Precalculus, Chapter 7, 7.3, Section 7.3, Problem 46

You may use the reduction method to solve the system, hence, you may multiply the first equation by 3, such that:
3(x + y + z + w) = 3*6
3x + 3y + 3z + 3w = 18
You may now add the equation 3x + 3y + 3z + 3w = 18 to the third equation -3x + 4y + z + 2w= 4 , such that:
3x + 3y + 3z + 3w - 3x + 4y + z + 2w= 18 + 4
7y + 4z + 5w = 22
Adding the first equation to the second yields:
3x + 4y + z = 6
Adding the second equation to the last yields:
3x + 5y - z = 0
Adding the resulted equations yields:
6x + 9y = 6 => 2x + 3y = 2
Multiply the second equation by 2 and add it to the third, such that:
x + 10y + z = 4
Add this equation to the 3x + 5y - z = 0 , such that:
3x + 5y - z + x + 10y + z = 0 + 4
4x + 15y = 4
Consider a system formed by equations 4x + 15y = 4 and 2x + 3y = 2, such that:
-2*(2x + 3y) + 4x + 15y = -4 + 4
-4x - 6y + 4x + 15y = 0
9y = 0 => y = 0
You may replace 0 for y in equation 2x + 3y = 2 , such that:
2x + 0 = 2 => x = 1
You may also replace 1 for x and 0 for y in equation 2x + 3y - w = 0 , such that:
2 - w = 0 => -w = -2 => w = 2
You may also replace 1 for x, 0 for y and 2 for w in equation x + y + z + w = 6 , such that:
1 + 0 + z + 2 = 6 => z = 6 - 3 => z = 3
Hence, evaluating the solution to the given system, yields that x =1, y = 0, z = 3, w = 2.