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4 years ago

N(17+x)=34x-r solve for x

Mathematics

1 answer:

77julia77 [94]4 years ago

6 0

Answer:

$x=\frac{-17n-r}{(n-34)}$

Step-by-step explanation:

Given:

$n(17+x)=34x-r$

Taking the terms of x to one side.

$17n+nx=34x-r\\$

$nx-34x=-17n-r\\\\x(n-34)=-17n-r\\\\x=\frac{-17n-r}{(n-34)}$

The value of x is:

$x=\frac{-17n-r}{(n-34)}$

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The ____ property allows you to regroup terms using parentheses when adding or multiplying

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Answer:

"associative"

Step-by-step explanation:

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but the associative property does NOT allow you to change the grouping to ...

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A City's population rose 3% one year, 0.08 the next year, and by 2/50 the next year. Order these increases from least to greates

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I want to be a teacher some day and i know this question----------What is -15 -(-8)=? remember a negitive plus a negitive equals

irina [24]

Answer is -7 because -15+8=-7

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4 years ago

Seven less than the product of a number and three equal to eleven

Anon25 [30]

Answer:

7-(3x)=11

Step-by-step explanation:

4 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