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Answers
Step-by-step explanation:
7)
Given equation is 4/(s-3) + (s/2) = -2
=> [(4×2)+s(s-3)]/2(s-3) = -2
=> (8+s²-3s)/(2s-6) = -2
=> 8+s²-3s = -2(2s-6)
=> 8+s²-3s = -4s+12
=> s²-3s+8+4s-12 = 0
=> s²+s-4 = 0
The standard form is s²+s-4 = 0
8)
Given equation is 2t/(t-5) + 1/(t-3) = 3
=> [2t(t-3)+1(t-5)]/(t-5)(t-3) = 3
=> (2t²-6t+t-5)/(t²-5t-3t+15) = 3
=> (2t²-5t-5)/(t²-8t+15) = 3
=> 2t²-5t-5 = 3(t²-8t+15)
=> 2t²-5t-5 = 3t²-24t+45
=>3t²-24t+45-2t²+5t+5 = 0
=> t²-19t+50 = 0
The standard form is t²-19t+50 = 0
9)
Given equation is (8/y) = 2/(y-2) +1
=> (8/y)- 2/(y-2) = 1
=> [8(y-2)-2y]/(y-2)y = 1
=> (8y-16-2y)/(y²-2y) = 1
=> (6y-16)/(y²-2y) = 1
=> y²-2y = 6y-16
=> y²-2y -6y+16 = 0
=> y²-8y+16 = 0
The standard form is y²-8y+16 = 0
10)
Given equation is 1/(x+2) - 1/(x-2) = 1
=> [(x-2)-(x+2)]/(x+2)(x-2) = 1
=> (x-2-x-2)/(x²-4) = 1
=> (-2-2)/(x²-4) = 1
=> -4/(x²-4) = 1
=> x²-4 = -4
=> x²-4+4 = 0
=> x² +0 = 0
=> x² = 0
The standard form is x² = 0 or x²+0x+0 = 0
Used formulae:-
The standard quadratic equation is ax²+bx+c = 0