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# Adding And Subtracting Rational Expressions

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1. Simplify the expression. 5x2x­1­2xx+2 a. x + 12(2x ­ 1)(x + 2) b. x(x + 12)(2x ­ 1)(x + 2) c. x(3x + 12)(2x ­ 1)(x + 2) d. x(x ­ 12)(2x ­ 2)(x + 1) Solution :­ 5x2x ­ 1­2xx + 2 [Original expression.] The LCD is (2x ­ 1)(x + 2). To rewrite 5x2x ­ 1 with a denominator of (2x ­ 1)(x + 2), multiply the numerator and denominator by (x + 2). To rewrite 2xx + 2 with a denominator of (2x ­ 1)(x + 2), multiply the numerator and denominator by (2x ­ 1).
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Expressions
1.Simplifytheexpression.5x2x12xx+2
a. x+12(2x1)(x+2)
b. x(x+12)(2x1)(x+2)
c. x(3x+12)(2x1)(x+2)
d. x(x12)(2x2)(x+1)
Solution:5x2x12xx+2[Originalexpression.]
TheLCDis(2x1)(x+2).
by(x+2).
by(2x1).

5x(x+2)(2x1)(x+2)2x(2x1)(x+2)(2x-1)[RewriteusingLCD.]
=5x2+10x(2x1)(x+2)4x22x(x+2)(2x-1)[Simplifynumerators.]
=5x2+10x4x2+2x(2x1)(x+2)[Subtract.]
=x2+12x(2x1)(x+2)[Combineliketerms.]
=x(x+12)(2x1)(x+2)[Factor.]
So,thedifferenceisx(x+12)(2x1)(x+2).
2.Simplifytheexpression.x+3x+4xx+5
a. 15(x+4)(x+5)
b. 4x+15(x+4)(x+5)
c. 2(x+2)(x+4)(x+5)
d. 8x+15(x+4)(x+5)
Solution:x+3x+4xx+5[Originalexpression.]
TheLCDis(x+4)(x+5).
by(x+5).

(x+4).

(x+3)(x+5)(x+4)(x+5)(x)(x+4)(x+5)(x+4)[RewriteusingLCD.]
=x2+8x+15(x+4)(x+5)x2+4x(x+4)(x+5)[Simplifynumerators.]
=x2+8x+15x24x(x+4)(x+5)[Subtract.]
=4x+15(x+4)(x+5)[Combineliketerms.]
So,thedifferenceis4x+15(x+4)(x+5).
3.Simplifytheexpression.3xx+32xx1
a.x(x9)(x+3)(x-1)
b.x(x+3)x1
c. x(x+9)(x+3)(x-1)
d. xx1
Solution:3xx+32xx1[Originalexpression.]
numeratoranddenominatorby(x-1).
(x+3).

3x(x1)(x+3)(x1)2x(x+3)(x+3)(x-1)[RewriteusingLCD.]

=3x23x(x+3)(x1)2x2+6x(x+3)(x-1)[Simplifynumerators.]
=3x23x2x26x(x+3)(x-1)[Subtract.]
=x29x(x+3)(x-1)[Combineliketerms.]
=x(x9)(x+3)(x-1)[Factor.]
So,thedifferenceisx(x9)(x+3)(x1).
4.Simplifytheexpression.3xx5+x4x+3
a. 4(x2+5x)(x+3)(x-5)
b. 2(x2+5)(x+3)(x5)
c. 4(x2+5)(x+3)(x-5)
d. 2(x+1)(x5)(x+3)
Solution:3xx5+x4x+3[Originalexpression.]
TheLCDis(x5)(x+3).
(x+3).
by(x5).

3x(x+3)(x5)(x+3)+(x4)(x5)(x+3)(x-5)[RewriteusingLCD.]

=3x2+9x(x5)(x+3)+x24x5x+20(x+3)(x-5)[Simplifynumerators.]
=4x2+20(x+3)(x-5)[Combineliketerms.]
=4(x2+5)(x+3)(x-5)[Factor.]
So,thesumis4(x2+5)(x+3)(x5).
5.Simplify:10x2x28100+9x(x10)x28100
a. 1x+90
b. xx90
c. 1x-90
d. Xx+90
Solution:10x2x28100+9x(x10)x2-8100[Originalexpression.]
=10x29x2+90xx2-8100[Usedistributivepropertyofmultiplicationoversubtraction.]
=x2+90xx2-8100[Combinetheliketerms.]

=x(x+90)(x90)(x+90)[Factorthenumeratorandthedenominator.]

=xx-90[Divideoutthecommonfactors.]
overthe_________.
a. sumofthenumerators
b. sumofthedenominators
c. commondenominator
d. commonnumerator
resultoverthecommondenominator.
7.Simplify:1(x+1)+(x2)(x+1)
a. (x1)(x+1)
b. 1x+1
c. (x2)(x+1)
d. x(x+1)
resultoverthecommondenominator.
1(x+1)+(x2)(x+1)[Originalexpression.]
=(x1)(x+1)[Combinetheliketerms.]

8.Simplify:x2+3(3x4)x24(3x4)
a. 73x4
b. 83x4
c. 73+x4
d. 73x4
Solution:Tosubtractrationalexpressionswithlikedenominators,subtractthenumeratorsand
writetheresultoverthecommondenominator.
x2+3(3x4)x24(3x4)[Originalexpression.]
=x2+3(x24)3x4[Subtractthenumerators.]
=x2+3x2+43x4[Usedistributiveproperty.]
=73x4[Combinetheliketerms.]
a. 4-xx
b. 5xx
c. 5x3x
d. 5x2x
resultoverthecommondenominator.

4x+1xx[Originalexpression.]

=5xx[Simplify.]
10.Simplify:10xx+5+5x+5
a. 15xx+5
b. xx+5
c. 15+xx+5
d. 15X+5
resultoverthecommondenominator.
10xx+5+5x+5[Originalexpression.]
=15xx+5[Combineliketerms.]

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