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# there is a dart board with 50 darts and he 152 points every bulls eye = 7 pts and if not a bulls eye its 2 pts how many bulls eyes did he make

The wording on the question is a little confusing, but I think I get what you're saying. Someone has thrown 50 darts at a dart board where a bullseye is worth 7 points and anything else is worth 2 points. The person has 152 points, and the question is to find out how many bullseyes were made.

I think the numbers here are a bit off, which might be why you're having trouble. I'll walk through how I would solve the problem using equations and explain.

There are 2 types of results from throwing a dart, bullseye or not. We don't know how many of either, so we'll let b=number of bullseyes and n = number of non-bullseyes.

Since there are 50 darts, we know that: b + n = 50

The total number of points is 152: bullseye points + non-bullseye points. All the bullseye points would be 7*b (7b) and non-bullseye points would be 2*n (2n). So we could also write 7b + 2n = 152

Now we have the system of equations
b + n   = 50
7b + 2n = 152

I'm going to use substitution to solve this problem. Look at the first equation: if we subtract b from both sides, we see that n = 50-b. Substitute (50-b) in place of n in the 2nd equation.

7b + 2(50-b) = 152
7b + 100 - 2b = 152 (distribute)
5b +100 = 152 (combine like terms)
5b = 52 (subtract 100 from both sides)
b = 10.4 (divide both sides by 5)

Does this answer make sense? Can you have 10.4 bullseyes? Nope! Now, if there were 51 darts thrown instead, everything works out rather nicely.

Let x=# bullseyes

y=# 2-point shots

z=# missed shots

Solve so that both of the following equations are TRUE

#1 x + y + z = 50               where x, y, z are positive integers

#2 7x + 2y + 0z = 152        where x, y, z are positive integers

What is needed are all points (x,y,z) that meet the above criteria.

Maximum values:

x:     7x <= 152                 where x is a positive EVEN number

x <= 21                  why? because of the property of integers; an odd + even integer = odd

And odd x odd = odd, but 7x must be even in order for 7x+2y+0z = 152.

y:      2y <= 152                 where y is a positive number

y <= 76                   but there’s an upward limit of 50!

y <= 50                   because of #1 equation x + y + z = 50

Now let’s remove z as a variable from #1 and #2 above.

#1a    x + y <= 50

#2a  7x + 2y = 152 (subtract 1a from 2a)

------------------

6x + y <= 102

y <= -6x + 102

Substitute into #1a

x + -6x + 102 <= 50

52 <= 5x

10.2 <= x <= 21    where x is a positive EVEN number (remember max above?)

valid values of x are {12, 14, 16, 18, 20}

Solve #2a for y.

7x + 2y = 152

2y = 152 + -7x

y = 76 + 7x/2

now find y as f(x) = -76 + 7x/2 for all valid values of x

f(12) = -76 + 42 = 34

f(14) = 27

f(16) = 20

f(18) = 13

f(20) = 6

valid values of y are {34, 27, 20, 13, 6}

Now find z as f(x,y) = 50 – x – y (#1 solved for z)

f(12,34) = 4

f(14,27) = 9

f(16,20) = 14

f(18,13) = 19

f(20,6) = 24

(12,34,4)       12 bullseyes, 34 2-pointers, 4 misses

(14,27,9)       14 bullseyes, 27 2-pointers, 9 misses

(16,20,14)     16 bullseyes, 20 2-pointers, 14 misses

(18,13,19)     18 bullseyes, 13 2-pointers, 19 misses

(20,6,24)       20 bullseyes, 6 2-pointers, 24 misses

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