−1[a−3{b−4(a−¯¯¯¯¯¯¯¯¯¯¯¯b−8)+4a}+10] equals
−a+15b+106
−a+15b−106
−a−15b−106
a+15b−106
−1[a−3{b−4(a−¯¯¯¯¯¯¯¯¯¯¯¯b−8)+4a}+10]
= −1[a−3{b−4(a−b+8)+4a}+10]
= −1[a−3{b−4a+4b−32+4a}+10]
= −1[a−3{5b−32}+10]
= −1[a−15b+96+10]
= −1[a−15b+106]
= −a+15b−106
Solve the following
(a) p+4=15
(b) m−9=4
(c) x+2x+1=4x+6
(d) 9a−3a−4a−6=0
−1[a−3{b−4(a−¯¯¯¯¯¯¯¯¯¯¯¯b−8)+4a}+10] equals
Evaluate: (iii) (8a+15b)−(3b−7a)
P(5, r) = 2P(6,r - 1) R = ?