Astore sells small notebooks for $6 and large notebooks for $12. if a student buys 6 notebooks and spends $60, how many of each size did he buy? he bought __ nothing small notebooks and __ nothing large notebooks.

The student was able to buy 2 small notebooks and 4 large notebooks for a total of $60.

Assume that the small notebooks are x and the large notebooks are y.

The first expression would be:

Number of small notebooks bought + Number of large bought = 6

x + y = 6

The small books cost $6 and the large ones cost $12. A total of $60 was spent on both. As an expression, it will be:

(Number of small books x price of small books) + (Number of large books x price of large books) = Total cost

6x + 12y = 60

Express x in terms of y to solve:

x + y = 6

x = 6 - y

Solve the equation to get y:

6x + 12y = 60

(6 x (6 - y)) + 12y = 60

36 - 6y + 12y = 60

6y = 60 - 36

y = 24 / 6

y = 4 large notebooks

Number of small notebooks is:

= 6 books bought - 4 large books

= 2 small notebooks

In conclusion, the student bought 2 small notebooks and 4 large notebooks

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  5.   y=6; de=7; fg=8

  6.   x=4.2; jl=14.2; mn=13

step-by-step explanation:

when chords cross inside a circle, the product of the segment lengths is a constant.

5. dh×eh = fh×gh

  y = gh = dh×eh/fh = 3×4/2 . . . divide by the coefficient of gh

  y = 6

of course, the total chord length is the sum of the lengths of its segments.

  de = dh +eh = 3+4 = 7

  fg = fh +gh = 2+6 = 8

6. this problem works the same way as the previous one.

  x = lk = mk×nk/jk = 6×7/10 = 4.2

  jl = jk +lk = 10 +4.2 = 14.2

  mn = mk +nk = 6 +7 = 13

comment on chords and secants

you can think of the points h (problem 5) and k (problem 6) as points where the chords meet. a chord is part of a secant line, a line that intersects the circle in two points. one can also consider these points (h or k) to be the points where the respective secant lines meet.

then, the product that is a constant is the product of the distance from the meeting point (h or k) to one circle intersection with the distance from that meeting point to the other circle intersection. for problem 5, the constant is the product hd×he = hf×hg.

it turns out that this rule regarding the products of lengths to points of intersection is also true when the secants intersect outside the circle.

that is, you only need to remember one rule to work both kinds of problems.