A variety of equations arise from time to time that involve the sums and differences of rational expressions. The solution of these almost always begins by multiplying both sides of the equation by well chosen factors that will 'clear' the denominators.
Here are three examples:
Example 1
Solve $\frac{x}{x-5}+\frac{10}{x+5}=\frac{4}{x^2-25}$xx−5+10x+5=4x2−25
Observe that $x^2-25=\left(x-5\right)\left(x+5\right)$x2−25=(x−5)(x+5), and so the strategy becomes clear. We need to multiply both sides of the equation by $x^2-25$x2−25 to clear the denominators.
| $\frac{x}{x-5}+\frac{10}{x+5}$xx−5+10x+5 | $=$= | $\frac{4}{x^2-25}$4x2−25 |
| $\left(x-5\right)\left(x+5\right)\frac{x}{\left(x-5\right)}+\left(x-5\right)\left(x+5\right)\frac{10}{\left(x+5\right)}$(x−5)(x+5)x(x−5)+(x−5)(x+5)10(x+5) | $=$= | $\left(x-5\right)\left(x+5\right)\frac{4}{x^2-25}$(x−5)(x+5)4x2−25 |
| $x\left(x+5\right)+10\left(x-5\right)$x(x+5)+10(x−5) | $=$= | $4$4 |
| $x^2+15x-54$x2+15x−54 | $=$= | $0$0 |
| $\left(x+18\right)\left(x-3\right)$(x+18)(x−3) | $=$= | $0$0 |
| $\therefore$∴ $x$x | $=$= | $3,-18$3,−18 |
Example 2
Solve $\frac{48}{x^2-8x}-\frac{80}{x^2-64}=0$48x2−8x−80x2−64=0
Here we need to notice that the factor $\left(x-8\right)$(x−8) is common to both denominator expressions. Thus multiplying both sides of the equation by $x\left(x-8\right)\left(x+8\right)$x(x−8)(x+8) will clear the denominators.
| $\frac{48}{x^2-8x}-\frac{80}{x^2-64}$48x2−8x−80x2−64 | $=$= | $0$0 |
| $\frac{48}{x\left(x-8\right)}-\frac{80}{\left(x-8\right)\left(x+8\right)}$48x(x−8)−80(x−8)(x+8) | $=$= | $0$0 |
| $x\left(x-8\right)\left(x+8\right)\frac{48}{x\left(x-8\right)}-x\left(x-8\right)\left(x+8\right)\frac{80}{\left(x-8\right)\left(x+8\right)}$x(x−8)(x+8)48x(x−8)−x(x−8)(x+8)80(x−8)(x+8) | $=$= | $0$0 |
| $48\left(x+8\right)-80x$48(x+8)−80x | $=$= | $0$0 |
| $32x$32x | $=$= | $384$384 |
| $\therefore$∴ $x$x | $=$= | $12$12 |
Example 3
Find all real solutions to the equation $x^{-4}+7x^{-2}-144=0$x−4+7x−2−144=0. Are there any complex solutions?
To find the real solutions we could proceed as follows:
Re-write the equation as $\frac{1}{x^4}+\frac{7}{x^2}-144=0$1x4+7x2−144=0, and then multiply both sides by $x^4$x4. This would lead to the new equation $1+7x^2-144x^4=0$1+7x2−144x4=0.
Then by setting $u=x^2$u=x2, and rearranging, we establish the quadratic equation given by $144u^2-7u-1=0$144u2−7u−1=0. We solve this as follows:
| $144u^2-7u-1$144u2−7u−1 | $=$= | $0$0 |
| $\left(16u+1\right)\left(9u-1\right)$(16u+1)(9u−1) | $=$= | $0$0 |
| $\therefore$∴ $u$u | $=$= | $-\frac{1}{16},\frac{1}{9}$−116,19 |
For $u=x^2=-\frac{1}{16}$u=x2=−116, there is no real solution because, for all $x$x, $x^2\ge0$x2≥0.
For $u=x^2=\frac{1}{9}$u=x2=19, we have that $x=\pm\frac{1}{3}$x=±13.
NOTE:
A slightly simpler alternative approach to the problem is to immediately set $u=x^{-2}$u=x−2 so that the equation $x^{-4}+7x^{-2}-144=0$x−4+7x−2−144=0 changes to $u^2+7u-144=0$u2+7u−144=0.
Factoring, we have $\left(u+16\right)\left(u-9\right)=0$(u+16)(u−9)=0 and therefore $u=-16,9$u=−16,9.
Setting $u=x^{-2}=-16$u=x−2=−16, and $u=x^{-2}=9$u=x−2=9, we again arrive at the same conclusion that the real solutions are given by $x=\pm\frac{1}{3}$x=±13.
If we admit complex solutions, then we can solve the equation $x^2=-\frac{1}{16}$x2=−116 as follows:
| $x^2$x2 | $=$= | $-\frac{1}{16}$−116 |
| $16x^2+1$16x2+1 | $=$= | $0$0 |
| $16x^2-i^2$16x2−i2 | $=$= | $0$0 |
| $\left(4x+i\right)\left(4x-i\right)$(4x+i)(4x−i) | $=$= | $0$0 |
| $\therefore$∴ $x$x | $=$= | $\pm\frac{i}{4}$±i4 |
Some more examples
Question 1
Find all solutions to $\frac{x}{x-2}+\frac{1}{x+2}=\frac{8}{x^2-4}$xx−2+1x+2=8x2−4.
Question 2
Solve $\frac{3}{x^2-2x}-\frac{5}{x^2-4}=0$3x2−2x−5x2−4=0.
Question 3
Find all complex solutions to $x^{-4}+8x^{-2}-9=0$x−4+8x−2−9=0.