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3 years ago

What is the solution to the system of linear equations?

Mathematics

2 answers:

qwelly [4]3 years ago

6 0

Answer:

The answer is $(x,y)=(3,1)$

Step-by-step explanation:

Firstly, we have to eliminate a variable, x or y. <u>It doesn't matter which one</u>.

For that, we orderer the variables in the same way for both equations.

$-x+y=-2\\2*x+y=7$

Then, we need to have the same coefficient in the variable that we want to eliminate, but with different sign

Choosing the x variable:

$2*(-x+y)=2*(-2)\\-2*x+2*y=-4\\\\1*(2*x+y)=1*(7)\\2*x+y=7$

Then we add both equations:

$-2*x+2*y+2*x+y=-4+7\\3*y=3\\y=\frac{3}{3}\\ y=1$

Then we replace the value of "<em>y</em>" in whatever equation:

$-x+1=-2\\-x=-2-1\\-x=-3\\x=3$

Finally, the answer is $(x,y)=(3,1)$

malfutka [58]3 years ago

4 0

The answer of your question is the third one :(3,1)

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. Sara teaches aerobics classes as a part-time job. The amount of money she earned is shown in the table below. Based on the inf

nikitadnepr [17]

Answer:

She should earn 225 dollars.

Step-by-step explanation:

I know this because 50 ÷ 4 = 12.5 and 12.5 × 18 = 225.

3 0

2 years ago

What’s the probability of getting each card out of a deck?

VMariaS [17]

Answer:

a. 1/13

b. 1/52

c. 2/13

d. 1/2

e. 15/26

f. 17/52

g. 1/2

Step-by-step explanation:

a. In a deck of cards, there are 4 suits and each of them has a 7. Therefore, the probability of drawing a 7 is:

P(7) = 4/52 = 1/13

b. There is only one 6 of clubs, therefore, the probability of drawing a 6 of clubs is:

P(6 of clubs) = 1/52

c. There 4 fives (one for each suit) and 4 queens in a deck of cards. Therefore, the probability of drawing a five or a queen is:

P(5 or Q) = P(5) + P(Q)

= 4/52 + 4/52

= 1/13 + 1/13

P(5 or Q) = 2/13

d. There are 2 suits that are black. Each suit has 13 cards. Therefore, there are 26 black cards. The probability of drawing a black card is:

P(B) = 26/52 = 1/2

e. There are 2 suits that are red. Each suit has 13 cards. Therefore, there are 26 red cards. There are 4 jacks. Therefore:

P(R or J) = P(R) + P(J)

= 26/52 + 4/52

= 30/52

P(R or J) = 15/26

f. There are 13 cards in clubs suit and there are 4 aces, therefore:

P(C or A) = P(C) + P(A)

= 13/52 + 4/52

P(C or A) = 17/52

g. There are 13 cards in the diamonds suit and there are 13 in the spades suit, therefore:

P(D or S) = P(D) + P(S)

= 13/52 + 13/52

= 26/52

P(D or S) = 1/2

6 0

3 years ago

Plzzzz help right answer gets a brainly!

masha68 [24]

Answer: A, The line starts at 9 and ends at 29

5 0

3 years ago

Read 2 more answers

Please help. It’s a picture

Korolek [52]

$260

I don’t know what model/formula you are supposed to be using.

But what I did first was calculated what 30% of 2700$ is.
2700 x .3 = 810

So it depreciates $810 per year.

$810 x. 3 years = 2430

2700 - 2430 = 260

In three years, the laptop will be worth $260

6 0

3 years ago

8 freshmen, 9 sophomores, 9 juniors, and 7 seniors are eligible to be on a committee.

loris [4]

Answer: 11113200

Step-by-step explanation:

We know that , the number of combination of choosing r things from n things is given by :-

$^nC_r=\dfrac{n!}{r!(n-r)!}$

Then , the number of ways to choose a dancing committee if it is to consist of 4 freshmen, 5 sophomores, 2 juniors, and 3 seniors :-

$^8C_4\times ^9C_5\times^9C_2\times^7C_3$

$=\dfrac{8!}{(8-4)!4!}\times\dfrac{9!}{(9-5)!5!}\times\dfrac{9!}{(9-2)!2!}\times\dfrac{7!}{3!(7-3)!}\\\\ =\dfrac{8\times7\times6\times5\times4!}{4!4!}\times\dfrac{9\times8\times7\times6\times5!}{4!5!}\times\dfrac{9\times8\times7!}{7!2!}\times\dfrac{7\times6\times5\times4!}{3!4!}\\\\=11113200$

∴ The number of ways to choose a dancing committee if it is to consist of 4 freshmen, 5 sophomores, 2 juniors, and 3 seniors = 11113200

6 0

3 years ago