Question

Daca x+y=15 si y+z=23 , determinati valoarea numarului natural n , in urmatoarele cazuri :

a) n = x+2y+z ;
b) n = 2x+3y+z ;
c) n = 3x+5y+2z ;
d) n = 4x+7y+3z ;

VA ROG FOARTE MULT !!!!!!!!!

Answer 1

x+y=15
y+z=23

n=x+2y+z
n=15+23
n=38
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n=2x+3y+z
n=30+23
n=53
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n=3x+5y+2z
n=45+46
n=91
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n=4x+7y+3z
n=15x4+3x23
n=60+69
n=129
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Answer 2

a) x+2y+z=(x+y)+(y+z)=15+23=28
b)2x+3y+z=2(x+y)+(y+z)=2*15+23=30+23=53
c)3x+5y+2z=3(x+y)+2(y+z)=3*15+2*23=45+46=91
d)4x+7y+3z=4(x+y)+3(y+z)=4*15+3*23=60+69=129