Find the values of p and q for which the following system of equations has infinitely many solutions:
2x + 3y = 7
(p + q + 1)x + (p + 2q + 2)y = 4(p + q) + 1

p = 3

q = 2

Step-by-step explanation:

they have infinitely many solutions, if they are basically the same equations. in other words, when they can be combined/reduced into a statement of tautology, a statement that is always true, no matter what.

like 15 = 15 or 8x = 8x

now we need to aim for creating the same (or multiples of the same) factors in the original equation 2 as in the original equation 1.

in other words we find the p and q, where we get the same ratios between the factors of squadron 1 and equation 2.

2/(p+q+1) = 3/(p+2q+2) = 7/(4p +4q +1)

p+q+1 = 2/3 × (p+2q+2) = 2/7 × (4p + 4q + 1)

3p + 3q + 3 = 2p + 4q + 4 = 6/7 × (4p+4q+1)

21p + 21q + 21 = 14p + 28q + 28 = 24p + 24q + 6

7p - 7q = 7

p - q = 1

p = 1 + q

14(1+q) + 28q + 28 = 24(1+q) + 24q + 6

14 + 14q + 28q + 28 = 24 + 24q + 24q + 6

42 + 42q = 30 + 48q

12 = 6q

q = 2

p = 1 + 2 = 3

step-by-step explanation: you take each of the numbers and divide by the amount then you take the range

step-by-step explanation:

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