-2
If 2x+y= 15, 2y+z= 25 and 2z+x =26 ,what is value of z?
- A4
- B7
- C9
- D12
Answer not set
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0
CCommunity
Solution
Equation (i) : $2x+y=15$
Equation (ii) : $2y+z=25$
Equation (iii) : $2z+x=26$
Adding the three equations, => $3x+3y+3z=15+25+26$
=> $3(x+y+z)=66$
=> $x+y+z=\frac{66}{3}=22$ ------------(iv)
Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$
=> $z-x=7$ -------------(v)
Adding equations (v) and (iii), => $3z=26+7=33$
=> $z=\frac{33}{3}=11$
=> Ans - (E)
Equation (i) : $2x+y=15$
Equation (ii) : $2y+z=25$
Equation (iii) : $2z+x=26$
Adding the three equations, => $3x+3y+3z=15+25+26$
=> $3(x+y+z)=66$
=> $x+y+z=\frac{66}{3}=22$ ------------(iv)
Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$
=> $z-x=7$ -------------(v)
Adding equations (v) and (iii), => $3z=26+7=33$
=> $z=\frac{33}{3}=11$
=> Ans - (E)
0
CCommunity
Solution
Equation (i) : $2x+y=15$
Equation (ii) : $2y+z=25$
Equation (iii) : $2z+x=26$
Adding the three equations, => $3x+3y+3z=15+25+26$
=> $3(x+y+z)=66$
=> $x+y+z=\frac{66}{3}=22$ ------------(iv)
Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$
=> $z-x=7$ -------------(v)
Adding equations (v) and (iii), => $3z=26+7=33$
=> $z=\frac{33}{3}=11$
=> Ans - (E)
Equation (i) : $2x+y=15$
Equation (ii) : $2y+z=25$
Equation (iii) : $2z+x=26$
Adding the three equations, => $3x+3y+3z=15+25+26$
=> $3(x+y+z)=66$
=> $x+y+z=\frac{66}{3}=22$ ------------(iv)
Subtracting equation (i) from (iv), => $(z)+(x-2x)+(y-y)=(22-15)$
=> $z-x=7$ -------------(v)
Adding equations (v) and (iii), => $3z=26+7=33$
=> $z=\frac{33}{3}=11$
=> Ans - (E)