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masha68 [24]
3 years ago
8

2x+3y = -33 3x+6y=-57 using elimination

Mathematics
1 answer:
ella [17]3 years ago
4 0

\left\{\begin{array}{ccc}2x+3y = -33&\text{multiply both sides by 3}\\3x+6y=-57 &\text{multiply both sides by (-2)}\end{array}\right\\\\\underline{\left\{\begin{array}{ccc}6x+9y=-99\\-6x-12y=114 \end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad-3y=15\qquad\text{divide both sides by (-3)}\\.\qquad y=-5\\\\\\\text{Put the value of y to the first equation}\\\\2x+3(-5)=-33\\2x-15=-33\qquad\text{add 15 to both sides}\\2x=-18\qquad\text{divide both sides by 2}\\x=-9

Answer:\ x=-9\ and\ y=-5\to(-9,\ -5)

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Given that €1=£0.72 how much is €410 in £ what is the £ to € exchange rate
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Find the perimeter of the right triangle. If necessary, round to the nearest tenth.
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7 0
3 years ago
The prices of commodities X,Y,Z are respectively x, y, z, rupees per unit. Mr. A purchases 4 units of Z and sells 3 units of X a
liubo4ka [24]

Answer:

(x,y,z)=(1477, 1464, 1437)

Step-by-step explanation:

Consider the selling of the units positive earning and the purchasing of the units negative earning.

<h3>Case-1:</h3>
  • Mr. A purchases 4 units of Z and sells 3 units of X and 5 units of Y
  • Mr.A earns Rs6000

So, the equation would be

3x  +  5y - 4z = 6000

<h3>Case-2:</h3>
  • Mr. B purchases 3 units of Y and sells 2 units of X and 1 units of Z
  • Mr B neither lose nor gain meaning he has made 0₹

hence,

2x   - 3y  +  z = 0

<h3>Case-3:</h3>
  • Mr. C purchases 1 units of X and sells 4 units of Y and 6 units of Z
  • Mr.C earns 13000₹

therefore,

- x    + 4y  +  6z = 13000

Thus our system of equations is

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

<u>Solving </u><u>the </u><u>system </u><u>of </u><u>equations</u><u>:</u>

we will consider elimination method to solve the system of equations. To do so ,separate the equation in two parts which yields:

\begin{cases}3x  +  5y - 4z = 6000\\2x   - 3y  +  z = 0\end{cases}\\\begin{cases}2x   - 3y  +  z = 0\\ - x    + 4y  +  6z = 13000\end{cases}

Now solve the equation accordingly:

\implies\begin{cases}11x-7y=6000\\-13x+22y=13000\end{cases}

Solving the equation for x and y yields:

\implies\begin{cases}x= \dfrac{223000}{151}\\\\y= \dfrac{221000}{151}\end{cases}

plug in the value of x and y into 2x - 3y + z = 0 and simplify to get z. hence,

\implies z= \dfrac{217000}{151}

Therefore,the prices of commodities X,Y,Z are respectively approximately 1477, 1464, 1437

6 0
2 years ago
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