How do you solve 2b(x-1)= a(bx+5)?

2xb-2b=abx+5a
2xb-abx=5a+2b
x(2b-ab)=5a+2b
x=(5a+2b)/(2b-ab)

2 b x - 2 b = a b x + 5 a

( 2 b - a b ) x = 5 a + 2 b

x = ( 5 a + 2 b ) / ( 2 b - a b )

2b(x-1)= a(bx+5)
2bx - 2 = abx + 5a
2bx - abx = 2 + 5a
x(2b - ab) = 2 + 5a
x = 2 + 5a / 2b - ab

2b(x - 1) = a(bx + 5)
x(2b - ab) = 5a + 2b
x = (5a + 2b)/[b(2 - a)]

Howdy renn, & I sure hope all is well. :)

As for your question,

2b(X -- 1)= a(bX + 5)

First, we expand, by multiplying the terms together. Remember that multiplication is distributive. So a(b + c) = ab + ac. So we have:

2bX -- 2b = abX + 5a

To solve for X, we must first isolate it. So we subtract abX from & add 2b to both sides of the equation. We get:

(Note: Remember that any operation must be performed on both sides of an equation, to maintain the equality. An equation is like a scale. Regardless whether the scale is empty or has the same amount of weight in both trays, it would be level. Now, if you wanted to add weights to it (or remove some of the ones already in the trays), but still keep the scale level, then you can only do that by adding (or removing) the same amount of weights to (or from) both trays. The same is true for equations, except that it also applies to multiplication, division, exponents (like when you square something or take its square root), as well as logarithms & natural logarithms (Ln), which you'll learn in more advanced grades of math, probably grade 12, if you have it.)

2bX -- abX -- 2b + 2b = abX -- abX + 5a + 2b

Simplifying yields:

2bX -- abX = 5a + 2b

Since X is the common factor among the left-hand terms, we factor it out. We get:

(2b -- ab)X = 5a + 2b

Finally, to solve for X, we need to get rid of the coefficient (the number or expression being multiplied by the X, in this case, it's the expression (2b -- ab).). So we divide both sides by (2b -- ab). We get:

[(2b -- ab)X]/(2b -- ab) = (5a + 2b)/(2b -- ab)

Simplifying yields:

X = (5a + 2b)/(2b -- ab)

I sure hope that was as clear & as helpful as you needed it to be. :) Good luck, take care, & have a great day & a great weekend. :)

Cheers! :)

2b(x - 1) = a(bx + 5)
2b*x - 2b*1 = a*bx + a*5
2bx - 2b = abx + 5a
2bx - abx = 5a + 2b
x(2b - ab) = 5a + 2b
x = (5a + 2b)/[b(2 - a)]

Just use your available algebra rules to rearrange it until you get x alone. The first thing you need to go is distribute the two multiplication operations, because x will never be alone if it's inside a set of parenthesis.
2b(x-1)= a(bx+5)
2bx - 2b = abx + 5a [distribution, twice]
2bx - abx - 2b = abx - abx + 5a [subtract abx from both sides, so all the x terms are on the same side]
2bx - abx - 2b + 2b = 0 + 5a + 2b [add 2b to both sides to get the x terms alone]
2bx - abx = 5a + 2b [simplify]
x(2bx/x - abx/x) = 5a + 2b [multiply left side by x/x = 1, to factor out the x from the two terms]
x(2b - ab) = 5a + 2b [simplify, x/x = 1]
x(2b - ab)/(2b - ab) = (5a + 2b)/(2b - ab) [divde both sides by 2b - ab to get x alone]
x = (5a + 2b)/(2b - ab) [simpify]

2xb-2b=abx+5a
2xb-abx=5a+2b
x(2b-ab)=5a+2b
x=(5a+2b)/(2b-ab)
x= (5a+2b)/ b(2-a)

X = 5a-2/2b-a