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Get live Maths 1-on-1 Classs - Class 6 to 12
Last updated at March 22, 2023 by Teachoo
Ex 3.3, 1 Solve the following pair of linear equations by the substitution method. (v) β2 π₯+β3 π¦=0 β3 π₯ββ8 π¦=0 β2 π₯+β3 π¦=0 β3 π₯ββ8 π¦=0 From (1) β2 π₯+β3 π¦=0 β2 π₯=ββ3 π¦ x = (ββπ π)/βπ Substituting x in (2) β3 π₯ββ8 π¦=0 β3 ((ββ3 π¦)/β2)ββ8 π¦ = 0 (β3 Γ (ββ3 π¦))/β2ββ8 π¦=0 (β3π¦)/β2ββ8 π¦=0 β3y β β8 π¦ Γ β2= β2 Γ 0 β3y β β(8 Γ 2) π¦=0 β3y β β16 π¦=0 β3y β β(4 Γ 4) π¦=0 β3y β β42 π¦=0 β3y β 4y = 0 β7y = 0 y = 0/(β7) y = 0 Putting the value of y in (1) β2 π₯+β3 π¦=0 β2 π₯+β3 Γ 0=0 β2 π₯+0=0 β2 π₯=0 x = 0/β2 x = 0 Hence, x = 0, y = 0