Question

Solve for
xx
x
x
.
n(17+x)=34x−rn(17+x) = 34x-r
n(17+x)=34x−r
n, left parenthesis, 17, plus, x, right parenthesis, equals, 34, x, minus, r
x=x=
x=
x, equals

Answer 1

Given equation : n(17+x)=34x−r.

We need to solve it for x.

Distributing n over (17+x) on left side, we get

17n + nx  = 34x - r.

Adding r on both sides, we get

17n+r + nx  = 34x - r+r.

17n + r + nx = 34x.

Subtracting nx from both sides, we get

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

17n + r = 34x -nx.

Factoring out gcf x on right side, we get

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

Dividing both sides by (34-n), we get

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

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

Therefore, final answer is $x=\frac{(17x + r)}{(34-n)}.$

Answer 2

Answer:

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

Step-by-step explanation:

We have to solve the equation for the value of x.

n(17 + x) = 34x - r

17n + nx = 34x - r [By Distributive property ]

17n + nx - 17n = 34x - r - 17n [subtracting 17n from both the sides of the equation]

nx = 34x - r - 17n

nx - 34x = 34x - r - 17n - 34x [Subtracting 34x from both the sides of the equation]

x(n - 34) = -(r + 17n) [Division by (n - 34) on both the sides of the equation]  

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