
Get live Maths 1-on-1 Classs - Class 6 to 12
Last updated at March 23, 2023 by Teachoo
Ex 2.5, 2 Solve the following linear equations. π/2β3π/4+5π/6=21 π/2β3π/4+5π/6=21 (6π β 3(3π) + 2(5π))/12=21 (6π β 9π + 10π)/12=21 (6π + π)/12=21 7π/12=21 7n = 21 Γ 12 n = (21 Γ 12)/7 n = 36 L.H.S π/2β 3π/4+5π/6 =36/2 β(3 Γ 36)/4+(5 Γ 36)/6 =18β27+30 =21 = R.H.S β΄ LHS = RHS Hence Verified.