Solving Systems of Equations

Question

In the 2019 football season, the Miami Dolphins scored a total of 295 points; the points were earned with 7-point touchdowns,3-point field goals, and 2-point safeties. They scored 56 times during the season and scored 9 more touchdowns than field goals. How many touchdowns, field goals, and safeties did they score?

Answer 1

7T + 3F + 2S = 295

T=9+F or F =T-9

T + F + S = 56, multiply by 2 and subtract from the 1st equation

7T+3F + 2S = 295

2T +2F + 2S =112

which gives

5T + F = 183, replace F by T-3 to get

5T +T - 9 = 183

6T = 192

T = 192/6 = 32 touchdowns

F=T-9 = 32-9 = 23 field goals

S = 56-32-23= 1 safety

7(32) = 224

3(23) = 69

2(1) = 2

224+69+2 = 295 points total

Answer 2

So to begin this problem, we need to set a variable for each way to score. I'm going to use:

x=touchdown

y=field goal

z=safety

They scored 56 times which means the amount of touchdowns, field goals, and safety's = 56 or x+y+z = 56

A total of 295 points was scored. A touchdown is 7 points, a field goal is 3 and a safety is 2. This results in the equation 295 = 7x+3y+2z

As well as those two equations, we know there was 9 more touchdowns than field goals which means

x = y+9

Using substitution we can rewrite x+y+z = 56 as y+9+y+z = 56 which can also be rewritten as 2y+z = 47.

With a bit of algebraic rearrangement, we can get that z = 47-2y.

By plugging in the equations x = y+9, and z = 47-2y into the equation 295 = 7(y+9)+3y+2(47-2y)

Distributing then results in 295 = 7y+63+3y+94-4y, and by combining like terms it is found that 295 = 6y+157.

Subtract 157 from both sides and find that 138 = 6y. Divide by 6 to solve for y. I found that y = 23.

Plug the y value into the equation x = y+9 to find that x is 32.

Plug in both the x and y values into x+y+z = 56 to find that z = 1.

So the answer is that the Miami Dolphins had 32 touchdowns, 23 field goals, and 1 safety in the 2019 season.