Solve 2(x1)-15=5(x4)+53=

$$2(x1)-15=5(x4)+53=$$

Solve for x_1

$$x_{1}=\frac{5x_{4}}{2}+34$$

Steps for Solving Linear Equation

Add 15 to both sides.

$$2x_{1}=5x_{4}+53+15$$

Add 53 and 15 to get 68.

$$2x_{1}=5x_{4}+68$$

Divide both sides by 2.

$$\frac{2x_{1}}{2}=\frac{5x_{4}+68}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x_{1}=\frac{5x_{4}+68}{2}$$

Divide 5x_{4}+68 by 2.

$$x_{1}=\frac{5x_{4}}{2}+34$$

Solve for x_4

$$x_{4}=\frac{2\left(x_{1}-34\right)}{5}$$

Steps for Solving Linear Equation

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

$$5x_{4}+53=2x_{1}-15$$

Subtract 53 from both sides.

$$5x_{4}=2x_{1}-15-53$$

Subtract 53 from -15 to get -68.

$$5x_{4}=2x_{1}-68$$

Divide both sides by 5.

$$\frac{5x_{4}}{5}=\frac{2x_{1}-68}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x_{4}=\frac{2x_{1}-68}{5}$$