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secondary 3 | E Maths
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Pls help asap ty
Date Posted: 1 year ago
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(6–4x)/5+(x+3)/2 = [2(6–4x)+5(x+3)]/10 = (12–8x+5x+15)/10 =(27–3x)/10
i)
To be negative,
27–3x < 0
x > 9
ii)
To be ≤ –3
(27–3x)/10 ≤ 3
27–3x ≤ 30
x ≥ –1
i)
To be negative,
27–3x < 0
x > 9
ii)
To be ≤ –3
(27–3x)/10 ≤ 3
27–3x ≤ 30
x ≥ –1
why is the 3 positive in the second step?
is it not -30?
Sry it is supposed to be –3
ii)
To be ≤ –3
(27–3x)/10 ≤ –3
27–3x ≤ –30
x ≥ 19
To be ≤ –3
(27–3x)/10 ≤ –3
27–3x ≤ –30
x ≥ 19
thanks it’s +19 right
Yep
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