In Triangle ABC. 2A-2B+3C=100 , A+5B-2C=360.Find the angles.. Please, Its Urgent,,PLSPLS

We know that the sum of angles of triangle is 180 degree.
$$\begin{gathered} A+B+C = 180 ...\left(\mathrm{i}\right) \\ 2\mathrm{A}-2\mathrm{B}+3\mathrm{C} = 100 ...\left(\mathrm{ii}\right) \\ \mathrm{A}+5\mathrm{B}-2\mathrm{C} = 360 ...\left(\mathrm{iii}\right) \end{gathered}$$
Subtracting (iii) from (i) we get;
$$\begin{gathered} A+B+C-A-5B+2C = 180-360 \\ \Rightarrow -4B+3C = -180 \\ \Rightarrow 4B-3C = 180 ...\left(\mathrm{iv}\right) \end{gathered}$$
Multiplying (i) by 2 and then subtracting (ii) from (i) we get;
$$\begin{gathered} 2A+2B+2C-2A+2B-3C = 360-100 \\ \Rightarrow 4B-C = 260 ...\left(\mathrm{v}\right) \end{gathered}$$
Now subtracting (v) from (iv) we get;
$$\begin{gathered} 4B-3C-4B+C = 180-260 \\ \Rightarrow -2C = -80 \\ \Rightarrow C = 40 \end{gathered}$$
putting the value of C in (v) we get;
$$\begin{gathered} 4B-40 = 260 \\ \Rightarrow 4B = 300 \\ \Rightarrow B = 75 \end{gathered}$$
So from (i) we have;
$A = 180-75-40 = 180-115 = 65$
Therefore $\angle \mathrm{A} = 65°; \angle \mathrm{B} = 75° \mathrm{and} \angle \mathrm{C} = 40°$.