As you say, for the first equation either x = 0 or (2x+1) = 0. If x is 0, the second equation becomes a value other than 0 so solve for 2x-1 = 0 => x = -1/2.
Use that value for x in the second equation and the first part (x+ 1/2) becomes 0; hence both equations are satisfied for a unique value of x.
(You can do the same thing backwards, since for the second part of the second equation, x = - 3/2; using that value in the first equation makes it a non zero value.)
Quicker way of doing Q98:
As you say, for the first equation either x = 0 or (2x+1) = 0. If x is 0, the second equation becomes a value other than 0 so solve for 2x-1 = 0 => x = -1/2.
Use that value for x in the second equation and the first part (x+ 1/2) becomes 0; hence both equations are satisfied for a unique value of x.
(You can do the same thing backwards, since for the second part of the second equation, x = - 3/2; using that value in the first equation makes it a non zero value.)
polkadottedjupiter 1 month ago
quicker way of doing Q97:
Since 2+16=18 we know that
144 = 18x thus 8 = x
8*16= 128
rsiddi01 1 year ago