# `2x + 2z = 2, 5x + 3y = 4, 3y - 4z = 4` Solve the system of linear equations and check any solutions algebraically.

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You may use the substitution method to solve the system, hence, you need to use the first equation to write x in terms of z, such that:

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

You may now replace 1 - z for x in equation `5x + 3y = 4,` such that:

`5(1 - z) + 3y = 4 =>5 - 5z + 3y = 4 => - 5z + 3y = -1`

You may use the third equation, `3y - 4z = 4` , along with `-5z + 3y = -1 ` equation, such that:

`3y = 4 + 4z`

Replace 4 + 4z for 3y in equation -5z + 3y = -1, such that:

`-5z + 4 + 4z = -1 => -z = -1 - 4 => -z = -5 => z = 5`

You may replace 5 for z in equation `3y = 4 + 4z:`

`3y = 4 + 4*5 => 3y = 24 => y = 8`

You may replace 5 for z in equation `x = 1 - z:`

`x = 1 - 5 => x = -4`

**Hence, evaluating the solution to the given system, yields **`x = -4, y = 8, z = 5.`