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 Ex 34, 1 (Elimination)Solve the following pair of linear equations by the elimination method and the substitution method (i) x y = 5 and 2x – 3y = 4 x y = 5 2x – 3y = 4Find an answer to your question 2xxy/6=2 x2xy/3=1 by elimination method plz help fast rinkusingh rinkusingh English Secondary School answered 2xxy/6=2Step1 The first step is to multiply or divide both the linear equations with a nonzero number to get a common coefficient of any one of the variables in both equations Step2 Add or subtract Maths Guru Please Help Me Solve This Simultaneous Equations For My Young Photo Education 2 Nigeria How to solve two equations using elimination

 Factorise (x^2y^2z^2)^2 4x^2y^2 2 See answers please give me answer Advertisement AdvertisementX 22 x y y 2z 2 = xy 2z 2 ∵ a 22 a b b 2 = (ab) 2 = (xyz) (xy z) ∵ a 2b 2 = ab a b Hence, the factors of x 22 x y y 2z 2 are xyz and xy zMathf(y) = x(y^2 z^2) y(z^2 x^2) z(y^2 x^2) = (xz)y^2 (z^2 x^2)y xz(xz)/math by letting mathf(y) = 0 /math and solving for its two roots mathr_1/math , mathr_2/math 2nd Year Maths Factors Factoring By Grouping How To Factorise 2xy Xz 2y Z Fully R Homeworkhelp Solve (x^2-y^2-z^2)p+2xyq=2xz