Question 98(iii)
Solve the following:
49×z−37−3×10×z−5(z≠0)
49×z−37−3×10×z−6=(7)2×z−37−3×10z−5=(7)2+3×z−3+510=(7)5z210=7510z2
Solve the inequation:
3z−5≤z+3<5z−9; z∈R.
49x2yz27y2z3=?