Solve 2x+a/b-x-b/a=3ax+(a-b)^2/ab

$$\frac { 2 x + a } { b } - \frac { x - b } { a } = \frac { 3 a x + ( a - b ) ^ { 2 } } { a b }$$

Solve for a

$$a=\frac{bx}{2b-x}$$

$$x\neq 0\text{ and }b\neq 0\text{ and }x\neq 2b$$

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ab, the least common multiple of b,a,ab.

$$a\left(2x+a\right)-b\left(x-b\right)=3ax+\left(a-b\right)^{2}$$

Use the distributive property to multiply a by 2x+a.

$$2ax+a^{2}-b\left(x-b\right)=3ax+\left(a-b\right)^{2}$$

Use the distributive property to multiply b by x-b.

$$2ax+a^{2}-\left(bx-b^{2}\right)=3ax+\left(a-b\right)^{2}$$

To find the opposite of bx-b^{2}, find the opposite of each term.

$$2ax+a^{2}-bx+b^{2}=3ax+\left(a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$2ax+a^{2}-bx+b^{2}=3ax+a^{2}-2ab+b^{2}$$

Subtract 3ax from both sides.

$$2ax+a^{2}-bx+b^{2}-3ax=a^{2}-2ab+b^{2}$$

Combine 2ax and -3ax to get -ax.

$$-ax+a^{2}-bx+b^{2}=a^{2}-2ab+b^{2}$$

Subtract a^{2} from both sides.

$$-ax+a^{2}-bx+b^{2}-a^{2}=-2ab+b^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$-ax-bx+b^{2}=-2ab+b^{2}$$

Add 2ab to both sides.

$$-ax-bx+b^{2}+2ab=b^{2}$$

Add bx to both sides.

$$-ax+b^{2}+2ab=b^{2}+bx$$

Subtract b^{2} from both sides.

$$-ax+2ab=b^{2}+bx-b^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$-ax+2ab=bx$$

Combine all terms containing a.

$$\left(-x+2b\right)a=bx$$

The equation is in standard form.

$$\left(2b-x\right)a=bx$$

Divide both sides by -x+2b.

$$\frac{\left(2b-x\right)a}{2b-x}=\frac{bx}{2b-x}$$

Dividing by -x+2b undoes the multiplication by -x+2b.

$$a=\frac{bx}{2b-x}$$

Variable a cannot be equal to 0.

$$a=\frac{bx}{2b-x}\text{, }a\neq 0$$

Solve for b

$$b=\frac{ax}{2a-x}$$

$$x\neq 0\text{ and }a\neq 0\text{ and }x\neq 2a$$

Steps for Solving Linear Equation

Variable b cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by ab, the least common multiple of b,a,ab.

$$a\left(2x+a\right)-b\left(x-b\right)=3ax+\left(a-b\right)^{2}$$

Use the distributive property to multiply a by 2x+a.

$$2ax+a^{2}-b\left(x-b\right)=3ax+\left(a-b\right)^{2}$$

Use the distributive property to multiply b by x-b.

$$2ax+a^{2}-\left(bx-b^{2}\right)=3ax+\left(a-b\right)^{2}$$

To find the opposite of bx-b^{2}, find the opposite of each term.

$$2ax+a^{2}-bx+b^{2}=3ax+\left(a-b\right)^{2}$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-b\right)^{2}.

$$2ax+a^{2}-bx+b^{2}=3ax+a^{2}-2ab+b^{2}$$

Add 2ab to both sides.

$$2ax+a^{2}-bx+b^{2}+2ab=3ax+a^{2}+b^{2}$$

Subtract b^{2} from both sides.

$$2ax+a^{2}-bx+b^{2}+2ab-b^{2}=3ax+a^{2}$$

Combine b^{2} and -b^{2} to get 0.

$$2ax+a^{2}-bx+2ab=3ax+a^{2}$$

Subtract 2ax from both sides.

$$a^{2}-bx+2ab=3ax+a^{2}-2ax$$

Combine 3ax and -2ax to get ax.

$$a^{2}-bx+2ab=ax+a^{2}$$

Subtract a^{2} from both sides.

$$-bx+2ab=ax+a^{2}-a^{2}$$

Combine a^{2} and -a^{2} to get 0.

$$-bx+2ab=ax$$

Combine all terms containing b.

$$\left(-x+2a\right)b=ax$$

The equation is in standard form.

$$\left(2a-x\right)b=ax$$

Divide both sides by -x+2a.

$$\frac{\left(2a-x\right)b}{2a-x}=\frac{ax}{2a-x}$$

Dividing by -x+2a undoes the multiplication by -x+2a.

$$b=\frac{ax}{2a-x}$$

Variable b cannot be equal to 0.

$$b=\frac{ax}{2a-x}\text{, }b\neq 0$$