Algebraic fractions 1
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1) Algebraic fractions can be added, subtracted, multiplied, and divided. The key is to find common factors in the numerators and denominators. 2) For example, when adding fractions, find the common denominator and add only the numerators. When multiplying, multiply the numerators and denominators. 3) Algebraic fractions can also be used to solve equations by simplifying the fractions and then using inverse operations like addition/subtraction to isolate the variable.
Algebraic fractions 1
- 3. Addition bd bcad d c b a + =+ y x x 2 3 + Simplify )2( ))(()2(3 yx xxy + = xy xy 2 6 2 + =
- 4. Subtraction bd bcad d c b a − =− p t p 2 2 − Simplify )2)(( ))(()2)(2( pp tpp − = 2 2 4 p ptp − = 2 2 )4( p tp − = p t 2 )4( − = Can you see an alternative, quicker method here?
- 5. Subtraction bd bcad d c b a − =− 4 33 2 12 + − + xxSimplify )4)(2( )33(2)12(4 +−+ = xx 8 6648 −−+ = xx 8 22 − = x 8 )1(2 − = x 4 4 1− = x
- 6. Multiplication bd ac d c b a =× 5 5 3 + − × x xxSimplify )5(3 )5)(( + − = x xx )5(3 52 + − = x xx
- 9. Example: 2 4 1 2 + + − xx )2)(1( +− = xx The denominators don’t share any factors )2(2 +x )1(4 −+ x )2)(1( 6 +− = xx x
- 10. Example : 2 )1( 4 1 2 − + − x x x The 1st denominator is a factor of the 2nd )2)(1( 26 +− − = xx x )2)(1( )13(2 +− − = xx x 2 )1( − = x )1(2 −x x4+
- 11. The denominators share a factor Example : 23 3 34 2 22 +− − +− xxxx We have to find the factors )2( 3 )3( 2 − − − = xx)1( −x )1( −x )2)(3)(1( −−− = xxx 42 −x 93 +− x )2)(3)(1( 5 −−− +− = xxx x )2)(3( −− = xx)1( −x )2(2 −x )3(3 −− x
- 12. − −+ − = )2)(1( 12 xx x Example : 6 3 2 12 22 −+ + − −− − xx x xx x )3)(2)(1( +−+ = xxx 352 2 −+ xx )34( 2 ++− xx )3)(2)(1( 62 +−+ −+ = xxx xx )3)(2)(1( +−+ = xxx )3)(2( 3 +− + xx x )3)(2)(1( +−+ = xxx 352 2 −+ xx 342 −−− xx )3)(2)(1( )2)(3( +−+ −+ = xxx xx1 1 1 + = x )3)(12( +− xx )1)(3( ++− xx )3)(2)(1( +−+ = xxx
- 13. Equations Once you know how to simplify algebraic fractions, you Can solve equations containing them For example 7 43 2 =− xx 3 5 2 2 1 = + + + xx Click an equation to see it solved
- 15. Solve 3 5 2 2 1 = + + + xx 3 )5)(2( )2)(2()1)(5( = +++ ⇒ xx 3 10 4255 = +++ ⇒ xx 3 10 97 = + ⇒ x 3097 =+⇒ x 217 =⇒ x 3=x
- 16. Short Quiz : Express each of the following as a single fraction in its simplest form. 4 5 3 2 + + − xx 1. )4)(3( )3(5)4(2 +− −++ = xx xx )4)(3( 15582 +− −++ = xx xx )4)(3( 77 +− − = xx x )4)(3( )1(7 +− − = xx x