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andreev551 [17]
3 years ago
5

Brianna wants to buy a digital camera for a photography class. One store offers the camera for $50 down and a payment plan of $2

0 per month. The payment plan for a second store is described by y 15x+ 80, where y is the total cost in dollars and x is the number of months. Which camera is cheaper when the camera is paid off in 12 months? Explain.
Mathematics
1 answer:
Archy [21]3 years ago
5 0

Step-by-step explanation:

To answer this question, we have to use the same equation for each payment plan:

1. y = 20x + 50

2. y = 15x + 80

If x = 12, answer to the first equation would be y = 290

20 x 12 = 240   and   240 + 50 = 290

And the answer to the second equation would be y = 260

15 x 12 = 180   and   180 + 80 = 260

So, the second payment plan is cheaper by 30 dollars!

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2 years ago
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We are given with an equation in <em>variable y</em> and we need to solve for <em>y</em> . So , now let's start !!!

We are given with ;

{:\implies \quad \sf \dfrac{33}{2}+\dfrac{3y}{5}=\dfrac{7y}{10}+15}

Take LCM on both sides :

{:\implies \quad \sf \dfrac{165+6y}{10}=\dfrac{7y+150}{10}}

<em>Multiplying</em> both sides by <em>10</em> ;

{:\implies \quad \sf \cancel{10}\times \dfrac{165+6y}{\cancel{10}}=\cancel{10}\times \dfrac{7y+150}{\cancel{10}}}

{:\implies \quad \sf 165+6y=7y+150}

Can be <em>further written</em> as ;

{:\implies \quad \sf 7y+150=165+6y}

Transposing <em>6y </em>to<em> LHS</em> and <em>150</em> to<em> RHS </em>

{:\implies \quad \sf 7y-6y=165-150}

{:\implies \quad \bf \therefore \quad \underline{\underline{y=15}}}

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