7.4: Solving Rational Equations
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In Section 3 of Chapter 2, we showed that the most efficient way to solve an equation containing fractions was to first clear the fractions by multiplying both sides of the equation by the least common denominator. For example, given the equation
12x+13=14
we would first clear the fractions by multiplying both sides by 12.
12[12x+13]=[14]126x+4=3
This procedure works equally well when the the denominators contain a variable.
Example 7.4.1
Solve for x: 1−2x=3x2
Solution
The common denominator is x2. We begin by clearing fractions, multiplying both sides of the equation by x2.
1−2x=3x2Original equation.x2[1−2x]=[3x2]x2Multiply both sides by x2.
Now we use the distributive property.
x2[1]−x2[2x]=[3x2]x2Distribute x2.
Now we cancel common factors and simplify.
x2−2x=3Cancel. Simplify.
The resulting equation is nonlinear (x is raised to a power larger than 1). Make one side zero, then factor.
x2−2x−3=0Nonlinear. Make one side zero.(x−3)(x+1)=0Factor.
Use the zero product property to complete the solution. Either the first factor is zero or the second factor is zero.
x−3=0x=3
or
x+1=0x=−1
Hence, the solutions are x=−1 and x=3.
Check. Substitute −1 for x, then 3 for x in the original equation and simplify.
1−2x=3x21−2(−1)2=3(−1)21+2=33=3
and
1−2x=3x21−2(3)=3(3)21−23=3913=13
Note that both result in true statements, showing that both x=−1 and x=3 check in the original equation.
Exercise 7.4.1
Solve for x: 1−6x=−8x2
- Answer
-
2,4
Example 7.4.2
Solve for x: 6−22x2=29x
Solution
The common denominator is x2.
6−22x2=29xOriginal equation.x2[6−22x2]=[29x]x2Multiply both sides by x2x2[6]−x2[22x2]=[29x]x2Distribute x26x2−22=29xCancel and Simplify.
This last equation is nonlinear. Make one side zero.
6x2−29x−22=0
The integer pair 4 and −33 has product ac=−132 and sum b=−29. Break up the middle term into a sum using this pair, then factor by grouping.
6x2+4x−33x−22=02x(3x+2)−11(3x+2)=0(2x−11)(3x+2)=0
Finally, use the zero product property to write:
2x−11=02x=11x=112
or
3x+2=03x=−2x=−23
Check: Let’s check these solutions with our calculators. Enter 11/2, push the STO► button, push the X,T,θ,n button and the ENTER key (see the calculator screen on the left in Figure 7.4.1). Next, enter the left-hand side of the equation as 6−22/X2 and press ENTER. Enter the right-hand side of the equation as 29/X and press ENTER. The results are the same (see the calculator screen on the left in Figure 7.4.1 ). This verifies that 11/2 is a solution of 6−22/x2=29/x.
The calculator screen on the right in Figure 7.4.1 shows a similar check of the solution x=−2/3.

Exercise 7.4.2
Solve for x: 7x2+8=−30x
- Answer
-
−1/4,−7/2
Solving Rational Equations with the Graphing Calculator
Let’s use the graphing calculator to solve an equation containing rational expressions.
Example 7.4.3
Consider the following equation: 2−9x=5x2Solve the equation algebraically, then solve the equation graphically using your graphing calculator. Compare your solutions.
Solution
Algebraic solution: First, an algebraic approach. Multiply both sides of the equation by the common denominator x2.
2−9x=5x2Original equation.x2[2−9x]=[5x2]x2Multiply both sides by x2x2[2]−x2[9x]=[5x2]x2Distribute x22x2−9x=5Cancel and Simplify.
The last equation is nonlinear. Make one side zero.
2x2−9x−5=0Make one side zero.
The integer pair −10 and 1 have product equaling ac=−10 and sum equaling b=−9. Break up the middle term using this pair, then factor by grouping.
2x2−10x+x−5=0−10x+x=−9x2x(x−5)+1(x−5)=0Factor by grouping.(2x+1)(x−5)=0Factor out x−5
Now use the zero product property to write:
2x+1=02x=−1x=−12
or
x−5=0x=5
Hence, the solutions are x=−1/2 and x=5.
Graphical solution: We could load each side of the equation separately, then use the intersect utility to find where the graphs intersect. However, in this case, it’s a bit easier to make one side of the equation zero, draw a single graph, then note where the graph crosses the x-axis.
2−9x=5x2Original equation.2−9x−5x2=0Make one side zero.
Load the left-hand side of the equation into Y1 as 2−9/X−5/X∧2 (see the image on the left in Figure 7.4.2), then select 6:ZStandard from the ZOOM menu to produce the image at the right in Figure 7.4.2.

Next, the solutions of
2−9x−5x2=0
are found by noting where the graph of y=2−9x−5x2 cross the x-axis. Select 2:zero from the CALC menu. Use the arrow keys to move the cursor to the left of the first x-intercept, then press ENTER to set the “Left bound.” Next, move the cursor to the right of the first x-intercept, then press ENTER to set the “Right bound.” Finally, leave the cursor where it is and press ENTER to set your “Guess.” The calculator responds with the result shown in the figure on the left in Figure 7.4.3.
Repeat the zero-finding procedure to capture the coordinates of the second x-intercept (see the image on the right in Figure 7.4.3).

Reporting the solution on your homework: Duplicate the image in your calculator’s viewing window on your homework page. Use a ruler to draw all lines, but freehand any curves.
- Label the horizontal and vertical axes with x and y, respectively (see Figure 7.4.4).
- Place your WINDOW parameters at the end of each axis (see Figure 7.4.4).
- Label the graph with its equation (see Figure 7.4.4).
- Drop dashed vertical lines through each x-intercept. Shade and label the x-values of the points where the dashed vertical line crosses the x-axis. These are the solutions of the equation 2−9/x−5/x2=0 (see Figure 7.4.4).

Thus, the calculator is reporting that the solutions of 2−9/x−5/x2=0 are x=−0.5 and x=5, which match the algebraic solutions x=−1/2 and x=5.
Exercise 7.4.3
Solve the equation 2+5x=12x2 both algebraically and graphically, then compare your solutions.
- Answer
-
−4,3/2
Numerical Applications
Let’s apply what we’ve learned to an application.
Example 7.4.4
The sum of a number and its reciprocal is 41/20. Find the number.
Solution
In the solution, we address each step of the Requirements for Word Problem Solutions.
- Set up a variable dictionary: Let x represent the unknown number.
- Set up an equation: If the unknown number is x, then its reciprocal is 1/x. Thus, the “sum of a number and its reciprocal is 41/20” becomes:x+1x=4120
- Solve the equation: Clear the fractions by multiplying both sides by 20x, the least common denominator.x+1x=4120Model equation.20x[x+1x]=[4120]20xMultiply both sides by 20x20x[x]+20x[1x]=[4120]20xDistribute 20x.20x2+20=41xCancel and simplify.The equation is nonlinear. Make one side zero.20x2−41x+20=0Make one side zero.The integer pair −16 and −25 has product ac=400 and sum b=−41. Break up the middle term in the last equation into a sum of like terms using this pair, then factor by grouping.20x2−16x−25x+20=0−16x−25x=−41x4x(5x−4)−5(5x−4)=0Factor by grouping.(4x−5)(5x−4)=0Factor out 5x−4.We can now use the zero product property to write: 4x−5=04x=5x=54or5x−4=05x=4x=45
- Answer the question: There are two possible numbers, 5/4 and 4/5.
- Look back: The sum of the unknown number and its reciprocal is supposed to equal 41/20. The answer 5/4 has reciprocal 4/5. Their sum is:54+45=1620+2520=4120Thus, 5/4 is a valid solution. The second answer 4/5 has reciprocal 5/4, so it is clear that their sum is also 41/20. Hence, 4/5 is also a valid solution.
Exercise 7.4.4
The sum of a number and its reciprocal is 53/14. Find the number.
- Answer
-
2/7,7/2