Question

If x+3y+5z = 200 and x+4y+7z = 225. Then find the value of x+y+z​

Answer 1

Answer:

In mathematical terms:

x+y+z=a(x+3y+5z)+b(x+4y+7z)

x+y+z=ax+3ay+5az+bx+4by+7bz

x+y+z=(a+b)x+(3a+4b)y+(5a+7b)

a+b=1

3a+4b=1

5a+7b=1

We solve this system of 3 linear equations with 2 unknowns.

From the first:

a=1−b

Substituting in the second:

3(1−b)+4b=1

3+b=1

b=−2

a=1−b=1−(−2)=1+2=3

We check the third one: 5∗3+7∗(−2)=15–14=1

Then:

x+y+z = a*200 + b*225 = 3*200 -2*225 = 600 - 450 = 150

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