Lesson 1 · from §69 of the 1910 text
Clearing an Equation of Fractions
Clearing of fractions
An equation with fractions in it is solved by first getting rid of the fractions. Multiply every term on both sides by the L.C.M. of the denominators. Both members stay equal (equals multiplied by equals give equals), and every denominator cancels. This is called clearing the equation of fractions.
Example. Solve \dfrac27\left(9 + \dfrac{5x}{3}\right) - \dfrac{3(2x - 1)}{5} + \dfrac{5x}{3} = 2x.
Perform the indicated multiplication first: \frac{18}{7} + \frac{10x}{21} - \frac{6x - 3}{5} + \frac{5x}{3} = 2x. The L.C.M. of 7, 21, 5, and 3 is 105. Multiply each term by 105 and cancel: 270 + 50x - (6x - 3)\cdot 21 + 175x = 210x. 270 + 50x - 126x + 63 + 175x = 210x. Combining like terms, 99x + 333 = 210x, so -111x = -333 and x = 3.
Check: \frac27\left(9 + \frac{15}{3}\right) - \frac{3(6 - 1)}{5} + \frac{15}{3} = 4 - 3 + 5 = 6, and 2x = 6. ✓
Watch the minus signs
A fraction bar acts like parentheses around its numerator. When a fraction has a minus sign in front of it, keep the numerator in parentheses after you multiply: -\dfrac{6x - 3}{5}\cdot 105 is -21(6x - 3) = -126x + 63, not -126x - 63. This is where most mistakes happen.
Rule. Free the equation of any parentheses it may contain. Find the L.C.M. of the denominators of the fractions and multiply each fraction and each integral term of the equation by it, using cancellation wherever possible. Transpose and solve as usual.
Two special cases
- Sometimes x drops out entirely and you're left with something true, like 0 = 0. Then every number is a root: the equation is really an identity. Type
any. - If x drops out and you're left with something false, like 0 = 5, no number is a root. Type
none.
Letters other than x may appear as known numbers. Treat them exactly like numbers: in \frac{c}{3x} - \frac{c}{5x} = \frac1{15}, multiplying by 15x gives 5c - 3c = x, so x = 2c.
Try one: solve \dfrac{x - 1}{3} - \dfrac{x - 3}{5} = 2.
Quick check
\displaystyle \frac{x-1}{3}-\frac{x-3}{5}=2
Exercises
How to type math
Use ^ for powers (a^2), / for fractions ((a+b)/2), sqrt(x) for square roots, and cbrt(x) for cube roots. You can leave out the multiplication sign: 3ab means 3 \cdot a \cdot b. Mixed numbers like 4 1/2 work too. The preview under each box shows how your answer was read.
Solve each equation and check. Type the root, like x = 3, -5/2, or 2c. If every number is a root, type any; if none is, type none.
\displaystyle \frac{x}{2}+\frac{x}{3}=10
\displaystyle \frac{4}{3}x+\frac{2}{5}x=5\tfrac{1}{5}
\displaystyle \frac{x+5}{4}-\frac{2x+4}{9}=1
\displaystyle \frac{2x+3}{5}-\frac{1}{3}(x-3)=2
\displaystyle \frac{4x+2}{11}-\frac{1}{5}(x+5)=0
\displaystyle \frac{3}{4}(x+1)-\frac{5x-7}{6}=\frac{7}{3}
\displaystyle \frac{x+6}{10}+\frac{3}{2}(x+4)=-3
\displaystyle \frac{5x-12}{6}-\frac{4}{11}(2x-7)=\frac{1}{3}
\displaystyle 2x-1-\frac{12x-7}{6}-\frac{1}{6}=0
\displaystyle \frac{10x-7}{6}+\frac{5}{2}\left(\frac{2}{5}-x\right)=\frac{15x-11}{3}
\displaystyle \frac{5x}{6}-\frac{1}{2}-\frac{3}{8}\left(x-\frac{5}{3}\right)+\frac{7}{32}=0
\displaystyle \frac{5}{x}+\frac{4}{3}=\frac{9}{x}
\displaystyle \frac{9x}{4}-\frac{3}{4}-\frac{3x-7}{3}+4=-\frac{17}{24}
\displaystyle \frac{c}{3x}-\frac{c}{5x}=\frac{1}{15}
\displaystyle \frac{1}{2x}-\frac{13}{24}=\frac{8}{3x}
\displaystyle \frac{2a-3x}{6a}+\frac{5a-2x}{5a}+\frac{41}{30}=0
\displaystyle \frac{x}{a}-\frac{1}{3}(a-3x)+\frac{19a}{3}=-6
\displaystyle \frac{cx}{3}-\frac{n}{5}(3x-5cn)=cn\left(\frac{2c}{3}-\frac{n}{5}\right)
\displaystyle 2x-b-\frac{b}{a}(3x-4b)+2a=\frac{2(2a^2-b^2)}{a}
\displaystyle (x+5)(x-6)=x\left(x-\frac{5}{2}\right)
\displaystyle \left(x+\frac{3}{4}\right)\left(3+\frac{x}{2}\right)=\frac{x}{2}(x-5)+8\tfrac{1}{8}
\displaystyle \left(x-\frac{1}{2}\right)\left(x+\frac{3}{7}\right)=(x-1)(x+2)+1\tfrac{11}{14}
\displaystyle \left(x-\frac{3}{2}\right)\left(x+\frac{2}{3}\right)-\left(x-\frac{1}{3}\right)^2-1\tfrac{1}{9}=0
\displaystyle \left(x+\frac{2}{5}\right)^2-\left(x-\frac{1}{2}\right)\left(x+\frac{1}{5}\right)+\frac{31}{50}=0