All categories

3 years ago

Solve for y: 2x - y = 3 PLEASE GIVE STEPS!!

Mathematics

1 answer:

Studentka2010 [4]3 years ago

6 0

Answer:

I think its 2×3 - 3 = 3

I think

You might be interested in

How to answer this on descending powers of x please help pretty please

zheka24 [161]

Answer: 25x² - 64

<u>Step-by-step explanation:</u>

(5x + 8)(5x - 8)

= 5x(5x - 8) + 8(5x - 8)

= 25x² - 40x + 40x - 64

= 25x² + 0x - 64

= $\large\boxed{25x^2-64}$

NOTE: descending powers means the biggest exponent goes first, then the next biggest exponent, etc.

For example: x⁴ then x³ then x² then x then the number (which is actually x⁰)

4 0

4 years ago

NOOOOOOOO LINNNKKKKSSSSSSSS!!!!!!

mr Goodwill [35]

39 ………………………………………………………………

7 0

3 years ago

Read 2 more answers

One side of a square flower bed is 22 feet long. How many plants are needed if they are to be spaced 22 inches apart around the

Juli2301 [7.4K]

(22 ft)*(12 in/ft)/(22 in) = 12 . . . . plants per side

12*4 = 48 plants are needed for the outside border.

4 0

4 years ago

I found a place that will give me 20% discount if I spend over $50. My pill is $75. How much money will I save?

olasank [31]

75*.2=15
you would save $15

hope that helps

5 0

3 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

4 years ago