A fraction becomes 4
5
, if 2 is added to both numerator and denominator, if
however 4 is subtracted from both numerator and denominator, then the
fraction becomes 1
2
. The algebraic representation of situation is
) 5 − 4 + 2 = 0 , − = 0 ) 5 − 4 + 2 = 0 , 2 − − 4 = 0
) + 4 = 0 , + 2 = 0 ) ℎ
Answers
Step-by-step explanation:
(52)(−2)(3)+(72)(22)+(−3)(−5)(42)
(5p^{2})(-2q)(3q)+(7p^{2})(2q^{2})+(-3p)(-5p)(4q^{2})(5p2)(−2q)(3q)+(7p2)(2q2)+(−3p)(−5p)(4q2)
52(−2)(3)+(72)(22)+(−3)(−5)(42)
−302+(72)(22)+(−3)(−5)(42)
-30p^{2}{\color{#c92786}{q}}{\color{#c92786}{q}}+(7p^{2})(2q^{2})+(-3p)(-5p)(4q^{2})−30p2qq+(7p2)(2q2)+(−3p)(−5p)(4q2)
−3022+(72)(22)+(−3)(−5)(42
52(−2)⋅3+(72)(22)+(−3)(−5)(42)
{\color{#c92786}{5}}p^{2}\left({\color{#c92786}{-2}}\right)q \cdot {\color{#c92786}{3}}q+(7p^{2})(2q^{2})+(-3p)(-5p)(4q^{2})5p2(−2)q⋅3q+(7p2)(2q2)+(−3p)(−5p)(4q2)
−302+(72)(22)+(−3)(−5)(42)
STEP1:Equation at the end of step 1
((((5•(a2))•(b3))-(2a•(b2)))+((3•(b3))•(a2)))-5ab2
STEP 2 :
Equation at the end of step2:
((((5•(a2))•(b3))-(2a•(b2)))+(3b3•a2))-5ab2
STEP 3 :
Equation at the end of step3:
((((5•(a2))•(b3))-2ab2)+3a2b3)-5ab2
STEP 4 :
Equation at the end of step4:
(((5a2 • b3) - 2ab2) + 3a2b3) - 5ab2
STEP5:
STEP6:Pulling out like terms
6.1 Pull out like factors :
8a2b3 - 7ab2 = ab2 • (8ab - 7)
Final result :
ab2 • (8ab - 7)
Answer:
i don't know sorry I will try next don't take seriously