Given - x + 3Y = 5, find the value of 5 (x-3y) &# - 8 (x-3y) - 5 | Math enthusiast | Mathematical art

Given - x + 3Y = 5, find the value of 5 (x-3y) &# - 8 (x-3y) - 5

Because - x + 3Y = 5, so x = 3y-5
5(3y-5-3y)2—8(3y-5-3y)-5
=125+40-5
=160

|X-3 | + | 2y-8 | + | Z-2 | = 0 to find the value of x-3y-z

|x-3|+|2y-8|+|z-2|=0
Then x-3 = 0, 2y-8 = 0, Z-2 = 0
x=3,y=4,z=2
x-3y-z
=3-3×4-2
=3-12-2
=-11

Given (x ^ 2 + 3Y ^ 2) ^ 2-2 (x ^ 2 + 3Y ^ 2) - 8 = 0, find the value of x ^ 2 + 3Y ^ 2

(x^2+3y^2)^2-2(x^2+3y^2)-8=0
Let a = x ^ 2 + 3Y ^ 2
a^2-2a-8=0
(a-4)(a+2)=0
a=4,a=-2
Because x ^ 2 + 3Y ^ 2 〉 = 0
So a = - 2 is left out
x^2+3y^2=4

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