Ax+by=12
2x+8y=60
in the system of equations above, a and b are constants. if the system has infinitely many solutions, what is the value of a/b?

If the system of equations has infinitely many solutions, we know they must coincide each other. If two equations intersect at a point, we know they have a single solution at the intersection. If they are parallel with different vertical shifts, we know they have no solutions for their intersection. But if they are the same lines, they must have infinitely many solutions (or intersections)
We can write these two equations as a set of linear equations



If we divide Equation 2 by 12 (the right hand side of equation 1), we find the solution is 5. Therefore, we need to divide equation 2 by 5 to get the values in relation to equation 1.




This is in the same form as the first equation. This means we can see the values for A and B will be 2/5 and 8/5 respectively.

Now all we need to do is divide A/B to find the solution to your problem.


[tex]\frac{2}{8}=\frac{1}{4}[\tex]

The answer is 49/16 or 3 1/16

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt