Question

4. Solve the following system of equations.

50𝑥+20𝑦=1800 10𝑥+3𝑦=300

Answer 1

$\begin{cases} 50x+20y=1800, \\ 10x+3y=300; \end{cases} \implies$

$\begin{cases} 50x+20y=1800, \\ 50x+15y=1500; \end{cases} \implies$

$\begin{cases} 50x+20y=1800, \\ 5y=300; \end{cases}\implies$

$\begin{cases} 50x+20y=1800, \\ y=60; \end{cases}\implies$

$\begin{cases} 50x+1200=1800, \\ y=60; \end{cases} \implies$

$\begin{cases} x=12, \\ y=60. \end{cases}$