To find the number of CD on sale, we can set an equation.
Let y be the number if CD.
As she bought 40%, the amount she bought= 40%y
The equation:
40%y = 80
0.4y = 80
y = 80÷0.4
y=320
Therefore there were originally 320 CDs.
Hope it helps!
Answer:
First choice.
Step-by-step explanation:
You could plug in the choices to see which would make all the 3 equations true.
Let's start with (x=2,y=-6,z=1):
2x+y-z=-3
2(2)+-6-1=-3
4-6-1=-3
-2-1=-3
-3=-3 is true so the first choice satisfies the first equation.
5x-2y+2z=24
5(2)-2(-6)+2(1)=24
10+12+2=24
24=24 is true so the first choice satisfies the second equation.
3x-z=5
3(2)-1=5
6-1=5
5=5 is true so the first choice satisfies the third equation.
We don't have to go any further since we found the solution.
---------Another way.
Multiply the first equation by 2 and add equation 1 and equation 2 together.
2(2x+y-z=-3)
4x+2y-2z=-6 is the first equation multiplied by 2.
5x-2y+2z=24
----------------------Add the equations together:
9x+0+0=18
9x=18
Divide both sides by 9:
x=18/9
x=2
Using the third equation along with x=2 we can find z.
3x-z=5 with x=2:
3(2)-z=5
6-z=5
Add z on both sides:
6=5+z
Subtract 5 on both sides:
1=z
Now using the first equation along with 2x+y-z=-3 with x=2 and z=1:
2(2)+y-1=-3
4+y-1=-3
3+y=-3
Subtract 3 on both sides:
y=-6
So the solution is (x=2,y=-6,z=1).
Graph it on the Ti-84 and use the trace function to find the intersection.
The answer is
(-3,3)
Well, you would need to compensate for the cost of the banquet hall adding an addition $700 to the goal of $1000.
If you need to raise at least $1700 you can write this inequality.
Let x represent the number of tickets sold.
15x≥1700
x≥ 114 (rounded to the nearest whole number because you can't sell half a ticket)
So, at least 114 tickets need to be sold.
