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Determine the intercepts of the following lines.
1. x/3+y/5=1
2.2x-4y=13

Sagot :

Riza1idn
[tex]1.\\ A \ linear \ equation \ is \ an \ equation \ for \ a \ straight \ line :\\\\y=ax+b\\\\ \frac{x}{3} +\frac{y}{5} =1\ \ | \ multiply \ each \ term \ by \ 5 \\\\\frac{5}{3}x +y=5 \ \ |\ subtract\ \frac{5}{3}x\ to\ both\ sides \\\\y=-\frac{5}{3}x +5[/tex]

[tex]the \ intersection \ with \ y-axis\\ x=0 \ \to\ y=-\frac{5}{3} \cdot 0 +5=5 \\\\the \ intersection \ with \ x-axis\\ y=0 \ \to\ 0=-\frac{5}{3}x +5\ \ | \ add \ \frac{5}{3}x\ to\ both\ sides \\\\ \frac{5}{3}x=5\ \ | \ multiply\ both\ sides\ by\ \frac{3}{5}\\\\x=3[/tex]


 
[tex]2.\\\\2x-4y=13 \ \ |\ subtract\ 2x\ to\ both\ sides\\\\-4y=-2x+ 13 \ \ | \ divide \ each \ term \ by \ (-4) \\ \\y= \frac{2}{4}x-\frac{13}{4}\\\\y= \frac{1}{2}x-\frac{13}{4}[/tex]

[tex]the \ intersection \ with \ y-axis\\ x=0 \ \to\ y= \frac{1}{2} \cdot 0-\frac{13}{4}=-\frac{13}{4}=-3\frac{1}{4} \\\\the \ intersection \ with \ x-axis\\ y=0 \ \to\ 0= \frac{1}{2}x-\frac{13}{4} \ \ | \ add \ \frac{13}{4} \ to\ both\ sides \\\\ \frac{1}{2}x= \frac{13}{4}\ \ | \ multiply\ both\ sides\ by\ 2\\\\x= \frac{13}{2}=6\frac{1}{2}[/tex]


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