Solve for d
d=x\left(3y-4\right)
x\neq 0
Solve for x
\left\{\begin{matrix}x=-\frac{d}{4-3y}\text{, }&d\neq 0\text{ and }y\neq \frac{4}{3}\\x\neq 0\text{, }&y=\frac{4}{3}\text{ and }d=0\end{matrix}\right.
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d-3yx+x\times 4=0
Multiply both sides of the equation by x.
d+x\times 4=3yx
Add 3yx to both sides. Anything plus zero gives itself.
d=3yx-x\times 4
Subtract x\times 4 from both sides.
d=3yx-4x
Multiply -1 and 4 to get -4.
d-3yx+x\times 4=0
Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by x.
-3yx+x\times 4=-d
Subtract d from both sides. Anything subtracted from zero gives its negation.
\left(-3y+4\right)x=-d
Combine all terms containing x.
\left(4-3y\right)x=-d
The equation is in standard form.
\frac{\left(4-3y\right)x}{4-3y}=-\frac{d}{4-3y}
Divide both sides by -3y+4.
x=-\frac{d}{4-3y}
Dividing by -3y+4 undoes the multiplication by -3y+4.
x=-\frac{d}{4-3y}\text{, }x\neq 0
Variable x cannot be equal to 0.
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