Question

One year there was a total of 53

Answer 1

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 "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

$\bf \begin{cases} c+n=53 \\\\ \boxed{n}=3c+1\\ --------------\\ c+n=53\implies c+\boxed{3c+1}=53 \end{cases}$

solve for "c", to see how many commercial launches were there

what about "n"?  well, n = 3c + 1

Answer 2

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.