All categories

2 years ago

Does (2,0) work as a solution to the systems of equations to lines 3x+y=6 and 3x-y=6

Mathematics

1 answer:

fenix001 [56]2 years ago

7 0

Answer:

Yes

Step-by-step explanation:

To see if (2,0) works as a solution to the systems of equations, we plug in the values of x and y and simplify. If the results are equal, then (2,0) is a solution.

3x + y = 6:

$3(2) + 0 = 6$
$6 + 0 = 6$
$6 = 6$
(2,0) is a solution to this equation.

3x - y = 6:

$3(2) - 0 = 6$
$6 - 0 = 6$
$6 = 6$
(2,0) is a solution to this equation.

Therefore, the answer is yes, it does work as a solution.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

You might be interested in

PLEASE HELP!

fenix001 [56]

Answer:

Step-by-step explanation:

No because (1,2) falls on the line, and the line is a dotted line, which means that any point on the line is not a solution to the inequality.

5 0

3 years ago

jenna has 4 double fudge brownies to share with 5 people including yourself how can jenna share the brownies equally.

Oliga [24]

Answer:

4÷5=0.8 or 4/5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Can someone help me with E please I don’t know what operation to use

gtnhenbr [62]

The first question asks who sold more rolls. So start with figuring out how many Christie sold.
5 total - 1 2/3 left = 3 1/3 sold
you can convert the numbers to improper fractions with the same denominator. Like this:
5 x (3/3) - (3+2)/3
15/3 - 5/3 = 10/3
10/3 = 3 1/3

So now we know Christie sold more because 3 1/3 dozen is more than 2 1/2 dozen.

The part asks how many more.. Subtract the amounts the two girls sold.

3 1/3 - 2 1/2
10/3 x (2/2) - 5/2 x (3/3)
20/6 - 15/6 = 5/6

Christie sold 5/6 dozen more rolls. A dozen is 12 rolls so if you wanted to go further you just multiply 12 x 5/6 = 10 rolls

4 0

2 years ago

Prove the divisibility of the following numbers:

ratelena [41]

Make use of prime factorizations:

$16^5+2^{15}=(2^4)^5+2^{15}=2^{20}+2^{15}$

Both terms have a common factor of $2^{15}$ :

$16^5+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33$

- - -

The second one is not true! We can write

$15^7+5^{13}=(3\cdot5)^7+5^{13}=3^7\cdot5^7+5^{13}$

Both terms have a common factor of $5^7$ :

$15^7+5^{13}=5^7\left(3^7+5^6\right)$

Since $30=5\cdot6$ , and $5\mid5^7$ , we'd still have to show that $5^6(3^7+5^6)$ is a multiple of 6. This is impossible, because $6=3\cdot2$ and there is no multiple of 2 that can be factored out.