In order to have infinitely many solutions with linear equations/functions, the two equations have to be the same; In accordance, we can say: (2p + 7q)x = 4x [1] (p + 8q)y = 5y [2] 2q - p + 1 = 2 [3] All we have to do is choose two equations and solve them simultaneously (The simplest ones for what I'm doing and hence the ones I'm going to use are [3] and [2]): Rearrange in terms of p: p + 8q = 5 [2] p = 5 - 8q [2] p + 2 = 2q + 1 [3] p = 2q - 1 [3] Now equate rearranged [2] and [3] and solve for q: 5 - 8q = 2q - 1 10q = 6 q = 6/10 = 3/5 = 0.6
Now, substitute q-value into rearranges equations [2] or [3] to get p: p = 2(3/5) - 1 p = 6/5 - 1 p = 1/5 = 0.2
