135 
…‹7.5Š Equations with Rational Expressions.

„Š Please solve the following equations.


ƒ‰Solve,Š2ˆ‹5ˆŠΪΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΏ
#ˆˆˆΔŠ=ŠΔΔΔΔΔ .‰ ³ˆˆŠ ¨
ˆˆˆyˆy - 3ˆ ³ˆŠ2(y - 3)‹=‹5y
ˆˆˆˆˆˆ³ˆŠ 2y - 6Š=‹5y
ˆˆŠ 2ˆˆŠ 5Š ³ˆˆ- 6Š=‹3y
ˆ‹y(y - 3)ωĊ=Šy(y - 3)ωΔΔΔΔΔ‹³ˆˆ‹yŠ=‹-2
ˆˆŠ yˆˆ‹y - 3‹ΩˆŠy can not be 0 or 3



‚Sˆˆˆˆ‹2ˆ5
ˆŠ To solve the equation,Š Δ‹=‹ΔΔΔΔΔ‹, first multiply
ˆˆˆˆˆ‹y‰y - 3
both sides of the equation by the least common denominator, y(y - 3).

ˆˆˆˆ2ˆˆ‰ 5
ˆˆŠy(y - 2) ω Ċ=Šy(y - 3) ω ΔΔΔΔΔ
ˆˆˆˆyˆˆŠ y - 3

Next reduce each side of the equation by canceling like factors.

ˆˆˆ‹2(y - 3)Š=Š5y

Finally, clear the paren‡s and solve the equation for y.

ˆˆˆŠ 2y - 6Š=Š5y
ˆˆˆˆ- 6Š=Š3y
ˆˆˆˆ‹yŠ=Š-2

"y" can not be 0 or 3 because ‡ numbers would produce zeros in the
denominators of the original equation.

‹1ˆˆŠ x + 3‰x - 4ˆ5
ˆˆŠ Solve,Š ΔΔΔΔΔ‹+‹ΔΔΔΔĊ=ŠΔ .
ˆˆˆˆ‹2ˆ‹3ˆ‹6

ˆ‰4ˆˆ 5ˆˆ‹5ˆˆ3
ˆ A)‹ΔˆŠ B)‹ΔˆŠ C)‹ΔĈŠD)‹Δ
ˆ‰5ˆˆ 7ˆˆ 12ˆˆ5


Š x + 3Šx - 4‰5ˆ‰ ΪΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΏ
#ˆŠΔΔΔΔΔ + ΔΔΔΔΔ‹=‹Δˆ‰ ³ˆˆŠ ¨
ˆ‰2ˆ3ˆ6ˆ‰ ³‹3(x + 3) + 2(x - 4)‹=‹5
ˆˆˆˆˆˆ³‰ 3x + 9 + 2x - 8‹=‹5
ˆ(x + 3)‰(x - 4)ˆ‹5‰ ³ˆˆ 5x + 1‹=‹5
‰6ωΔΔΔΔΔΔΔ + 6ωΔΔΔΔΔΔĊ=Š6ωΔ‰ ³ˆˆ‰5x‹=‹4
ˆŠ2ˆŠ 3ˆ‰6‰Δوˆ‰ x‹=‹4/5

€‹A

‹2ˆˆŠ a + 5‰a - 3ˆ 7
ˆˆŠ Solve,Š ΔΔΔΔΔ‹-‹ΔΔΔΔĊ=ŠΔΔ .
ˆˆˆˆ‹2ˆ‹5ˆ‹10

ˆ‰2ˆˆˆˆˆˆˆ3
ˆ A)‹ΔˆŠ B)‹5ˆŠ C)‹-8ˆŠD)‹- ΔΔ
ˆ‰3ˆˆˆˆˆˆˆ10


Š a + 5Ša - 3‰ 7ˆ‰ΪΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΏ
#ˆŠΔΔΔΔΔ - ΔΔΔΔΔ‹=‹ΔĈ‰³ˆˆŠ ¨
ˆ‰2ˆ5ˆ10ˆ‰³‹5(a + 5) - 2(a - 3)‹=‹7
ˆˆˆˆˆˆ³‰5a + 25 - 2a + 6‹=‹7
ˆ(a + 5)‰ (a - 3)ˆŠ 7Š³ˆˆ3a + 31‹=‹7
Š 10ωΔΔΔΔΔΔΔ - 10ωΔΔΔΔΔΔĊ=Š10ωΔĊ³ˆˆŠ3a‹=‹-24
ˆŠ2ˆ‰5ˆ‰ 10‹ΔوˆŠ a‹=‹-8

€‹C

‹3ˆˆ‰ 6ˆŠ3ˆ‹14
ˆˆŠ Solve,Š ΔΔΔΔΔ‹+‹ΔΔΔΔΔĊ=ŠΔΔ .
ˆˆˆˆz - 2‰3z - 6ˆ 3

ˆˆˆˆ7ˆˆˆˆŠ7
ˆ A)‹6ˆŠ B)‹ΔˆŠ C)‹-12ˆ‹D)‹Δ
ˆˆˆˆ3ˆˆˆˆŠ2


‰ 6ˆ 3ˆ 14ˆ‰ ΪΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΏ
#ˆŠΔΔΔΔΔ + ΔΔΔΔΔΔΔ‹=‹ΔĈ‰ ³ˆˆŠ¨
ˆŠz - 2Š3(z -2)‰ 3ˆ‰ ³‹3ω6 + 3Š=Š14(z - 2)
ˆˆˆˆˆˆŠ³Š18 + 3Š=Š14z - 28
ˆŠ6ˆˆ‹3ˆ‰14Š³ˆ49Š=Š14z
3(z -2)ωΔΔΔΔ + 3(z -2)ωΔΔΔΔΔΔΔ = 3(z -2)ΔĊ³ˆ zŠ=Š49/14
#ˆ z -2ˆŠ 3(z -2)ˆŠ 3‹Δو z = 7/2,‹z Ÿ 2

€‹D

‹4ˆˆŠ 7x - 15
ˆˆŠ Solve,Š ΔΔΔΔΔΔĊ=Š2x.
ˆˆˆˆŠx

ˆˆˆˆŠ 3ˆˆŠ2
ˆ A)‹14ˆŠB)‹-5, ĈC)‹3,‹- Δ‰ D)‹-7
ˆˆˆˆŠ 2ˆˆŠ5


ˆ‹7x - 15ˆˆ‰ ΪΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΔΏ
#ˆˆ ΔΔΔΔΔΔĊ =Š 2xˆ‹³ˆˆŠ¨
#ˆˆŠ xˆˆˆ‹³‹7x - 15Š=Š2x
#ˆˆˆˆˆˆ³‹2x - 7x + 15‹=‹0
ˆˆ 7x - 15ˆˆ‰ ³‹(x + 5)(2x - 3)‹=‹0
ˆŠ x ω ΔΔΔΔΔΔĊ =Š x ω x‰ ³‹x + 5 = 0 ³ 2x - 3 = 0
#ˆˆŠ xˆˆˆ‹Ω‹x = -5Š ³ x = 3/2, x Ÿ 0


€‹B

‹5ˆˆ‰ 4ˆ‹6ˆ‰24
ˆˆŠ Solve,Š ΔΔΔΔΔ‹-‹ΔΔΔΔĊ=Š ΔΔΔΔΔΔ
#ˆˆˆˆr - 3‰r + 3ˆ r - 9


ˆ A)‹3ˆ B)‹No solutionŠ C)‹-3ˆ‹D)‹6



Š Multiplying the equation by the least common denominator,
ˆŠ(r + 3)(r - 3), gives the following equation.
Š(r + 3)ω4 - (r - 3)ω6‹=‹24ˆŠ Since r can not equal 3, (this
ˆ4r + 12 - 6r + 18‹=‹24ˆŠ would put zero in the denomi-
ˆˆ‹-2r + 30‹=‹24ˆŠ nator of the original equa-
ˆˆˆ-2r‹=‹-6ˆŠ tion) there is no solution to
#ˆˆˆ‹r‹=‹3, r Ÿ 3‰this equation. ΔΔΔΔΔΔΔΔΔΔΔ



€‹B