Answer:
The answer is E
Step-by-step explanation:
Before we find the value of 5, let's find x by using the elimination process for the two given equation.
2(3x + y + z = 14) → 6x + 2y + 2z = 28
3x + 2y + 2z = 19
-6x + 2y + 2z = 28
-3x = -9
x = 3
Now plug in 3 into both of the given equations, which will give you your new equation, then subtract them:
3(3) + 2y + 2z = 19 → 9 + 2y + 2z = 19
3(3) + y + z = 14 → - 9 + y + z = 14
Our answer is: y + z = 5
Our value is 5
Optional: If you want to make sure it's 5, plug-in any random numbers for y and z that equals 5. For example: 3 + 2, 1 + 4, 5 + 0, 6 + (-1), etc...
Once you choose the numbers you can plug them into the two new given equations. For me, I decided to let y = 1 and z = 4:
9 + 2(1) + 2(4) = 19 9 + (1) + (4) = 14
9 + 2 + 8 = 19 9 + 1 + 4 = 14
19 = 19 14 = 14