Solve x^4+x^3+1+2+y+a=b

$$x ^ { 4 } + x ^ { 3 } + 1 + 2 + y + a = b$$

Solve for a

$$a=-\left(x^{4}+x^{3}+y-b+3\right)$$

Solution Steps

Add 1 and 2 to get 3.

$$x^{4}+x^{3}+3+y+a=b$$

Subtract x^{4} from both sides.

$$x^{3}+3+y+a=b-x^{4}$$

Subtract x^{3} from both sides.

$$3+y+a=b-x^{4}-x^{3}$$

Subtract 3 from both sides.

$$y+a=b-x^{4}-x^{3}-3$$

Subtract y from both sides.

$$a=b-x^{4}-x^{3}-3-y$$

Solve for b

$$b=x^{4}+x^{3}+y+a+3$$

Solution Steps

Add 1 and 2 to get 3.

$$x^{4}+x^{3}+3+y+a=b$$

Swap sides so that all variable terms are on the left hand side.

$$b=x^{4}+x^{3}+3+y+a$$