Solve {l}{80lambda+160y=4}{x+3y=0.1}

$$\left. \begin{array} { l } { 80 \lambda + 160 y = 4 } \\ { x + 3 y = 0.1 } \end{array} \right.$$

Solve for x, y

$$x=\frac{3\lambda }{2}+0.025$$

$$y=-\frac{\lambda }{2}+0.025$$

Steps Using Substitution

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$160y+80\lambda =4,3y+x=0.1$$

Pick one of the two equations which is more simple to solve for y by isolating y on the left hand side of the equal sign.

$$160y+80\lambda =4$$

Subtract 80\lambda from both sides of the equation.

$$160y=4-80\lambda$$

Divide both sides by 160.

$$y=-\frac{\lambda }{2}+\frac{1}{40}$$

Substitute \frac{1}{40}-\frac{\lambda }{2} for y in the other equation, 3y+x=0.1.

$$3\left(-\frac{\lambda }{2}+\frac{1}{40}\right)+x=0.1$$

Multiply 3 times \frac{1}{40}-\frac{\lambda }{2}.

$$-\frac{3\lambda }{2}+\frac{3}{40}+x=0.1$$

Subtract \frac{3}{40}-\frac{3\lambda }{2} from both sides of the equation.

$$x=\frac{3\lambda }{2}+\frac{1}{40}$$

The system is now solved.

$$y=-\frac{\lambda }{2}+\frac{1}{40},x=\frac{3\lambda }{2}+\frac{1}{40}$$