Holt.Blue
Back To Class Notes Menu
Section 7.2: Adding and Subtracting Rational Expressions


Arithmetic of Rational Expressions: Rational expressions have their own arithmetic (addition, subtraction, multiplication, and division).

In the last section we learned how to multiply and divide rational expressions. Today, we are going to learn how to add and subtract.













Adding Rational Expressions

Recall: remember how we add two fractions? Of course you do! Get a common denominator and then add. For example: $$\displaystyle \frac{1}{6}+\displaystyle \frac{2}{15}=$$ Great News! The SAME RULE APPLIES to rational expressions.

Example: $\displaystyle \frac{1}{(x-3)(x+2)} + \displaystyle \frac{x}{(x+2)(x+5)}$

$$ \begin{array}{cll} \displaystyle \frac{1}{(x-3)(x+2)} + \displaystyle \frac{x}{(x+2)(x+5)} &= \displaystyle \frac{1}{(x-3)(x+2)}\cdot \color{magenta}{1} + \displaystyle \frac{x}{(x+2)(x+5)}\cdot \color{magenta}{1} &\mbox{multiply by 1}\\ &= \displaystyle \frac{1}{(x-3)(x+2)}\cdot\frac{\color{magenta}{x+5}}{\color{magenta}{x+5}} + \displaystyle \frac{x}{(x+2)(x+5)}\cdot\frac{\color{magenta}{x-3}}{\color{magenta}{x-3}} &\mbox{turn each 1 into a fancy 1}\\ &= \displaystyle \frac{1\cdot \color{magenta}{(x+5)}}{(x-3)(x+2)\color{magenta}{(x+5)}} + \displaystyle \frac{x\color{magenta}{(x-3)}}{(x+2)(x+5)\color{magenta}{(x-3)}} &\mbox{multiply}\\ &= \displaystyle \frac{1\cdot (x+5)+x(x-3)}{(x-3)(x+2)(x+5)} &\mbox{add numerators over common denominator}\\ &= \displaystyle \frac{x+5+x^2-3x}{(x-3)(x+2)(x+5)} &\mbox{simplify}\\ &= \displaystyle \frac{x^2-2x+5}{(x-3)(x+2)(x+5)} &\mbox{simplify}\\ \end{array} $$















Adding Rational Expressions

Process: To add two rational expressions:

$1)$ Factor the denominators of each summand.

$2)$ Multiply each expression by "fancy ones" to get a common denominator.

$3)$ Add numerators

$4)$ Simplify and reduce if necessary.











Adding Rational Expressions

Examples

$\displaystyle \frac{ 2 z +3}{z^2 -7 z +6}+\displaystyle \frac{ 6 z -4}{z^2 -13 z +42}$

$$ \begin{array}{cll} \displaystyle \frac{ 2 z +3}{z^2 -7 z +6}+\displaystyle \frac{ 6 z -4}{z^2 -13 z +42} &= \displaystyle \frac{ 2 z +3}{(z-1)(z-6)}+\displaystyle \frac{ 6 z -4}{(z-7)(z-6)} &\mbox{factor denominators}\\ &= \displaystyle \frac{ (2 z +3)\color{magenta}{(z-7)}}{(z-1)(z-6)\color{magenta}{(z-7)}}+\displaystyle \frac{ (6 z -4)\color{magenta}{(z-1)}}{(z-7)(z-6)\color{magenta}{(z-1)}} & \mbox{multiply by fancy 1s}\\ &= \displaystyle \frac{ (2 z +3)(z-7)+(6 z -4)(z-1)}{(z-1)(z-6)(z-7)} & \mbox{add over common denominator}\\ &= \displaystyle \frac{ 2z^2-11z-21+6z^2-10z+4}{(z-1)(z-6)(z-7)} & \mbox{simplify numerator}\\ &= \displaystyle \frac{ 8z^2-21z-17}{(z-1)(z-6)(z-7)} & \mbox{simplify numerator}\\ \end{array} $$



$\displaystyle \frac{ 6 b +2}{b^2 -5 b +4}+\displaystyle \frac{ 5 b +5}{ b -4}$

$$ \begin{array}{cll} \displaystyle \frac{ 6 b +2}{b^2 -5 b +4}+\displaystyle \frac{ 5 b +5}{ b -4} &= \displaystyle \frac{ 6 b +2}{(b-1)(b-4)}+\displaystyle \frac{ 5 b +5}{ b -4} &\mbox{factor denominators}\\ &= \displaystyle \frac{ 6 b +2}{(b-1)(b-4)}+\displaystyle \frac{ (5 b +5)\color{magenta}{(b-1)}}{ (b -4)\color{magenta}{(b-1)}} & \mbox{multiply by fancy 1s}\\ &= \displaystyle \frac{6b+2+(5b+5)(b-1)}{(b-1)(b-4)} & \mbox{add over common denominator}\\ &= \displaystyle \frac{6b+2+5b^2-5}{(b-1)(b-4)} & \mbox{simplify numerator}\\ &= \displaystyle \frac{5b^2+6b-3}{(b-1)(b-4)} & \mbox{simplify numerator}\\ \end{array} $$



$\displaystyle \frac{ 7 v -6}{v^2 + v }+\displaystyle \frac{ 2 v -7}{- v -1}$

$$ \begin{array}{cll} \displaystyle \frac{ 7 v -6}{v^2 + v }+\displaystyle \frac{ 2 v -7}{- v -1} &= \displaystyle \frac{ 7 v -6}{v(v+1)}+\displaystyle \frac{ 2 v -7}{- (v+1)} &\mbox{factor denominators}\\ &= \displaystyle \frac{ 7 v -6}{v(v+1)}-\displaystyle \frac{ 2 v -7}{v+1} &\mbox{negative in denominator makes whole term negative}\\ &= \displaystyle \frac{ 7 v -6}{v(v+1)}-\displaystyle \frac{ (2 v -7)\color{magenta}{v}}{(v+1)\color{magenta}{v}} & \mbox{multiply by fancy 1s}\\ &= \displaystyle \frac{7v-6-(2v-7)v}{v(v+1)} & \mbox{subtract over common denominator}\\ &= \displaystyle \frac{7v-6-(2v^2-7v)}{v(v+1)} & \mbox{simplify numerator}\\ &= \displaystyle \frac{7v-6-2v^2+7v}{v(v+1)} & \mbox{simplify numerator (distribute minus to BOTH terms!)}\\ &= \displaystyle \frac{-2v^2+14v-6}{v(v+1)} & \mbox{simplify numerator}\\ \end{array} $$













Subtracting Rational Expressions

Recall: Remember how we subtract two fractions? It's not difficult if you know how to add fractions. Get a common denominator and then subtract. For example: $$\displaystyle \frac{1}{6}-\displaystyle \frac{2}{15}=$$ Great News! The SAME RULE APPLIES to rational expressions.

Example: $\displaystyle \frac{1}{(x-3)(x+2)} - \displaystyle \frac{x}{(x+2)(x+5)}$

$$ \begin{array}{cll} \displaystyle \frac{1}{(x-3)(x+2)} - \displaystyle \frac{x}{(x+2)(x+5)} &= \displaystyle \frac{1}{(x-3)(x+2)}\cdot \color{magenta}{1} - \displaystyle \frac{x}{(x+2)(x+5)}\cdot \color{magenta}{1} &\mbox{multiply by 1}\\ &= \displaystyle \frac{1}{(x-3)(x+2)}\cdot\frac{\color{magenta}{x+5}}{\color{magenta}{x+5}} - \displaystyle \frac{x}{(x+2)(x+5)}\cdot\frac{\color{magenta}{x-3}}{\color{magenta}{x-3}} &\mbox{turn each 1 into a fancy 1}\\ &= \displaystyle \frac{1\cdot \color{magenta}{(x+5)}}{(x-3)(x+2)\color{magenta}{(x+5)}} - \displaystyle \frac{x\color{magenta}{(x-3)}}{(x+2)(x+5)\color{magenta}{(x-3)}} &\mbox{multiply}\\ &= \displaystyle \frac{1\cdot (x+5)-x(x-3)}{(x-3)(x+2)(x+5)} &\mbox{SUBTRACT numerators over common denominator}\\ &= \displaystyle \frac{x+5-(x^2-3x)}{(x-3)(x+2)(x+5)} &\mbox{simplify}\\ &= \displaystyle \frac{x+5-x^2+3x}{(x-3)(x+2)(x+5)} &\mbox{simplify (distribute minus to BOTH terms!)}\\ &= \displaystyle \frac{-x^2+4x+5}{(x-3)(x+2)(x+5)} &\mbox{simplify}\\ \end{array} $$















Dire Warning

In the above example, the minus sign must distribute to each term after getting our common denominator.













Subtracting Rational Expressions

Process: To subtract two rational expressions:

$1)$ Factor the denominators of the minuend and the subtrahend.

$2)$ Multiply each expression by "fancy ones" to get a common denominator.

$3)$ Subtract numerators. (Remember dire warning: the minus must distribute to each term in the subtrahend.)

$4)$ Simplify and reduce if necessary.











Subtracting Rational Expressions

Examples

$\displaystyle \frac{5 y +4}{y -2}-\displaystyle \frac{3 y -6}{4 y -7}$

$$ \begin{array}{cll} \displaystyle \frac{5 y +4}{y -2}-\displaystyle \frac{3 y -6}{4 y -7} &= \displaystyle \frac{(5 y +4)\color{magenta}{(4y-7)}}{(y -2)\color{magenta}{(4y-7)}}-\displaystyle \frac{(3 y -6)\color{magenta}{(y-2)}}{(4 y -7)\color{magenta}{(y-2)}} & \mbox{multiply by fancy 1s}\\ &= \displaystyle \frac{(5y+4)(4y-7)-(3y-6)(y-2)}{(y-2)(4y-7)} & \mbox{subtract over common denominator}\\ &= \displaystyle \frac{20y^2-19y-28-(3y^2-12y+12)}{(y-2)(4y-7)} & \mbox{simplify numerator}\\ &= \displaystyle \frac{20y^2-19y-28-3y^2+12y-12}{(y-2)(4y-7)} & \mbox{simplify numerator (distribute minus to BOTH terms!)}\\ &= \displaystyle \frac{17y^2-7y-40}{(y-2)(4y-7)} & \mbox{simplify numerator}\\ \end{array} $$



$\displaystyle \frac{6 t +7}{t^2-25}-\displaystyle \frac{4}{t+5}-\displaystyle \frac{7}{t-5}$

$$ \begin{array}{cll} \displaystyle \frac{6 t +7}{t^2-25}-\displaystyle \frac{4}{t+5}-\displaystyle \frac{7}{t-5} &= \displaystyle \frac{6 t +7}{(t+5)(t-5)}-\displaystyle \frac{4}{t+5}-\displaystyle \frac{7}{t-5} &\mbox{factor denominators}\\ &= \displaystyle \frac{6 t +7}{(t+5)(t-5)}-\displaystyle \frac{4\cdot \color{magenta}{(t-5)}}{(t+5)\color{magenta}{(t-5)}}-\displaystyle \frac{7\cdot\color{magenta}{(t+5)}}{(t-5)\color{magenta}{(t+5)}} & \mbox{multiply by fancy 1s}\\ &= \displaystyle \frac{6t+7-4(t-5)-7(t+5)}{(t+5)(t-5)} & \mbox{perform operations over common denominator}\\ &= \displaystyle \frac{6t+7-4t+20-7t-35}{(t+5)(t-5)} & \mbox{simplify numerator (distribute minus to BOTH terms!)}\\ &= \displaystyle \frac{-5t-8}{(t+5)(t-5)} & \mbox{simplify numerator}\\ \end{array} $$