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Degger [83]
3 years ago
12

Solve using substitution y=x+6 and y=-2x-3

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
5 0
Y=x+6
y=-2x-3
=> x+6 = -2x-3
=> x+2x = -6-3
=> 3x= -9
=> x= -3 
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Use the elimination method to solve the system of equations.
ExtremeBDS [4]

Answer:

Option D. (8, – 4)

Step-by-step explanation:

3x + 4y = 8 ..... (1)

x – y = 12.... (2)

To solve the above equation by elimination method, do the following:

Step 1:

Multiply equation 1 by the coefficient of x in equation 2 i.e 1.

Multiply equation 2 by the coefficient of x in equation 1 i.e 3. This is illustrated below:

1 × Equation 1

1 × (3x + 4y = 8)

3x + 4y = 8 ...... (3)

3 × Equation 2

3 × ( x – y = 12)

3x – 3y = 36......(4)

Step 2:

Subtract equation 3 from equation 4. This is illustrated below:

. 3x – 3y = 36

– (3x + 4y = 8)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

– 7y = 28

Divide both side by the coefficient of y i.e –7

y = 28/–7

y = – 4

Step 3:

Substitute the value of y into any of the equation to obtain the value of x. In this case, we shall substitute the value of y into equation 2 as shown below:

x – y = 12

y = –4

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4 0
4 years ago
is picking out some movies to rent, primarily interested in horror films and comedies. He has narrowed down his selections to 15
tigry1 [53]

Answer: 31,365

Step-by-step explanation:

Given : The number of horror films = 15

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Then , the number of different combinations of 4 movies can he rent if he wants at least two comedies is given by :-

^{18}C_2\times ^{15}C_2+^{18}C_3\times ^{15}C_1+^{18}C_4\times ^{15}C_0\\\\=\dfrac{18!}{2!(18-2)!}\times\dfrac{15!}{2!(15-2)!}+\dfrac{18!}{3!(18-3)!}\times\dfrac{15!}{1!(15-1)!}+\dfrac{18!}{4!(18-4)!}\times\dfrac{15!}{0!(15-0)!}\\\\=16065+12240+3060=31365

Hence, the  number of different combinations of 4 movies can he rent if he wants at least two comedies = 31,365

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3 years ago
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Bezzdna [24]

Answer:

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Step-by-step explanation:

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QveST [7]

Answer:

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Step-by-step explanation:

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Answer:

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