Soalan 1:
$$ \text { Tuliskan persamaan garis lurus berikut dalam bentuk } $$ $$ \frac{x}{a}+\frac{y}{b}=1 \text { dan } y=m x+c $$
$$ \text { (a) } 3 x-4 y=24 $$
$$ \text { (b) } 7 x+2 y=28 $$
$$ \text { (c) } 5 x-3 y=15 $$
$$ \text { (d) }-2 x+3 y=9 $$
Penyelesaian:
(a)(i)
$$ \begin{gathered} 3 x-4 y=24 \\ \frac{3 x}{24}-\frac{4 y}{24}=\frac{24}{24} \\ \frac{x}{8}-\frac{y}{6}=1 \end{gathered} $$
(a)(ii)
$$ \begin{aligned} 3 x-4 y & =24 \\ -4 y & =-3 x+24 \\ \frac{-4 y}{(-4)} & =\frac{-3 x}{(-4)}+\frac{24}{(-4)} \\ y & =\frac{3}{4} x-6 \end{aligned} $$
(b)(i)
$$ \begin{gathered} 7 x+2 y=28 \\ \frac{7 x}{28}+\frac{2 y}{28}=\frac{28}{28} \\ \frac{x}{4}+\frac{y}{14}=1 \end{gathered} $$
(b)(ii)
$$ \begin{aligned} 7 x+2 y & =28 \\ 2 y & =-7 x+28 \\ \frac{2 y}{2} & =\frac{-7 x}{2}+\frac{28}{2} \\ y & =-\frac{7}{2} x+14 \end{aligned} $$
(c)(i)
$$ \begin{gathered} 5 x-3 y=15 \\ \frac{5 x}{15}-\frac{3 y}{15}=\frac{15}{15} \\ \frac{x}{3}-\frac{y}{5}=1 \end{gathered} $$
(c)(ii)
$$ \begin{aligned} 5 x-3 y & =15 \\ -3 y & =-5 x+15 \\ \frac{-3 y}{(-3)} & =\frac{-5 x}{(-3)}+\frac{15}{(-3)} \\ y & =\frac{5}{3} x-5 \end{aligned} $$
(d)(i)
$$ \begin{aligned} & -2 x+3 y=9 \\ & \frac{-2 x}{9}+\frac{3 y}{9}=\frac{9}{9} \\ & -\frac{2 x}{9}+\frac{y}{3}=1 \end{aligned} $$
(d)(ii)
$$ \begin{aligned} -2 x+3 y & =9 \\ 3 y & =2 x+9 \\ \frac{3 y}{3} & =\frac{2 x}{3}+\frac{9}{3} \\ y & =\frac{2}{3} x+3 \end{aligned} $$
$$ \text { Tuliskan persamaan garis lurus berikut dalam bentuk } $$ $$ \frac{x}{a}+\frac{y}{b}=1 \text { dan } y=m x+c $$
$$ \text { (a) } 3 x-4 y=24 $$
$$ \text { (b) } 7 x+2 y=28 $$
$$ \text { (c) } 5 x-3 y=15 $$
$$ \text { (d) }-2 x+3 y=9 $$
Penyelesaian:
(a)(i)
$$ \begin{gathered} 3 x-4 y=24 \\ \frac{3 x}{24}-\frac{4 y}{24}=\frac{24}{24} \\ \frac{x}{8}-\frac{y}{6}=1 \end{gathered} $$
(a)(ii)
$$ \begin{aligned} 3 x-4 y & =24 \\ -4 y & =-3 x+24 \\ \frac{-4 y}{(-4)} & =\frac{-3 x}{(-4)}+\frac{24}{(-4)} \\ y & =\frac{3}{4} x-6 \end{aligned} $$
(b)(i)
$$ \begin{gathered} 7 x+2 y=28 \\ \frac{7 x}{28}+\frac{2 y}{28}=\frac{28}{28} \\ \frac{x}{4}+\frac{y}{14}=1 \end{gathered} $$
(b)(ii)
$$ \begin{aligned} 7 x+2 y & =28 \\ 2 y & =-7 x+28 \\ \frac{2 y}{2} & =\frac{-7 x}{2}+\frac{28}{2} \\ y & =-\frac{7}{2} x+14 \end{aligned} $$
(c)(i)
$$ \begin{gathered} 5 x-3 y=15 \\ \frac{5 x}{15}-\frac{3 y}{15}=\frac{15}{15} \\ \frac{x}{3}-\frac{y}{5}=1 \end{gathered} $$
(c)(ii)
$$ \begin{aligned} 5 x-3 y & =15 \\ -3 y & =-5 x+15 \\ \frac{-3 y}{(-3)} & =\frac{-5 x}{(-3)}+\frac{15}{(-3)} \\ y & =\frac{5}{3} x-5 \end{aligned} $$
(d)(i)
$$ \begin{aligned} & -2 x+3 y=9 \\ & \frac{-2 x}{9}+\frac{3 y}{9}=\frac{9}{9} \\ & -\frac{2 x}{9}+\frac{y}{3}=1 \end{aligned} $$
(d)(ii)
$$ \begin{aligned} -2 x+3 y & =9 \\ 3 y & =2 x+9 \\ \frac{3 y}{3} & =\frac{2 x}{3}+\frac{9}{3} \\ y & =\frac{2}{3} x+3 \end{aligned} $$
Soalan 2:
Tuliskan persamaan garis lurus berikut dalam bentuk ax + by = c dan y = mx + c.
$$ \text { (a) } \frac{x}{4}+\frac{y}{3}=1 $$
$$ \text { (b) }-\frac{x}{3}+\frac{y}{6}=1 $$
$$ \text { (c) } \frac{3 x}{2}+\frac{y}{6}=1 $$
$$ \text { (d) } \frac{2 x}{3}-\frac{y}{4}=1 $$
Penyelesaian:
(a)(i)
$$ \begin{aligned} \frac{x}{4}+\frac{y}{3} & =1 \\ \frac{3 x+4 y}{4(3)} & =1 \\ 3 x+4 y & =12 \end{aligned} $$
(a)(ii)
$$ \begin{aligned} \frac{x}{4}+\frac{y}{3} & =1 \\ \frac{y}{3} & =-\frac{x}{4}+1 \\ \frac{(3) y}{3} & =-\frac{x(3)}{4}+1(3) \\ y & =-\frac{3}{4} x+3 \end{aligned} $$
(b)(i)
$$ \begin{aligned} -\frac{x}{3}+\frac{y}{6} & =1 \\ \frac{-6 x+3 y}{3(6)} & =1 \\ -6 x+3 y & =18 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} -\frac{x}{3}+\frac{y}{6} & =1 \\ \frac{y}{6} & =\frac{x}{3}+1 \\ \frac{(6) y}{6} & =\frac{x(6)}{3}+1(6) \\ y & =2 x+6 \end{aligned} $$
(c)(i)
$$ \begin{gathered} \frac{3 x}{2}+\frac{y}{6}=1 \\ \frac{18 x+2 y}{2(6)}=1 \\ 18 x+2 y=12 \\ 9 x+y=6 \end{gathered} $$
(c)(ii)
$$ \begin{aligned} \frac{3 x}{2}+\frac{y}{6} & =1 \\ \frac{y}{6} & =-\frac{3 x}{2}+1 \\ \frac{(6) y}{6} & =-\frac{3 x(6)}{2}+1(6) \\ y & =-9 x+6 \end{aligned} $$
(d)(i)
$$ \begin{gathered} \frac{2 x}{3}-\frac{y}{4}=1 \\ \frac{8 x-3 y}{3(4)}=1 \\ 8 x-3 y=12 \end{gathered} $$
(d)(ii)
$$ \begin{aligned} \frac{2 x}{3}-\frac{y}{4} & =1 \\ -\frac{y}{4} & =-\frac{2 x}{3}+1 \\ -\frac{(-4) y}{4} & =-\frac{2 x(-4)}{3}+1(-4) \\ y & =\frac{8}{3} x-4 \end{aligned} $$
Tuliskan persamaan garis lurus berikut dalam bentuk ax + by = c dan y = mx + c.
$$ \text { (a) } \frac{x}{4}+\frac{y}{3}=1 $$
$$ \text { (b) }-\frac{x}{3}+\frac{y}{6}=1 $$
$$ \text { (c) } \frac{3 x}{2}+\frac{y}{6}=1 $$
$$ \text { (d) } \frac{2 x}{3}-\frac{y}{4}=1 $$
Penyelesaian:
(a)(i)
$$ \begin{aligned} \frac{x}{4}+\frac{y}{3} & =1 \\ \frac{3 x+4 y}{4(3)} & =1 \\ 3 x+4 y & =12 \end{aligned} $$
(a)(ii)
$$ \begin{aligned} \frac{x}{4}+\frac{y}{3} & =1 \\ \frac{y}{3} & =-\frac{x}{4}+1 \\ \frac{(3) y}{3} & =-\frac{x(3)}{4}+1(3) \\ y & =-\frac{3}{4} x+3 \end{aligned} $$
(b)(i)
$$ \begin{aligned} -\frac{x}{3}+\frac{y}{6} & =1 \\ \frac{-6 x+3 y}{3(6)} & =1 \\ -6 x+3 y & =18 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} -\frac{x}{3}+\frac{y}{6} & =1 \\ \frac{y}{6} & =\frac{x}{3}+1 \\ \frac{(6) y}{6} & =\frac{x(6)}{3}+1(6) \\ y & =2 x+6 \end{aligned} $$
(c)(i)
$$ \begin{gathered} \frac{3 x}{2}+\frac{y}{6}=1 \\ \frac{18 x+2 y}{2(6)}=1 \\ 18 x+2 y=12 \\ 9 x+y=6 \end{gathered} $$
(c)(ii)
$$ \begin{aligned} \frac{3 x}{2}+\frac{y}{6} & =1 \\ \frac{y}{6} & =-\frac{3 x}{2}+1 \\ \frac{(6) y}{6} & =-\frac{3 x(6)}{2}+1(6) \\ y & =-9 x+6 \end{aligned} $$
(d)(i)
$$ \begin{gathered} \frac{2 x}{3}-\frac{y}{4}=1 \\ \frac{8 x-3 y}{3(4)}=1 \\ 8 x-3 y=12 \end{gathered} $$
(d)(ii)
$$ \begin{aligned} \frac{2 x}{3}-\frac{y}{4} & =1 \\ -\frac{y}{4} & =-\frac{2 x}{3}+1 \\ -\frac{(-4) y}{4} & =-\frac{2 x(-4)}{3}+1(-4) \\ y & =\frac{8}{3} x-4 \end{aligned} $$
Soalan 3:
$$ \text { Tuliskan persamaan garis lurus berikut dalam bentuk } a x+b y=c \text { dan } \frac{x}{a}+\frac{y}{b}=1 \text {. } $$
(a) y = 2x + 6
(b) y = 3x – 12
(c) y = –x + 5
(d) y = –2x – 4
Penyelesaian:
(a)(i)
$$ \begin{aligned} y & =2 x+6 \\ -2 x+y & =6 \end{aligned} $$
(a)(ii)
$$ \begin{aligned} y & =2 x+6 \\ -2 x+y & =6 \\ \frac{-2 x}{6}+\frac{y}{6} & =\frac{6}{6} \\ -\frac{x}{3}+\frac{y}{6} & =1 \end{aligned} $$
(b)(i)
$$ \begin{aligned} y & =3 x-12 \\ -3 x+y & =-12 \\ 3 x-y & =12 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} y & =3 x-12 \\ -3 x+y & =-12 \\ \frac{-3 x}{(-12)}+\frac{y}{(-12)} & =\frac{-12}{(-12)} \\ \frac{x}{4}-\frac{y}{12} & =1 \end{aligned} $$
(c)(i)
$$ \begin{aligned} y & =-x+5 \\ x+y & =5 \end{aligned} $$
(c)(ii)
$$ \begin{aligned} y & =-x+5 \\ x+y & =5 \\ \frac{x}{5}+\frac{y}{5} & =\frac{5}{5} \\ \frac{x}{5}+\frac{y}{5} & =1 \end{aligned} $$
(d)(i)
$$ \begin{aligned} y & =-2 x-4 \\ 2 x+y & =-4 \end{aligned} $$
(d)(ii)
$$ \begin{aligned} y & =-2 x-4 \\ 2 x+y & =-4 \\ \frac{2 x}{(-4)}+\frac{y}{(-4)} & =\frac{-4}{(-4)} \\ -\frac{x}{2}-\frac{y}{4} & =1 \end{aligned} $$
$$ \text { Tuliskan persamaan garis lurus berikut dalam bentuk } a x+b y=c \text { dan } \frac{x}{a}+\frac{y}{b}=1 \text {. } $$
(a) y = 2x + 6
(b) y = 3x – 12
(c) y = –x + 5
(d) y = –2x – 4
Penyelesaian:
(a)(i)
$$ \begin{aligned} y & =2 x+6 \\ -2 x+y & =6 \end{aligned} $$
(a)(ii)
$$ \begin{aligned} y & =2 x+6 \\ -2 x+y & =6 \\ \frac{-2 x}{6}+\frac{y}{6} & =\frac{6}{6} \\ -\frac{x}{3}+\frac{y}{6} & =1 \end{aligned} $$
(b)(i)
$$ \begin{aligned} y & =3 x-12 \\ -3 x+y & =-12 \\ 3 x-y & =12 \end{aligned} $$
(b)(ii)
$$ \begin{aligned} y & =3 x-12 \\ -3 x+y & =-12 \\ \frac{-3 x}{(-12)}+\frac{y}{(-12)} & =\frac{-12}{(-12)} \\ \frac{x}{4}-\frac{y}{12} & =1 \end{aligned} $$
(c)(i)
$$ \begin{aligned} y & =-x+5 \\ x+y & =5 \end{aligned} $$
(c)(ii)
$$ \begin{aligned} y & =-x+5 \\ x+y & =5 \\ \frac{x}{5}+\frac{y}{5} & =\frac{5}{5} \\ \frac{x}{5}+\frac{y}{5} & =1 \end{aligned} $$
(d)(i)
$$ \begin{aligned} y & =-2 x-4 \\ 2 x+y & =-4 \end{aligned} $$
(d)(ii)
$$ \begin{aligned} y & =-2 x-4 \\ 2 x+y & =-4 \\ \frac{2 x}{(-4)}+\frac{y}{(-4)} & =\frac{-4}{(-4)} \\ -\frac{x}{2}-\frac{y}{4} & =1 \end{aligned} $$