Question

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

Answer 1

{p, q, r} = {2, 5, -1}

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 = -2

p = q + r - 2

putting this equation in (i) at the place of p:

(q + r - 2) + q + 3r = 4

q + q + r + 3r = 4 + 2

2q + 4r = 6 .......(iv)

solve the equation (ii) in terms of r :

2q + 3r = 7

3r = 7 - 2q

r = \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 = - 10

q = \frac{10}{2}

q = 5

Now,

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} * 5

r = -1

Then, use the value of q and r to find the value of p:

p = 5 + (-1) -2

p = 2