Answer:
x = 5, y = -6, z = -2
Step-by-step explanation:
Given the following systems of linear equations:
Equation 1: x = y + 11
Equation 2: x + 2y - 4z = 1
Equation 3: 2x = 10
In order to use the substitution method, choose one of the given equation where the variable has a coefficient of 1. We could use Equation 1 and substitute its value into Equation 3:
Step 1:
Use Equation 1: x = y + 11 substitute into Equation 3 to solve for y:
2x = 10
2(y + 11) = 10
Distribute 2 into the parenthesis:
2y + 22 = 10
2y + 22 - 22 = 10 - 22
2y = -12
[tex]\frac{2y}{2} = \frac{-12}{2}[/tex]
y = -6
Step 2:
Substitute the value of y = -6 into Equation 1:
x = y + 11
x = -6 + 11
x = 5
Step 3:
Substitute the values of x = 5 and y = -6 into Equation 2 to solve for z:
x + 2y - 4z = 1
5 + 2(-6) - 4z = 1
5 - 12 - 4z = 1
-7 - 4z = 1
Add 7 to both sides:
-7 + 7 - 4z = 1 + 7
-4z = 8
Divide both sides by -4 to solve for z:
[tex]\frac{-4z}{-4} = \frac{8}{-4}[/tex]
z = -2
Verify whether the values for x, y, and z satisfy the three equations in the given linear system:
x = 5, y = -6, z = -2
Equation 1: x = y + 11
5 = -6 + 11
5 = 5 (True statement)
Equation 2: x + 2y - 4z = 1
5 + 2(-6) - 4(-2) = 1
5 - 12 + 8 = 1
13 - 12 = 1
1 = 1 (True statement)
Equation 3: 2x = 10
2(5) = 10
10 = 10 (True statement).
Therefore, x = 5, y = -6, z = -2 satisfy all the equations within the given system.
