All categories

3 years ago

Six times a number plus twice a second number is four. Twice the first number plus four times the second number is eighteen.

Mathematics

1 answer:

Debora [2.8K]3 years ago

8 0

Answer:

x=-1, y=5. (-1, 5). The first number is -1 and the second number is 5.

Step-by-step explanation:

6x+2y=4

2x+4y=18

---------------

-2(6x+2y)=-2(4)

2x+4y=18

-------------------------

-12x-4y=-8

2x+4y=18

----------------

-10x=10

x=10/-10

x=-1

6(-1)+2y=4

-6+2y=4

2y=4-(-6)

2y=4+6

2y=10

y=10/2

y=5

You might be interested in

What is the solution to 5(2x+6)=-4(-5-2x)+3

Gnesinka [82]

Answer:

$- \frac {7}{2}$

7 0

2 years ago

Carol is going to buy a purse she found. It is originally $120.00, but has been marked down 25%. When carol goes to buy the purs

german

$97.2

120*.75 =90 (discounted price)
90*.08 = 7.2 (amount sales tax)

90 + 7.2= 97.2

5 0

4 years ago

What is x/2=12? I need it for my homework thank you

sveta [45]

X/2 = 12
X = 12 * 2
X = 24

3 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

You walk east 720 feet to get from your house to class. After class, you walk southwest 1,044 feet to meet your friends for lunc

strojnjashka [21]

The direction you are traveling to get back home is North

4 0

3 years ago