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Pachacha [2.7K]

4 years ago

5

Joe and Willie are selling cheesecakes for a school fundraiser. Customers can buy New York style cheesecakes and chocolate marbl

e cheesecakes. Joe sold 4 New York style cheese cakes and 6 marble cheesecakes for a total of $74. Willie sold 5 New York style cheesecakes and 6 chocolate marble cheesecakes for a total of $79. What is the cost of each of one New York style cheesecake and one chocolate marble cheesecake?

1 answer:

Helen [10]4 years ago

7 0

Cost of each New York style cheesecake is 5 Dollars and chocolate marble cheesecake is 9 Dollars.

Step-by-step explanation:

Let,

New York styles cheesecakes = x

Chocolate marble cheesecakes = y

According to given statement;

4x + 6y = 74 Eqn 1

5x + 6y = 79 Eqn 2

$Subtracting\ Eqn\ 1\ from\ Eqn\ 2\\(5x+6y)-(4x+6y)=79-74\\5x+6y-4x-6y=5\\x=5$

Putting value of x in Eqn 1;

$4(5)+6y=74\\20+6y=74\\Subtracting\ 20\ from\ both\ sides\\20-20+6y=74-20\\6y=54\\Dividing\ both\ sides\ by\ 6\\\frac{6y}{6}=\frac{54}{6} \\ y=9$

Cost of each New York style cheesecake is 5 Dollars and chocolate marble cheesecake is 9 Dollars.

Keywords: Division, Linear Equations.

Learn more about linear equations at:

#LearnwithBrainly

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$\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$
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Integrate both sides of the first PDE with respect to $x$ ; then

$\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\dfrac{(x+y)^3}3+f(y)$

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$\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)$
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$\Psi(x,y)=\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3+C=C$

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$\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C$

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