We're given the system of equations:
1) [tex]\(2a + z = 4\)[/tex]
2) [tex]\(3a - 7z = 6\)[/tex]
We will solve this system using the method of elimination.
Step 1: Align the equations for elimination
First, we want to eliminate one of the variables. Let's aim to eliminate [tex]\(z\)[/tex]. To do this, we'll make the coefficients of [tex]\(z\)[/tex] in both equations equal (in magnitude) but opposite in sign.
The coefficients of [tex]\(z\)[/tex] in the first and second equations are 1 and -7, respectively. Let's find a common multiple of these coefficients. The least common multiple of 1 and 7 is 7.
Multiply the first equation by 7 to make the coefficient of [tex]\(z\)[/tex] match (but opposite) that of the second equation:
[tex]\[7(2a + z) = 7 \cdot 4\][/tex]
This gives us:
[tex]\[14a + 7z = 28\][/tex]
Step 2: Rewrite the system
Now we have:
1') [tex]\(14a + 7z = 28\)[/tex]
2) [tex]\(3a - 7z = 6\)[/tex]
Step 3: Add the equations
Adding these two equations will eliminate [tex]\(z\)[/tex]:
[tex]\[
(14a + 7z) + (3a - 7z) = 28 + 6
\][/tex]
Simplify:
[tex]\[14a + 3a + 7z - 7z = 28 + 6\][/tex]
This reduces to:
[tex]\[17a = 34\][/tex]
Step 4: Solve for [tex]\(a\)[/tex]
Divide both sides by 17:
[tex]\[a = \frac{34}{17} = 2\][/tex]
Step 5: Use [tex]\(a\)[/tex] to solve for [tex]\(z\)[/tex]
Now substitute [tex]\(a = 2\)[/tex] back into one of the original equations to solve for [tex]\(z\)[/tex]. We'll use the first equation:
[tex]\[2a + z = 4\][/tex]
Substitute [tex]\(a = 2\)[/tex]:
[tex]\[2(2) + z = 4\][/tex]
[tex]\[4 + z = 4\][/tex]
Subtract 4 from both sides:
[tex]\[z = 0\][/tex]
Step 6: Write the solution as an ordered pair
We've found [tex]\(a = 2\)[/tex] and [tex]\(z = 0\)[/tex]. So the solution to the system is:
[tex]\(\boxed{(2, 0)}\)[/tex]
Therefore, the correct answer is [tex]\((2, 0)\)[/tex].
