Question

One year there was a total of 53

commercial and noncommercial orbital launches worldwide. In​ addition, the number of noncommercial orbital launches was one

more than three times

the number of commercial orbital launches. Determine the number of commercial and noncommercial orbital launches.

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

commercial launches =13

non-commericial launches=40

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.