Solve {l}{alpha+beta+gamma=180^circ}{beta=4alpha}{gamma=alpha}

$$\left. \begin{array}{l}{ \alpha + \beta + \gamma = 180 ^ { \circ } }\\{ \beta = 4 \alpha }\\{ \gamma = \alpha }\end{array} \right.$$

Solve for α, β, γ

$$\alpha =30$$

$$\beta =120$$

$$\gamma =30$$

Short Steps Using Substitution

Reorder the equations.

$$\beta =4\alpha \alpha +\beta +\gamma =180 \gamma =\alpha$$

Substitute 4\alpha for \beta in the equation \alpha +\beta +\gamma =180.

$$\alpha +4\alpha +\gamma =180$$

Solve \alpha +4\alpha +\gamma =180 for \alpha .

$$\alpha =-\frac{1}{5}\gamma +36$$

Substitute -\frac{1}{5}\gamma +36 for \alpha in the equation \gamma =\alpha .

$$\gamma =-\frac{1}{5}\gamma +36$$

Solve \gamma =-\frac{1}{5}\gamma +36 for \gamma .

$$\gamma =30$$

Substitute 30 for \gamma in the equation \alpha =-\frac{1}{5}\gamma +36.

$$\alpha =-\frac{1}{5}\times 30+36$$

Calculate \alpha from \alpha =-\frac{1}{5}\times 30+36.

$$\alpha =30$$

Substitute 30 for \alpha in the equation \beta =4\alpha .

$$\beta =4\times 30$$

Calculate \beta from \beta =4\times 30.

$$\beta =120$$

The system is now solved.

$$\alpha =30 \beta =120 \gamma =30$$