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Valentin [98]
3 years ago
5

One year there was a total of 53

Mathematics
2 answers:
mojhsa [17]3 years ago
4 0
Hmm let's say 
c = commercial launches
n = non-commercial launches

we know the total of launches were 53 for that year, so
c + n = 53

we also know that, whatever "c" is, "n" is "<span>was one more than three times" that

so... three times "c", 3 * c or 3c
one more than that?
3c + 1

so.. whatever "c" is, n = 3c +  1

</span>\bf \begin{cases}&#10;c+n=53&#10;\\\\&#10;\boxed{n}=3c+1\\&#10;--------------\\&#10;c+n=53\implies c+\boxed{3c+1}=53&#10;\end{cases}
<span>
solve for "c", to see how many commercial launches were there

what about "n"?  well, n = 3c + 1</span>
hammer [34]3 years ago
3 0

Answer:

commercial launches =13

non-commericial launches=40

Step-by-step explanation:

If the total amount of commerical and non-commerical orbital launches were 53 we can write expression. Let n be the number of commercial and m be the number of non-commericial launches:

53=m+n (1)

If m is one more than 3 times more than n we can write an expression relating the two:

m=1+3*n  (2)

Therefore we can substitue equation 2 into 1 and solve for n:

53=(1+3*n)+n

n=52/4=13

Therefore can determine m:

m=1+3*(13)=40

The number of commerical launches were 13  and number of non-commerical launchers were 40.  

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