# Use substitution to solve 2x +y + z = 4, 2x - y - z = 1 and 2x - y + z = 3

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2x +y + z = 4 --- Equation 1

2x - y - z = 1 -- Equation 2

2x - y + z = 3 -- Equation 3

Adding 2 and 3, we get

4x - 2y = 4

=> 2y = 4x - 4 => y = 2x-2

substituting y = 2x-2 in equations 1 and 2, we get

2x + 2x - 2 + z = 4 => 4x + z = 6 -- Equation 4

2x -2x + 2 - z = 1 => z = 1

Substituting z = 1 in Equation 4, we get

4x + 1 = 6 => x = 5/4 = 1.25

and y = 2x-2 => y = 2*1.25 - 2 = 2.5 - 2 = 0.5

So the values are:

x = 1.25; y = 0.5; z = 1

The set of equations 2x +y + z = 4, 2x - y - z = 1 and 2x - y + z = 3 has to be solved for x, y and z.

Isolate y from 2x +y + z = 4, y = 4 - 2x - z

Substitute y = 4 - 2x - z in 2x - y - z = 1

=> 2x - 4 + 2x + z - z = 1

=> 4x = 5

=> x = 5/4

y = 4 - 2*(5/4) - z

=> y = 1.5 - z

Substitute x = 5/4 and y = 1.5 - z in 2x - y + z = 3

=> 2.5 - 1.5 + z + z = 3

=> 2z = 2

=> z = 1

y = 1.5 - z = 1.5 - 1 = 0.5

**The solution of the given set of equations is x = 1.25, y = 0.5 and z = 1**