How do you solve (x+5) - 2(4x-1)=0(x+5)2(4x1)=0?

1 Answer
Jul 13, 2017

See a solution process below:

Explanation:

First, expand the terms in the right parenthesis by multiply each term within the parenthesis by the term outside the parenthesis:

(x + 5) - color(red)(2)(4x - 1) = 0(x+5)2(4x1)=0

(x + 5) - (color(red)(2) xx 4x) + (color(red)(2) xx 1) = 0(x+5)(2×4x)+(2×1)=0

(x + 5) - 8x + 2 = 0(x+5)8x+2=0

x + 5 - 8x + 2 = 0x+58x+2=0

x - 8x + 5 + 2 = 0x8x+5+2=0

1x - 8x + 5 + 2 = 01x8x+5+2=0

(1 - 8)x + (5 + 2) = 0(18)x+(5+2)=0

-7x + 7 = 07x+7=0

Next, subtract color(red)(7)7 from each side of the equation to isolate the xx term while keeping the equation balanced:

-7x + 7 - color(red)(7) = 0 - color(red)(7)7x+77=07

-7x + 0 = -77x+0=7

-7x = -77x=7

Now, divide each side of the equation by color(red)(-7)7 to solve for xx while keeping the equation balanced:

(-7x)/color(red)(-7) = -7/color(red)(-7)7x7=77

(color(red)(cancel(color(black)(-7)))x)/cancel(color(red)(-7)) = 1

x = 1