Sandy had some money. If she bought 5 bangles and 7 rings, she would have $3 left. If she bought 7 bangles and 5 rings, she would need $5 more. If each bangle cost $8, how much did each ring cost?
+2498 b-bangles
r-rings
Sandy had some money:
some money is x
If she bought 5 bangles and 7 rings, she would have $3 left.
x-5b-7r=3
If she bought 7 bangles and 5 rings, she would need $5 more.
x-7b-5r=-5
So we have:
x-5b-7r=3
x-7b-5r=-5
let s minus it from each other
x-5b-7r-(x-7b-5r)=3-(-5)
2b-2r=8
If each bangle cost $8:
16-2r=8
r=4
Answer: 4
+2498 b-bangles
r-rings
Sandy had some money:
some money is x
If she bought 5 bangles and 7 rings, she would have $3 left.
x-5b-7r=3
If she bought 7 bangles and 5 rings, she would need $5 more.
x-7b-5r=-5
So we have:
x-5b-7r=3
x-7b-5r=-5
let s minus it from each other
x-5b-7r-(x-7b-5r)=3-(-5)
2b-2r=8
If each bangle cost $8:
16-2r=8
r=4
Answer: 4
+129850 Sandy had some money. If she bought 5 bangles and 7 rings, she would have $3 left. If she bought 7 bangles and 5 rings, she would need $5 more. If each bangle cost $8, how much did each ring cost?
Call the amount of money she had, x...and call the price of each ring, r......and we have that
5(8) + 7r = x - 3 → 43 + 7r = x
7(8) + 5r = x + 5 → 51 + 5r = x
Setting both equations equal, we have
43 + 7r = 51 + 5r subtract 43, 5r from both sides
2r = 8 divide both sides by 2
r = $4
