Question

A. Solve by eliminations :
1. 2x – y + z = 3
x + 3y – 2z = 11
3x – 2y + 4z = 1
3.
x = y – 2z
2y = x + 3z + 1
z = 2y – 2x - 3​

Answer 1

Answer:

Given,

x−2y+3z=11−−−−−(1)

3x+y−z=2−−−−−−(2)

5x+3y+2z=3−−−−−(3)

Multiplying (1) with 3 and then solving with (2),we get,

10z−7y=31−−−−−(4)

Multiplying (1) with 5 and then solving with (3),we get,

z−y=4−−−−−−(5)

Now,multiplying (5) with 7 and then solving with (4),we get,

=>z=1

Putting the value of z in (5),

y=−3

putting the value of z and y in (1),

x=2

Step-by-step explanation:

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