I believe it is meant to read 999 not qqq
Oh. Blame my eyesight. Then we have enough clues and don't need any guesses.
[MATH]n = \dfrac{7}{3} * p.[/MATH]
[MATH]m = \dfrac{1}{2} * n.[/MATH]
[MATH]m + n + p = t.[/MATH]
[MATH]t = 999.[/MATH]
Four variables, four facts. A general way to solve such problems is called substitution.
We can substitute from the fourth equation into the third giving us
[MATH]m + n + p = 999.[/MATH]
We can substitute from the first equation into the one immediately above.
[MATH]m + n + \dfrac{7}{3} * n = 999 \implies m + \dfrac{3}{3} * n + \dfrac{7}{3} * n = 999 = m + \dfrac{10}{3} * n = 999.[/MATH]
And we can substitute from the second equation into the one immediately above.
[MATH]\dfrac{1}{2} * n + \dfrac{10}{3} * n = 999.[/MATH]
Let's eliminate the fractions by multiplying both sides of the equation by 6.
[MATH]3n + 20n = 5994 \implies 23n = 5994.[/MATH]
Now divide both sides of that equation by 23 to get
[MATH]n = 260 + \dfrac{14}{23}.[/MATH]
[MATH]\therefore m = \dfrac{1}{2} * n = 130 + \dfrac{7}{23}.[/MATH]
[MATH]\text {And }p = \dfrac{7}{3} * n = \dfrac{7}{3} * (260 - 2) + \dfrac{7}{3} * 2 + \dfrac{7}{3} * \dfrac{14}{23} =[/MATH]
[MATH]7 * 86 + \dfrac{7}{3} * \dfrac{46}{23} + \dfrac{7}{3} * \dfrac{14}{3} = 602 + \dfrac{7}{3} * {60}{23} =[/MATH]
[MATH]602 + 7 * \dfrac{20}{ 23} = 602 + \dfrac{140}{23} = 602 + 6 + \dfrac{2}{23} = 608 + \dfrac{2}{23}.[/MATH]
That is a very silly answer in practical terms so let's check.
[MATH]130 + \dfrac{7}{23} + 260 + \dfrac{14}{23} + 608 + \dfrac{2}{23} = 998 + \dfrac{7 + 14 + 2}{23} = 999 \ \checkmark.[/MATH]
Name your variables. Write down your numerical facts in mathematical terms. Use substitution and simplify where you can. Check your work. That is algebra.