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3 years ago

У = 2х – 6у = х2 – 4х + 2

Mathematics

1 answer:

Dafna11 [192]3 years ago

4 0

Your answer would be (x,y)=(2,-2)

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Solve equation 3 - 8c = 35

NeTakaya

We would subtact 3 from both sides allowing the equation to simplify to -8c=32

We would then divide both sides by -8 to equal -4

The final answer C=-4

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Two automobiles start together from the same place and travel along the same route. The first averages 40 miles

natita [175]

Answer:

A. (55 x 5) - (40 x 5)

Step-by-step explanation:

You are solving how much miles (further along) would the second car be after 5 hours.

The first car averages 40 miles per hour. 5 hours later, it will have averaged about 200 miles in 5 hours (40 x 5 = 200).

The second car averages 55 miles per hour. 5 hours later, it will have averaged about 275 miles in 5 hours (55 x 5 = 275)

Subtract: 275 - 200 = 75

The second car would have averaged 75 more miles than the first car.

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3 years ago

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If x = 2y = 3 <br>find xy-y+ x × x

olga_2 [115]

Answer:

here ....x = 2 and y = 3

$= xy - y + x \times x$

$= 2 \times 3 - 3 + 2 + 2 \times 2$

$= 6 - 3 + 2 + 4$

$= 12 - 3$

$= 9$

4 0

2 years ago

Two problems here I need solved! I need every step, so please have that with your answers!!

Free_Kalibri [48]

QUESTION 1

We want to solve,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-6x+8}$

We factor the denominator of the fraction on the right hand side to get,

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x^{2}-4x - 2x+8}.$

This implies
$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{x(x-4) - 2(x - 4)}.$

$\frac{1}{(x-4)}+\frac{x}{(x-2)}=\frac{2}{(x-4)(x - 2)}$

We multiply through by LCM of
$(x-4)(x - 2)$

$(x - 2) + x(x-4) = 2$

We expand to get,

$x - 2 + {x}^{2} - 4x= 2$

We group like terms and equate everything to zero,

${x}^{2} + x - 4x - 2 - 2 = 0$

We split the middle term,

${x}^{2} + - 3x - 4 = 0$

We factor to get,

${x}^{2} + x - 4x- 4 = 0$

$x(x + 1) - 4(x + 1) = 0$

$(x + 1)(x - 4) = 0$

$x + 1 = 0 \: or \: x - 4 = 0$

$x = - 1 \: or \: x = 4$

But
$x = 4$
is not in the domain of the given equation.

It is an extraneous solution.

$\therefore \: x = - 1$
is the only solution.

QUESTION 2

$\sqrt{x+11} -x=-1$

We add x to both sides,

$\sqrt{x+11} =x-1$

We square both sides,

$x + 11 = (x - 1)^{2}$

We expand to get,

$x + 11 = {x}^{2} - 2x + 1$

This implies,

${x}^{2} - 3x - 10 = 0$

We solve this quadratic equation by factorization,

${x}^{2} - 5x + 2x - 10 = 0$

$x(x - 5) + 2(x - 5) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \: or \: x - 5 = 0$

$x = - 2 \: or \: x = 5$

But
$x = - 2$
is an extraneous solution

$\therefore \: x = 5$

7 0

3 years ago

У = 2х – 6у = х2 – 4х + 2​

У = 2х – 6у = х2 – 4х + 2