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White raven [17]
3 years ago
6

Consider the following system:

Mathematics
2 answers:
Goryan [66]3 years ago
4 0
You did not include the choices. However, I answered one that just included them. I've included the possible answers below and then the correct answers. 

<span>A multiple of Equation 1. 
B. The sum of Equation 1 and Equation 2 
C. An equation that replaces only the coefficient of x with the sum of the coefficients of x in Equation 1 and Equation 2. 
D. An equation that replaces only the coefficient of y with the sum of the coefficients of y in Equation 1 and Equation 2. 
E. The sum of a multiple of Equation 1 and Equation 2.

</span>A, B and E.

Adding and multiplying the terms allow them to keep working. However, you must make sure that each variable is changed each time. Not just one as in C and D. 
Shkiper50 [21]3 years ago
4 0

You did not include the choices. However, I answered one that just included them. I've included the possible answers below and then the correct answers. 


A multiple of Equation 1. 

B. The sum of Equation 1 and Equation 2 

C. An equation that replaces only the coefficient of x with the sum of the coefficients of x in Equation 1 and Equation 2. 

D. An equation that replaces only the coefficient of y with the sum of the coefficients of y in Equation 1 and Equation 2. 

E. The sum of a multiple of Equation 1 and Equation 2.


A, B and E.


Adding and multiplying the terms allow them to keep working. However, you must make sure that each variable is changed each time. Not just one as in C and D. 


Read more on Brainly.com - brainly.com/question/9662635#readmore

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Alborosie
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8 0
3 years ago
Question A volleyball team sold raffle tickets to raise money for the upcoming season. They sold three different types of ticket
m_a_m_a [10]

Answer:

24 premium tickets were sold.

Step-by-step explanation:

Let :

Deluxe ticket = x

Regular tickets = x + 78

Premium tickets = y

x + (x + 78) + y = 208

4x + 2(x+78) + 10y = 714

2x + y = 208 - 78

4x + 2x + 156 + 10y = 714

2x + y = 130 - - - - - (1)

6x + 10y = 558 - - - - (2)

Now we can solve the simultaneous equation using elimination method :

From (1)

y = 130 - 2x

Put y = 130 - 2x in (2)

6x + 10(130 - 2x) = 558

6x + 1300 - 20x = 558

- 14x = 558 - 1300

-14x = - 742

x = 742 / 14

x = 53

Put x = 53 in y = 130 - 2x

y = 130 - 2(53)

y = 130 - 106

y = 24

3 0
2 years ago
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