Solve z=x-y/4-y-x/3

$$z = \frac { x - y } { 4 } - \frac { y - x } { 3 }$$

Solve for x

$$x=\frac{12z}{7}+y$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 3 is 12. Multiply \frac{x-y}{4} times \frac{3}{3}. Multiply \frac{y-x}{3} times \frac{4}{4}.

$$z=\frac{3\left(x-y\right)}{12}-\frac{4\left(y-x\right)}{12}$$

Since \frac{3\left(x-y\right)}{12} and \frac{4\left(y-x\right)}{12} have the same denominator, subtract them by subtracting their numerators.

$$z=\frac{3\left(x-y\right)-4\left(y-x\right)}{12}$$

Do the multiplications in 3\left(x-y\right)-4\left(y-x\right).

$$z=\frac{3x-3y-4y+4x}{12}$$

Combine like terms in 3x-3y-4y+4x.

$$z=\frac{7x-7y}{12}$$

Divide each term of 7x-7y by 12 to get \frac{7}{12}x-\frac{7}{12}y.

$$z=\frac{7}{12}x-\frac{7}{12}y$$

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

$$\frac{7}{12}x-\frac{7}{12}y=z$$

Add \frac{7}{12}y to both sides.

$$\frac{7}{12}x=z+\frac{7}{12}y$$

The equation is in standard form.

$$\frac{7}{12}x=\frac{7y}{12}+z$$

Divide both sides of the equation by \frac{7}{12}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{7}{12}x}{\frac{7}{12}}=\frac{\frac{7y}{12}+z}{\frac{7}{12}}$$

Dividing by \frac{7}{12} undoes the multiplication by \frac{7}{12}.

$$x=\frac{\frac{7y}{12}+z}{\frac{7}{12}}$$

Divide z+\frac{7y}{12} by \frac{7}{12} by multiplying z+\frac{7y}{12} by the reciprocal of \frac{7}{12}.

$$x=\frac{12z}{7}+y$$

Solve for y

$$y=-\frac{12z}{7}+x$$

Steps for Solving Linear Equation

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 3 is 12. Multiply \frac{x-y}{4} times \frac{3}{3}. Multiply \frac{y-x}{3} times \frac{4}{4}.

$$z=\frac{3\left(x-y\right)}{12}-\frac{4\left(y-x\right)}{12}$$

Since \frac{3\left(x-y\right)}{12} and \frac{4\left(y-x\right)}{12} have the same denominator, subtract them by subtracting their numerators.

$$z=\frac{3\left(x-y\right)-4\left(y-x\right)}{12}$$

Do the multiplications in 3\left(x-y\right)-4\left(y-x\right).

$$z=\frac{3x-3y-4y+4x}{12}$$

Combine like terms in 3x-3y-4y+4x.

$$z=\frac{-7y+7x}{12}$$

Divide each term of -7y+7x by 12 to get -\frac{7}{12}y+\frac{7}{12}x.

$$z=-\frac{7}{12}y+\frac{7}{12}x$$

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

$$-\frac{7}{12}y+\frac{7}{12}x=z$$

Subtract \frac{7}{12}x from both sides.

$$-\frac{7}{12}y=z-\frac{7}{12}x$$

The equation is in standard form.

$$-\frac{7}{12}y=-\frac{7x}{12}+z$$

Divide both sides of the equation by -\frac{7}{12}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{7}{12}y}{-\frac{7}{12}}=\frac{-\frac{7x}{12}+z}{-\frac{7}{12}}$$

Dividing by -\frac{7}{12} undoes the multiplication by -\frac{7}{12}.

$$y=\frac{-\frac{7x}{12}+z}{-\frac{7}{12}}$$

Divide z-\frac{7x}{12} by -\frac{7}{12} by multiplying z-\frac{7x}{12} by the reciprocal of -\frac{7}{12}.

$$y=-\frac{12z}{7}+x$$