Question

# Solve{(x,+,3y,=,8),(2y,=,x,+,6):}

Systems of equations

Solve $$\displaystyle{\left\lbrace\begin{matrix}{x}&+&{3}{y}&=&{8}\\{2}{y}&=&{x}&+&{6}\end{matrix}\right.}$$

2020-12-28

$$\displaystyle{\left\lbrace\begin{matrix}{x}&+&{3}{y}&=&{8}\\{2}{y}&=&{x}&+&{6}\end{matrix}\right.}$$

Rewrite equation (2) in standart form NSK

2y – x = 6

5y = 14

Solve for y

$$\displaystyle{y}=\frac{14}{{5}}$$

Substitute $$\displaystyle{y}=\frac{14}{{5}}$$ into equation (1) and solve for x NSK

$$\displaystyle{x}={3}{\left(\frac{14}{{5}}\right)}={8}$$

$$\displaystyle{x}+\frac{42}{{5}}={8}$$

$$\displaystyle{x}={8}–\frac{42}{{5}}$$

$$\displaystyle{x}=\frac{{{40}–{42}}}{{5}}$$

$$\displaystyle{x}=-\frac{2}{{5}}$$

Thus, the solution is $$\displaystyle{\left(-\frac{2}{{5}};\frac{14}{{5}}\right)}$$