Each year a school has a concert of readings and songs.
In 1999 the concert has 3 readings and 9 songs.
It lasted 120 minutes.
In 2000 the concert has 5 readings and 5 songs.
It lasted 90 minutes.
In 2001 the school plans to have 5 readings and 7 songs.
Use simultaneous equations to estimate how long the concert is.
Call the time estimated for a reading X minutes and the estimated time for each song Y minutes.
Estimated time =...... Minutes
TY for anyhelp you can give
2007-03-18 06:07:04 · 7 answers · asked by ratrace667 1 in Science & Mathematics ➔ Mathematics
You have two equations:
3X + 9Y = 120
5X + 5Y = 90
I like to "clean up" equations whenever possible, so I notice the first equation can be divided by 3 and the second by 5. This makes the numbers smaller and since mathematicians are notoriously lazy, this is much better. Soooooo.....
X + 3Y = 40
X + Y = 18
Subtract the 2nd eq from the 1st to yield: 2Y = 22, or Y=11. Plug Y=11 back into one of your equations to find that X = 7.
Therefore, in 2001, the total number of minutes will be 5X + 7Y = 5*7 + 7*11 = 35 + 77 = 112 minutes.
2007-03-25 15:10:59 · answer #1 · answered by Kathleen K 7 · 0⤊ 0⤋
Let Xbe the time of the readings in minutes and Ybe the time of the songs in minutes
In 1999, we have
3X + 9Y= 120 (eq 1)
In 2000, we have
5X+ 5Y=90
This simplifies to
X+Y =18 (eq 2)
Multiply eq 2 by3
3X+3Y= 54 (eq 3)
Subtrac equation 3 from equation 1
3X- 3X+ 9Y-3Y =120 -54
6Y= 66
Y =66 /6 = 11 minutes
Putting Yinto equation 2
11 + X= 18
X= 18- 11 =7
In 2001
5X + 7Y= (5* 7) + (7*11) = 35 +77 = 112 minutes
2007-03-18 13:21:15 · answer #2 · answered by The exclamation mark 6 · 0⤊ 0⤋
3x+9y = 120
5x+5y = 90
15x+45y = 600
15x+15y = 270
30y = 330, so y= 11
5x+55 =90
5x = 45 so x = 9
5x+7y = 5*9 + 7 * 11 = 45+77= 122 minutes
2007-03-18 13:12:12 · answer #3 · answered by hustolemyname 6 · 0⤊ 1⤋
let song be 'x' minutes, readings be 'y' minutes!
then
120=9y+3x
90=5x+5y
solving these
x=7, y=11
so time fr yr 2001 = 5x+7y
=112minutes
2007-03-18 13:27:38 · answer #4 · answered by Second Newton... 2 · 0⤊ 0⤋
1999:
{3x+9y=120}*5
15x+45y=600...........eq(i)
2000:
{5x+5y=90}*3
15x+15y=270...........eq(ii)
Subtract eq(ii) from eq(i)
15x+45y=600
-(15x+15y=270)
you get
30y=330
or, y=11
Put y=11 in eq(i)
15x+45*11=600
15x+495=600
15x=105
x=7
therefore, in 2001,
estimated time =5x+7y
=5*7+7*11
=112 minutes
2007-03-18 13:17:15 · answer #5 · answered by Bubblez 3 · 0⤊ 1⤋
3x+9y=120
5x+5y=90
15x+45y=600
-15x-15y=-270
============
30y=330
y=11
5(11)+5y=90
55+5y=90
5y=35
x=7
5(10)+7(10)=120 minutes
2007-03-18 13:16:32 · answer #6 · answered by Anonymous · 0⤊ 1⤋
112 minutes
The prob is so simple i find it hard to give an explanation
2007-03-24 23:12:22 · answer #7 · answered by ♠ Author♠ 4 · 0⤊ 0⤋