Answer :
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Explanation:-
Given equations:
1. \( x - y = 4 \)
2. \( 2y + 3z = -2 \)
3. \( 3x + y + 2z = 1 \)
**Step 1: Solve Equation 1 for \( x \)**
From equation 1, rearrange to solve for \( x \):
\[ x = y + 4 \]
**Step 2: Substitute \( x \) from Step 1 into Equation 3**
Substitute \( x = y + 4 \) into equation 3:
\[ 3(y + 4) + y + 2z = 1 \]
\[ 3y + 12 + y + 2z = 1 \]
\[ 4y + 12 + 2z = 1 \]
\[ 4y + 2z = -11 \]
Answer:
x = 3/8
y = –29/4
z = 7/4
Solution:
x–y = 4
2y+3z = –2
3x+y+2z = 1
Add (1) and (2)
x+y+3z = 2
Solve (3) and (4)
2x–z = –1
Solve (1) and (2)
2x+3z = 6
Solve (5) and (6)
4z = 7
8x–7 = –4
8x = 7–4
8x = 3
Sub x in (1)
y = x–4
y = (3/8)–4
y = (3–32)/4
y = –29/4
Publisher: MSK™