, 15.05.2020rhaineandreirefuerzo

# 5x+3y+(⅓)z=100 x+y+z=100 Solve for (x, y,z)​

Any number or variable raised to 0 is always equal to 1 so

x^0 = 1
y^0 = 1
z^0 = 1

1+1+1=3

x = 20-(3/5)y-(1/15)z

y = (100/3)-(5/3)x-(1/9)z

z = 300-15x-9y

and

x = 100-y-z

y = 100-x-z

z = 100-x-y

Step-by-step explanation:

Solve for x:

Move the expression to the right:

5x+3y+(1/3)z = 100

5x = 100-3y-(1/3)z

Divide both sides by 5:

x = 20-(3/5)y-(1/15)z

Solve for y:

Move the expression to the right:

5x+3y+(1/3)z = 100

3y = 100-5x-(1/3)z

Divide both sides by 3:

y = (100/3)-(5/3)x-(1/9)z

Solve for z:

Multiply both sides by 3:

(5x+3y+(1/3)z = 100)3

15x+9y+z = 300

Move the variables to the right:

z = 300-15x-9y

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