Try it.

\(\frac{2A}{b}=h\ \text{for}\ b\)

Try it.

\(\frac{v+3}{w-1}=\frac{1}{2}\ \text{for}\ w\)

Solution

\(w=2v+7\)

Try it.

\(\frac{x+5}{2-y}=\frac{4}{3}\ \text{for}\ y\)

Try it.

\(a=\frac{b+3}{c-2}\ \text{for}\ c\)

Solution

\(c=\frac{b+3+2a}{a}\)

Try it.

\(m=\frac{n}{2-n}\ \text{for}\ n\)

Try it.

\(\frac{1}{p}+\frac{2}{q}=4\ \text{for}\ p\)

Solution

\(p=\frac{q}{4q-2}\)

Try it.

\(\frac{3}{s}+\frac{1}{t}=2\ \text{for}\ s\)

Try it.

\(\frac{2}{v}+\frac{1}{5}=\frac{3}{w}\ \text{for}\ w\)

Solution

\(w=\frac{15v}{10+v}\)

Try it.

\(\frac{6}{x}+\frac{2}{3}=\frac{1}{y}\ \text{for}\ y\)

Try it.

\(\frac{m+3}{n-2}=\frac{4}{5}\ \text{for}\ n\)

Solution

\(n=\frac{5m+23}{4}\)

Try it.

\(\frac{E}{c}={m}^{2}\ \text{for}\ c\)

Try it.

\(\frac{3}{x}-\frac{5}{y}=\frac{1}{4}\ \text{for}\ y\)

Solution

\(y=\frac{20x}{12-x}\)

Try it.

\(\frac{R}{T}=W\ \text{for}\ T\)

Try it.

\(r=\frac{s}{3-t}\ \text{for}\ t\)

Solution

\(t=\frac{3r-s}{r}\)

Try it.

\(c=\frac{2}{a}+\frac{b}{5}\ \text{for}\ a\)

Condensed — the full section is in OpenStax Elementary Algebra 2e.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Solve: \(\frac{1}{6}x+\frac{1}{2}=\frac{1}{3}.\)
If you missed this problem, review .

Rivela la risposta

\(x=-1\)
Solve: \({n}^{2}-5n-36=0.\)
If you missed this problem, review .

Rivela la risposta

\(n=9,\ n=-4\)
Solve the formula \(5x+2y=10\) for \(y.\)
If you missed this problem, review .

Rivela la risposta

\(y=\frac{10-5x}{2}\)
Solve: \(\frac{1}{x}+\frac{1}{3}=\frac{5}{6}.\)
Solve: \(\frac{1}{y}+\frac{2}{3}=\frac{1}{5}.\)

Rivela la risposta

\(y=-\frac{15}{7}\)
Solve: \(\frac{2}{3}+\frac{1}{5}=\frac{1}{x}.\)

Rivela la risposta

\(x=\frac{15}{13}\)

Solve: \(1-\frac{5}{y}=-\frac{6}{{y}^{2}}.\)

Rivela la risposta


Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation. The LCD is y².
Clear the fractions by multiplying both sides of the equation by the LCD.
Distribute.
Multiply.
Solve the resulting equation. First write the quadratic equation in standard form.
Factor.
Use the Zero Product Property.
Solve.
Check. We did not get 0 as an algebraic solution.
	The solution is \(y=2,\) \(y=3.\)

Solve: \(1-\frac{2}{x}=\frac{15}{{x}^{2}}.\)

Rivela la risposta

\(x=-3,x=5\)
Solve: \(1-\frac{4}{y}=\frac{12}{{y}^{2}}.\)

Rivela la risposta

\(y=-2,y=6\)

Solve: \(\frac{2}{x+2}+\frac{4}{x-2}=\frac{x-1}{{x}^{2}-4}.\)

Rivela la risposta


Note any value of the variable that would make any denominator zero.
Find the least common denominator of all denominators in the equation. The LCD is \((x+2)(x-2).\)
Clear the fractions by multiplying both sides of the equation by the LCD.
Distribute.
Remove common factors.
Simplify.
Distribute.
Solve.
Check: We did not get 2 or −2 as algebraic solutions.
	The solution is \(x=-1.\)

Solve: \(\frac{2}{x+1}+\frac{1}{x-1}=\frac{1}{{x}^{2}-1}.\)

Rivela la risposta

\(x=\frac{2}{3}\)
Solve: \(\frac{5}{y+3}+\frac{2}{y-3}=\frac{5}{{y}^{2}-9}.\)

Rivela la risposta

\(y=2\)

Solve: \(\frac{m+11}{{m}^{2}-5m+4}=\frac{5}{m-4}-\frac{3}{m-1}.\)

Rivela la risposta


Note any value of the variable that would make any denominator zero. Use the factored form of the quadratic denominator.
Find the least common denominator of all denominators in the equation. The LCD is \((m-4)(m-1).\)
Clear the fractions by multiplying both sides of the equation by the LCD.
Distribute.
Remove common factors.
Simplify.
Solve the resulting equation.
Check. The only algebraic solution was 4, but we said that 4 would make a denominator equal to zero. The algebraic solution is an extraneous solution.
	There is no solution to this equation.

Solve: \(\frac{x+13}{{x}^{2}-7x+10}=\frac{6}{x-5}-\frac{4}{x-2}.\)

Rivela la risposta

There is no solution.
Solve: \(\frac{y-6}{{y}^{2}+3y-4}=\frac{2}{y+4}+\frac{7}{y-1}.\)

Rivela la risposta

There is no solution.

Solve: \(\frac{y}{y+6}=\frac{72}{{y}^{2}-36}+4.\)

Rivela la risposta


Factor all the denominators, so we can note any value of the variable that would make any denominator zero.
Find the least common denominator. The LCD is \((y-6)(y+6).\)
Clear the fractions.
Simplify.
Simplify.
Solve the resulting equation.
Check.
	The solution is \(y=4.\)

Solve: \(\frac{x}{x+4}=\frac{32}{{x}^{2}-16}+5.\)

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\(x=3\)
Solve: \(\frac{y}{y+8}=\frac{128}{{y}^{2}-64}+9.\)

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\(y=7\)

Solve: \(\frac{x}{2x-2}-\frac{2}{3x+3}=\frac{5{x}^{2}-2x+9}{12{x}^{2}-12}.\)

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We will start by factoring all denominators, to make it easier to identify extraneous solutions and the LCD.
Note any value of the variable that would make any denominator zero.
Find the least common denominator. The LCD is \(12(x-1)(x+1).\)
Clear the fractions.
Simplify.
Simplify.
Solve the resulting equation.
Check. \(x=1\) and \(x=-1\) are extraneous solutions.
	The equation has no solution.

Solve: \(\frac{y}{5y-10}-\frac{5}{3y+6}=\frac{2{y}^{2}-19y+54}{15{y}^{2}-60}.\)

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There is no solution.
Solve: \(\frac{z}{2z+8}-\frac{3}{4z-8}=\frac{3{z}^{2}-16z-16}{8{z}^{2}+16z-64}.\)

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There is no solution.

Solve: \(\frac{4}{3{x}^{2}-10x+3}+\frac{3}{3{x}^{2}+2x-1}=\frac{2}{{x}^{2}-2x-3}.\)

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Factor all the denominators, so we can note any value of the variable that would make any denominator zero.	\(x\ne \text{-}1,x\ne \frac{1}{3},x\ne 3\)
Find the least common denominator. The LCD is \((3x-1)(x+1)(x-3).\)
Clear the fractions.

Simplify.
Distribute.
Simplify.

The only algebraic solution was \(x=3,\) but we said that \(x=3\) would make a denominator equal to zero. The algebraic solution is an extraneous solution.
	There is no solution to this equation.

Solve: \(\frac{15}{{x}^{2}+x-6}-\frac{3}{x-2}=\frac{2}{x+3}.\)

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There is no solution.
Solve: \(\frac{5}{{x}^{2}+2x-3}-\frac{3}{{x}^{2}+x-2}=\frac{1}{{x}^{2}+5x+6}.\)

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There is no solution.
For rational function, \(f(x)=\frac{2x-6}{{x}^{2}-8x+15},\) ⓐ find the domain of the function, ⓑ solve \(f(x)=1,\) and ⓒ find the points on the graph at this function value.

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ⓐ The domain of a rational function is all real numbers except those that make the rational expression undefined. So to find them, we will set the denominator equal to zero and solve.

\({x}^{2}-8x+15=0\)
Factor the trinomial. \((x-3)(x-5)=0\)
Use the Zero Product Property. \(x-3=0\ x-5=0\)
Solve. \(x=3\ x=5\)
The domain is all real numbers except \(x\ne 3,x\ne 5\text{.}\)
ⓑ

Substitute in the rational expression.
Factor the denominator.
Multiply both sides by the LCD,
\((x-3)(x-5).\)
Simplify.
Solve.
Factor.
Use the Zero Product Property.
Solve.
However, \(x=3\) is outside the domain of this function, so we discard that root as extraneous.
ⓒ The value of the function is 1 when \(x=7.\) So the points on the graph of this function when \(f(x)=1\) is \((7,1))\)
For rational function, \(f(x)=\frac{8-x}{{x}^{2}-7x+12},\) ⓐ find the domain of the function ⓑ solve \(f(x)=3\) ⓒ find the points on the graph at this function value.

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ⓐ The domain is all real numbers except \(x\ne 3\) and \(x\ne 4.\) ⓑ \(x=2,x=\frac{14}{3}\)
ⓒ \((2,3),(\frac{14}{3},3)\)
For rational function, \(f(x)=\frac{x-1}{{x}^{2}-6x+5},\) ⓐ find the domain of the function ⓑ solve \(f(x)=4\) ⓒ find the points on the graph at this function value.

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ⓐ The domain is all real numbers except \(x\ne 1\) and \(x\ne 5.\) ⓑ \(x=\frac{21}{4}\) ⓒ \((\frac{21}{4},4)\)
Solve:\(m=\frac{y-2}{x-3}\) for \(y.\)

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Note any value of the variable that would
make any denominator zero.
Clear the fractions by multiplying both sides of
the equation by the LCD, \(x-3.\)
Simplify.
Isolate the term with y.
Solve: \(m=\frac{y-5}{x-4}\)for \(y.\)

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\(y=mx-4m+5\)
Solve: \(m=\frac{y-1}{x+5}\) for \(y.\)

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\(y=mx+5m+1\)

Solve: \(\frac{1}{c}+\frac{1}{m}=1\) for c.

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Note any value of the variable that would make any denominator zero.
Clear the fractions by multiplying both sides of the equations by the LCD, cm.
Distribute.
Simplify.
Collect the terms with c to the right.
Factor the expression on the right.
To isolate c, divide both sides by \(m-1.\)
Simplify by removing common factors.
Notice that even though we excluded \(c=0,m=0\) from the original equation, we must also now state that \(m\ne 1.\)

Solve: \(\frac{1}{a}+\frac{1}{b}=c\) for a.

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\(a=\frac{b}{cb-1}\)
Solve: \(\frac{2}{x}+\frac{1}{3}=\frac{1}{y}\) for y.

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\(y=\frac{3x}{x+6}\)
\(\frac{1}{a}+\frac{2}{5}=\frac{1}{2}\)

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\(a=10\)
\(\frac{6}{3}-\frac{2}{d}=\frac{4}{9}\)
\(\frac{4}{5}+\frac{1}{4}=\frac{2}{v}\)

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\(v=\frac{40}{21}\)
\(\frac{3}{8}+\frac{2}{y}=\frac{1}{4}\)
\(1-\frac{2}{m}=\frac{8}{{m}^{2}}\)

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\(m=-2,m=4\)
\(1+\frac{4}{n}=\frac{21}{{n}^{2}}\)
\(1+\frac{9}{p}=\frac{-20}{{p}^{2}}\)

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\(p=-5,p=-4\)

Symbols used here

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Solve Rational Equations

Solve rational equations
Use rational functions
Solve a rational equation for a specific variable
Note any value of the variable that would make any denominator zero.
Find the least common denominator of
Clear the fractions by multiplying both sides of the equation by the LCD.
Solve the resulting equation.
Check:

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

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Parts of this page are adapted from OpenStax Elementary Algebra 2e (CC BY-NC-SA 4.0), OpenStax Intermediate Algebra 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

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