Solve the following systems of equations p+q+3r=4 ; 2q+3r=7 ; p-q-r=-2.

Answer:{p, q, r} = {2, 5, -1}Step-by-step explanation:p + q + 3r = 4   ....equation(i)2q + 3r = 7    ......equation(ii)p - q - r = -2     .......equation(iii)Solve equation (iii) in terms of p:p - q - r = -2p = q + r - 2  putting this equation in (i) at the place of p:(q + r - 2) + q + 3r = 4q + q + r + 3r = 4 + 22q + 4r = 6    .......(iv)solve the equation (ii) in terms of r :2q + 3r = 7 3r = 7 - 2qr = \frac{7}{3} - \frac{2q}{3}putting the value of r in equation (iv) :2q + 4(\frac{7}{3} - \frac{2q}{3}) = 6- \frac{2q}{3} = - \frac{10}{3}-2q = - 10q = \frac{10}{2}q = 5Now,r = \frac{7}{3} - \frac{2q}{3}use th value of q to find the value of r :r = \frac{7}{3} - \frac{2q}{3} * 5r = -1Then, use the value of q and r to find the value of p:p = 5 + (-1) -2 p = 2