How To Divide Fractional Algebraic Expressions 8 Steps

For example, you might see 2x2y−3x4{\displaystyle {\frac {\frac {2x^{2}}{y-3}}{\frac {x}{4}}}}. You can rewrite this as 2x2y−3÷x4{\displaystyle {\frac {2x^{2}}{y-3}}\div {\frac {x}{4}}}.

For example:2x2y−3÷x4{\displaystyle {\frac {2x^{2}}{y-3}}\div {\frac {x}{4}}}becomes2x2y−3×4x{\displaystyle {\frac {2x^{2}}{y-3}}\times {\frac {4}{x}}}

For example,2x2y−3×4x=4(2x2)x(y−3){\displaystyle {\frac {2x^{2}}{y-3}}\times {\frac {4}{x}}={\frac {4(2x^{2})}{x(y-3)}}}.

Remember that you cannot cancel out a single term (like y{\displaystyle y}) from a binomial (like y−3{\displaystyle y-3}). Also remember that if you have an x2{\displaystyle x^{2}} term in the numerator, and an x{\displaystyle x} term in the denominator, you can cancel out one x{\displaystyle x}, and the x{\displaystyle x} in the denominator disappears, and the x2{\displaystyle x^{2}} in the numerator becomes x{\displaystyle x}. For example, you can cancel an x{\displaystyle x} in the numerator and denominator in the expression 4(2x2)x(y−3){\displaystyle {\frac {4(2x^{2})}{x(y-3)}}}:4(2x2)x(y−3){\displaystyle {\frac {4(2x^{\cancel {2}})}{{\cancel {x}}(y-3)}}}4(2x)y−3{\displaystyle {\frac {4(2x)}{y-3}}}

For example, 4(2x)y−3=8xy−3{\displaystyle {\frac {4(2x)}{y-3}}={\frac {8x}{y-3}}}. So, 2x2y−3x4=4(2x)y−3=8xy−3{\displaystyle {\frac {\frac {2x^{2}}{y-3}}{\frac {x}{4}}}={\frac {4(2x)}{y-3}}={\frac {8x}{y-3}}}.

For example, if you are simplifying the expression y(x+3)4yx−2{\displaystyle {\frac {\frac {y(x+3)}{4}}{\frac {y}{x-2}}}}, after taking the reciprocal and combining terms, you end up with the expression y(x+3)(x−2)4y{\displaystyle {\frac {y(x+3)(x-2)}{4y}}}. First, cancel the y{\displaystyle y} in the numerator and denominator, then multiply the binomials using the FOIL method:y(x+3)(x−2)4y{\displaystyle {\frac {{\cancel {y}}(x+3)(x-2)}{4{\cancel {y}}}}}(x+3)(x−2)4{\displaystyle {\frac {(x+3)(x-2)}{4}}}x2−2x+3x−64{\displaystyle {\frac {x^{2}-2x+3x-6}{4}}}x2+x−64{\displaystyle {\frac {x^{2}+x-6}{4}}}

For example, if you are simplifying the expression 2x+4y2yx{\displaystyle {\frac {\frac {2x+4}{y}}{\frac {2y}{x}}}}, after taking the reciprocal and combining terms, you end up with the expression x(2x+4)2y(y){\displaystyle {\frac {x(2x+4)}{2y(y)}}}. First, factor out a 2 from 2x+4{\displaystyle 2x+4}. Then you can cancel a 2 from the numerator and denominator. Then, simplify the expression by completing the multiplication:(x)(2)(x+2)2y(y){\displaystyle {\frac {(x)(2)(x+2)}{2y(y)}}}(x)(2)(x+2)2y(y){\displaystyle {\frac {(x){\cancel {(2)}}(x+2)}{{\cancel {2}}y(y)}}}(x)(x+2)y(y){\displaystyle {\frac {(x)(x+2)}{y(y)}}}(x2+2x)y2{\displaystyle {\frac {(x^{2}+2x)}{y^{2}}}}

For example, if you have 2+3y5y2{\displaystyle {\frac {2+{\frac {3}{y}}}{\frac {5}{y^{2}}}}}, you would change the 2 into a fraction by multiplying it by yy{\displaystyle {\frac {y}{y}}}:2+3y5y2{\displaystyle {\frac {2+{\frac {3}{y}}}{\frac {5}{y^{2}}}}}2yy+3y5y2{\displaystyle {\frac {{\frac {2y}{y}}+{\frac {3}{y}}}{\frac {5}{y^{2}}}}}2y+3y5y2{\displaystyle {\frac {\frac {2y+3}{y}}{\frac {5}{y^{2}}}}}2y+3y÷5y2{\displaystyle {\frac {2y+3}{y}}\div {\frac {5}{y^{2}}}}2y+3y×y25{\displaystyle {\frac {2y+3}{y}}\times {\frac {y^{2}}{5}}}y2(2y+3)5y{\displaystyle {\frac {y^{2}(2y+3)}{5y}}}y2(2y+3)5y{\displaystyle {\frac {y^{\cancel {2}}(2y+3)}{5{\cancel {y}}}}}y(2y+3)5{\displaystyle {\frac {y(2y+3)}{5}}}2y2+3y5{\displaystyle {\frac {2y^{2}+3y}{5}}}