Suppose that Karen ordered 3 burgers and 5 sodas fo a total of 1841 calories. Jake ordered 5 burgers and 2 sodas for total 1770 calories. How many calories r in the burger??? How many are in the sodas?
2007-09-30 15:46:24 · 5 answers · asked by TriniBeauty. 3 in Science & Mathematics ➔ Mathematics
3b + 5s = 1841 eq 1
5b + 2s = 1770 eq 2
times eq 1 by 2 and eq 2 by 5
6b + 10s = 3682
25 b + 10s = 8850
Subtract 2 from 1
- 19b = - 5168
b = -5168/ -19
b = 272 calories
Substitute this value in the original equation
3b + 5s = 1841
3 (272) + 5s = 1841
816 + 5s = 1841
5s = 1025
s =1025/5
s = 205 calories
2007-09-30 15:55:51 · answer #1 · answered by Monah 3 · 0⤊ 0⤋
x = no. of calories in a burger
y = no. of calories in soda
3x + 5y = 1841 equation (1)
5x + 2y = 1770 equation (2)
Solve for y using equation (2)
5x + 2y = 1770
2y = 1770 - 5x
y = 1770 - 5x / 2 equation (3)
Substitute y in equation (1)
3x + 5y = 1841
3x + 5 (1770 - 5x/ 2) = 1841
3x + (8850 - 25x / 2) = 1841
Multiply both sides of the equation by (2)
6x + 8850 - 25x = 3682
-19x = -5168
x = 272 calories in the burger
Substitute the value of x in equation (3)
y = 1770 - 5x / 2
y = 1770 - 5(272) / 2
y = 1770 - 1360 / 2
y = 410 / 2
y = 205 calories in the soda
To check, substitute the value of x and y in equations (1 & 2)
3x + 5y = 1841 equation (1)
3 (272) + 5 (205) = 1841
816 + 1025 = 1841
5x + 2y = 1770 equation (2)
5(272) + 2(205) = 1770
1360 + 410 = 1770
2007-09-30 23:08:23 · answer #2 · answered by edith p 3 · 0⤊ 0⤋
Call x the calories in a burger and y the calories in a soda.
Then the problem tells us that
1) 3x + 5y = 1841 and that
2) 5x + 2y = 1770
Now we have a pair of 'simultaneous' equations that can be solved as follows:
multiply equation (1) by 2 and equation (2) by 5 to get
(3) 6x + 10y = 3682
(4) 25x + 10y = 8850
Now, subtract equation (3) from equation (4) to get
19x = 5168 and so
x = 272 calories in a burger. By now substituting 272 for x in equation (1) we get
6*272 + 10y = 3682
1632 + 10y = 3682
10y = 2050 and y = 205 calories in a soda.
You can check these by substituting them back into the original equations.
Doug
2007-09-30 23:02:05 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
x will = burgers and y will = sodas
3X + 5Y = 1841
5X + 2Y = 1770
(multiply the top by -2 and bottom by 5 to cancel out the y's)
-2(3X + 5Y = 1841) = -6X - 10Y = -3682
5(5X + 2Y = 1770) = 25X + 10Y = 8850
now add both equations together according to variables
-6X + 25X = 19X -10Y + 10Y = 0 -3682 + 8850= 5168
now form the equation
19X = 5168
divide by 19
X= 272
then go back and substitute X=272 back in the equation
so, 3(272) + 5Y = 1841
816 + 5Y = 1841
-816 -816
5Y = 1025
divide by 5
Y= 205
Therefore, the burger has 272 calories, and the soda has 205 calories
2007-09-30 22:59:00 · answer #4 · answered by angelgirl88 2 · 0⤊ 0⤋
3b+5s=1841
5b+2s=1770
multiply eqn 1 by 5 and eqn 2 by 3 so you get..
15b+25s=9205 and 15b+6s=5310
Then subtract the secon eqn from the first one:
15b+25s=9205
- 15b+6s=5310
and you get 0b+19s=3895 or 19s=3895
That means s=205
Since you have you substitue s=205 into one of the original equations and then solve for b.
2007-09-30 22:52:17 · answer #5 · answered by Anonymous · 0⤊ 0⤋