Question

z-3 = 2z - 5
Solve by systematic method of algebra. ​

Answer 1

please mark as brainliest

Answer 2

Answer:

z = 2

Explanation:

z−3=2z−5

z−3=2z−5At z=0, LHS=0−3=−3, RHS=0−5=−5

z−3=2z−5At z=0, LHS=0−3=−3, RHS=0−5=−5LHS



=RHS

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS=RHS

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS=RHSAt z=2, LHS=2−3=−1, RHS=2(2)−5=−1

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS=RHSAt z=2, LHS=2−3=−1, RHS=2(2)−5=−1LHS=RHS

=RHSAt z=1, LHS=1−3=−2, RHS=2(1)−5=−3LHS=RHSAt z=2, LHS=2−3=−1, RHS=2(2)−5=−1LHS=RHS∴z=2 is the root