how do you solve each varible
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2x+3y+5z=11 x-y+z=1 3x-4y-5z=10

This is a set of 3 simultaneous equations, which can be solved by elimination/substitution.

2x+3y+5z=11  --------------- (1)

x-y+z=1  ---------------------- (2)

3x-4y-5z=10  ---------------- (3)

Eliminate x from the three eqns by substituting x = 1 + y - z, from (2), into (1) and (3).

This gives us,

5y + 3z = 9   ---------------- (4)

y + 8z = -7  ----------------- (5)

Now eliminate y from the last two eqns by substituting y = -8z - 7, from (5), into (4).

This gives us,

33 + 37z = -11

37z = -44

z = -44/37

This also gives x = 174/37 and y = 93/37

Answer: x = 174/37, y = 93/37, z = -44/37

by Level 11 User (81.5k points)

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