Solve {l}{28-{(m+4*2)}=15093}{n=31}{o=n}{p=o}{q=p}{r=q}{s=r}{t=s}{text{Solvefor}utext{where}}{u=t}

$$\left. \begin{array} { l } { 28 - {(m + 4 \cdot 2)} = 15093 }\\ { n = 31 }\\ { o = n }\\ { p = o }\\ { q = p }\\ { r = q }\\ { s = r }\\ { t = s }\\ { \text{Solve for } u \text{ where} } \\ { u = t } \end{array} \right.$$

Solve for m, n, o, p, q, r, s, t, u

$$u=31$$

Solution Steps

Consider the first equation. Multiply 4 and 2 to get 8.

$$28-\left(m+8\right)=15093$$

To find the opposite of m+8, find the opposite of each term.

$$28-m-8=15093$$

Subtract 8 from 28 to get 20.

$$20-m=15093$$

Subtract 20 from both sides.

$$-m=15093-20$$

Subtract 20 from 15093 to get 15073.

$$-m=15073$$

Divide both sides by -1.

$$m=-15073$$

The system is now solved.

$$m=-15073 n=31 o=31 p=31 q=31 r=31 s=31 t=31 u=31$$