In the annual Best Infantry Unit Competition, 3 top units - Units A, B and C, are shortlisted for the eventual champion. The organizer decides to award $x$, $y$ and $z$ points for each NAPFA Gold Award, Shooting Marksmanship Badge and Army Proficiency Badge respectively and to declare the unit with the highest total points as the champion. With the emphasis of training of more marksman in the units, the organizer decides to have the points awarded for each Shooting Marksmanship Badge to be equal to the sum of points for an NAPFA Gold Award and an Army Proficiency Badge. The number of NAPFA Gold Awards, Shooting Marksmanship Badges and Army Proficiency Badges achieved by each unit are summarized in the following table.

	NAPFA Gold Awards	Shooting Marksmanship Badges	Army Proficiency Badges
Unit A	24	27	118
Unit B	29	21	124
Unit C	26	25	121

Upon computation of points, it is known that the unit with the most NAPFA Gold awards scores a total of 295 points and it falls behind the unit with the most Marksmanship Badges by 3 points. By solving a system of equations involving $x$, $y$ and $z$, determine the eventual champion of the competition.

1. Formulate an equation relating the points for each type of badge.

From the problem, the points for each Shooting Marksmanship Badge equals the sum of points for an NAPFA Gold Award and an Army Proficiency Badge:
$y = x + z$

2. Formulate an equation for the unit with the most NAPFA Gold awards.

From the table, Unit B has the most NAPFA Gold awards (29).
We are given that its total score is 295 points.
$29x + 21y + 124z = 295$

3. Formulate an equation for the unit with the most Marksmanship Badges.

From the table, Unit A has the most Marksmanship Badges (27).
Unit B falls behind Unit A by 3 points, meaning Unit A scores $295 + 3 = 298$ points.
$24x + 27y + 118z = 298$

4. Solve the system of equations for $x$, $y$, and $z$.

Substitute $y = x + z$ into the equations for Unit B and Unit A:

For Unit B:
$29x + 21(x + z) + 124z = 295$
$50x + 145z = 295 \quad \text{--- (1)}$

For Unit A:
$24x + 27(x + z) + 118z = 298$
$51x + 145z = 298 \quad \text{--- (2)}$

Subtracting (1) from (2) eliminates $z$:
$(51x - 50x) + (145z - 145z) = 298 - 295$
$x = 3$

Substitute $x = 3$ into (1):
$50(3) + 145z = 295$
$150 + 145z = 295$
$145z = 145 \implies z = 1$

Substitute $x$ and $z$ back to find $y$:
$y = 3 + 1 = 4$

5. Calculate the total score for Unit C and identify the champion.

Calculate the total points for Unit C using its tally from the table:
$\text{Score for Unit C} = 26x + 25y + 121z$
$= 26(3) + 25(4) + 121(1)$
$= 78 + 100 + 121 = 299$

Comparing the total scores:

• Unit A: 298 points
• Unit B: 295 points
• Unit C: 299 points

Since $299 > 298 > 295$, Unit C has the highest total points.