5 Adult tickets and 10 childrens tickets cost £120, whilst 2 adult tickets & 5 childrens tickets cost £54, find the cost of an adult & a childs ticklet. SHOW YOUR WORKING.
2007-02-26 06:50:29 · 7 answers · asked by lazarus 1 in Education & Reference ➔ Homework Help
5 A + 10 C = 120
2 A + 5 C = 54
- 2(2 A + 5C) = -2 * 54
- 4A -10 C = - 108
5 A + 10 C = 120
- 4A - 10 C = - 108
___________________
A + 0 = 12
A = 12
2 * 12 + 5 C = 54
24 + 5 C = 54
5 C = 54 - 24 = 30
C = 30/5
C = 6
2007-02-26 07:08:02 · answer #1 · answered by robertonereo 4 · 0⤊ 1⤋
Let x be the cost of adult ticket
y be the cost of child ticket
5x + 10y = 120 ; or x + 2y = 24; here x = 24-2y
2x + 5y = 54; 2 (24-2y) + 5y = 54; 48-4y+5y = 54; y=54-48 = 6
hence, y= 6; then x= 24-2(6) = 12
2007-03-06 10:31:33 · answer #2 · answered by PJA 4 · 0⤊ 0⤋
Drew is right. The key to this kind of problem is to solve one unknown in terms of the other and then substitute that value into the equation.
5a+10C=120
2a+5c=54, so 2a=54-5c, so a=(54-5c)/2; Substitute this value for "a" into the first equation:
5*(54-5c)/2 + 10c=120; divide to get rid of the fraction
2.5(54-5c)+10c=120; multiply to get figures out of parentheses
135-12.5c+10c=120
135-2.5c=120
135-120=2.5c
15=2.5c
c=6, so a children's ticket costs 6 pounds
Substitute that value into first equation to get the value of the adult ticket:
5a+10(6) = 120
5a+60 = 120
5a=120-60
5a=60
60/5
a=12, so an adult ticket is 12 pounds
540-30c=240; subtract 240 from ea side & add 30c to ea side
540-240=30c
300 = 30c; divide both sides by 30, so
2007-02-26 15:13:02 · answer #3 · answered by Lillian L 5 · 0⤊ 0⤋
x= adult ticket
y= children ticket
Equations: 5x+10y=120 & 2x+5y=54
5x+10y=120
-5x
10y= -5x+120
/10 /10
y= -1/2x+12
Substistute y with -1/2x+12 in equation two
2x+5(-1/2x+12)= 54
2x+ (-5/2x)+60=54
-60 -60
-1/2x= -6
Divide each side by -1/2
x= 12
Substitute x with 12 in first equation
5(12)+10y= 120
60+10y=120
-60 -60
10y=60
Divide each side by 10
y= 6
Adult tickets cost £12 and childrens tickets cost £6.
2007-02-26 18:13:56 · answer #4 · answered by Untitled 2 · 0⤊ 0⤋
need to change it to math language...have adult = x and children = y.
5x+10y=120
2x+5y= 54
solve for either x or y and then plug the answer into the other eqaution and you are set...
2007-02-26 14:59:20 · answer #5 · answered by Anonymous · 0⤊ 1⤋
5x+10y=120
2x+5y=54
x=24-2y
2(24-2y)+5y=54
y=6
x=12
2007-02-26 14:55:53 · answer #6 · answered by Drew 2 · 1⤊ 1⤋
5A + 10C = 120...............AND 2A + 5C = 54
........-10C = ..... - 10C
........ 5A .......= 120 -10C
............/5.............../5
............A..........= (120-10C)/5
............A..........= 24 - 2C
2A + 5C = 54
2(24-C) + ....5C......=.......54
48 - 2C + 5C = 54
48 +3C.................= 54
-48............................-48
.......3C..................=6
......./3....................../3
.......C....................= 2
CHILDRENS are $2
2A + 5C = 54
2A.........+.......5(2) = ......54
2A........+ 10 ..........= 54
..............-10...............-10
2A............................44
ADULTS=22
2007-02-26 16:43:56 · answer #7 · answered by Jennifer Anne 4 · 0⤊ 1⤋