\(standard \ linear \ equation\ :\\\\y=ax+b\\\\ \begin{cases} 4x+y=8\ \ |\ subtract\ 4x\ to\ both\ sides\\ x+3y=8 \ \ |\ subtract\ x\ to\ both\ sides \end{cases}\\\\\begin{cases} y=-4x+8 \\ 3y=-x+8\ \ | \ divide \ each \ term \ by \ 3 \end{cases}\)
\(\begin{cases} y=-4x+8 \\ y=-\frac{1}{3}x+\frac{8}{3} \end{cases}\\\\y=-4x+8\\ To \ find \ the \ x-axis \ intersection \ point, \\set \ y \ equal \ to \ zero \ and \ solve \ for \ x : \\ \\y=0 \ \to 0=-4x+8\\\\4x=8 \ \ | \ divide \ both \ sides\ by\ 4 \\\\x=2\\\\ point : \ \ (2,0)\)
\(To \ find \ the \ y-axis \ intersection \ point, \\set \ x \ equal \ to \ zero \ and \ solve \ for \ y : \\ \\x=0 \ \to y=-4 \cdot 0+8\\ y=8 \\ point: \ \ (0,8)\)
\(y=-\frac{1}{3}x+\frac{8}{3}\\\\ \ the \ x-axis \ intersection \ point \\ \\y=0 \ \to 0=-\frac{1}{3}x+\frac{8}{3}\\ \frac{1}{3}x=\frac{8}{3} \ \ | \ multiply\ both\ sides\ by\ 3 \\\\x=8 \\\\point: \ \ (8,0)\)
\(the \ y-axis \ intersection \ point \\ \\x=0 \ \to y=-\frac{1}{3} \cdot 0+\frac{8}{3} \\ y=\frac{8}{3} \\ point : \ \ (0,\frac{8}{3})\)
\(Answer :\\\\ \begin{cases} y=-4x+8 \ \ | \ multiply \ each \ term \ by \ (-3) \\ 3y=-x+8 \end{cases}\\\begin{cases} -3y=12x-24 \\ 3y=-x+8 \end{cases}\\+-------\\0=11x-16\\11x=16\ \ | \ divide \ both \ sides\ by\ 11\\x=\frac{16}{11}\)
\(y=-4 \cdot \frac{16}{11}+8 \\ y=- \frac{ 64}{11}+\frac{88}{11}\\y= \frac{24}{11}\\\\\begin{cases} x=\frac{16}{11} \\ y=\frac{24}{11} \end{cases}\)
