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lozanna [386]
3 years ago
14

Alana is having a party. She bought 3 rolls of streamers and 2 packages of balloons for $10.00. She realized she needed more sup

plies and went back to the store and bought 2 more rolls of streamers and 1 more package of balloons for $6.25. How much did each roll of streamers and each package of balloons cost?
Mathematics
2 answers:
notsponge [240]3 years ago
4 0
Streamers: 24 dollars Package of ballons: 6 dollars
Anvisha [2.4K]3 years ago
3 0

Answer:

Each roll of streamers is $2.50 and each package of balloons is $1.25.

Step-by-step explanation:

Let x represent the number of rolls of streamers and y represent the number of packages of balloons.

3 rolls of streamers, 3x, and 2 packages of balloons, 2y, cost 10.00; this gives us

3x + 2y = 10

2 rolls of streamers, 2x, and 1 package of balloons, 1y, cost 6.25; this gives us

2x + 1y = 6.25

This gives us the system of equations

\left \{ {{3x+2y=10} \atop {2x+1y=6.25}} \right.

To solve this, we will use elimination.  First we will make the coefficient of y the same.  We will do this by multiplying the bottom equation by 2:

\left \{ {{3x+2y=10} \atop {2(2x+1y=6.25)}} \right. \\\\\left \{ {{3x+2y=10} \atop {4x+2y=12.50}} \right.

We will eliminate y by subtracting the bottom equation from the top:

\left \{ {{3x+2y=10} \atop {-(4x+2y=12.5)}} \right. \\\\-1x=-2.5

Divide both sides by -1:

-1x/-1 = -2.5/-1

x = 2.5

Each roll of streamers costs $2.50.

Substituting this back into the second equation, we have

2(2.5) + y = 6.25

5 + y = 6.25

Subtract 5 from each side:

5 + y - 5 = 6.25 - 5

y = 1.25

Each package of balloons costs $1.25.

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The common factors between them are 1, 2, 4, but the greatest common factor between them is 4.


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Petes boat can travel 48 miles upstream in 4 hours. The return trip takes 3 hours. Find the speed of the boat in still water and
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So... hmm bear in mind, when the boat goes upstream, it goes against the stream, so, if the boat has speed rate of say "b", and the stream has a rate of "r", then the speed going up is b - r, the boat's rate minus the streams, because the stream is subtracting speed as it goes up

going downstream is a bit different, the stream speed is "added" to boat's
so the boat is really going faster, is going b + r

notice, the distance is the same, upstream as well as downstream
thus   \bf \begin{cases}
b=\textit{rate of the boat}\\
r=\textit{rate of the river}
\end{cases}\qquad thus
\\\\\\

\begin{array}{lccclll}
&distance&rate&time(hrs)\\
&----&----&----\\
upstream&48&b-r&4\\
downstream&48&b+4&3
\end{array}
\\\\\\

\begin{cases}
48=(b-r)(4)\to 48=4b-4r\\\\
\frac{48-4b}{-4}=r\\
--------------\\
48=(b+r)(3)\\
-----------------------------\\\\
thus\\\\
48=\left[ b+\left(\boxed{\frac{48-4b}{-4}}\right) \right] (3)
\end{cases}

solve for "r", to see what the stream's rate is

what about the boat's? well, just plug the value for "r" on either equation and solve for "b"
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