Soalan 3(g):
Selesaikan yang berikut.
Diberi √3a = 9b − 4c, hitung
(i) nilai a apabila b = 1/3 dan c = ½
(ii) nilai b apabila c = 3 dan a = 12
(iii) nilai c apabila a = 3 dan b = 3
Penyelesaian:
(g)(i)
$$ \begin{aligned} &\text { nilai } a \text { apabila } b=\frac{1}{3} \text { dan } c=\frac{1}{2}\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3 a} & =9\left(\frac{1}{3}\right)-4\left(\frac{1}{2}\right) \\ \sqrt{3 a} & =3-2 \\ \sqrt{3 a} & =1 \\ (\sqrt{3 a})^2 & =1^2 \\ 3 a & =1 \\ a & =\frac{1}{3} \end{aligned} \end{aligned} $$
(g)(ii)
$$ \begin{aligned} &\text { nilai } b \text { apabila } c=3 \text { dan } a=12\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3(12)} & =9 b-4(3) \\ \sqrt{36} & =9 b-12 \\ 6 & =9 b-12 \\ 18 & =9 b \\ 2 & =b \\ b & =2 \end{aligned} \end{aligned} $$
(g)(iii)
$$ \begin{aligned} &\text { nilai } c \text { apabila } a=3 \text { dan } b=3\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3(3)} & =9(3)-4 c \\ \sqrt{9} & =27-4 c \\ 3 & =27-4 c \\ 4 c & =24 \\ c & =6 \end{aligned} \end{aligned} $$
Selesaikan yang berikut.
Diberi √3a = 9b − 4c, hitung
(i) nilai a apabila b = 1/3 dan c = ½
(ii) nilai b apabila c = 3 dan a = 12
(iii) nilai c apabila a = 3 dan b = 3
Penyelesaian:
(g)(i)
$$ \begin{aligned} &\text { nilai } a \text { apabila } b=\frac{1}{3} \text { dan } c=\frac{1}{2}\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3 a} & =9\left(\frac{1}{3}\right)-4\left(\frac{1}{2}\right) \\ \sqrt{3 a} & =3-2 \\ \sqrt{3 a} & =1 \\ (\sqrt{3 a})^2 & =1^2 \\ 3 a & =1 \\ a & =\frac{1}{3} \end{aligned} \end{aligned} $$
(g)(ii)
$$ \begin{aligned} &\text { nilai } b \text { apabila } c=3 \text { dan } a=12\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3(12)} & =9 b-4(3) \\ \sqrt{36} & =9 b-12 \\ 6 & =9 b-12 \\ 18 & =9 b \\ 2 & =b \\ b & =2 \end{aligned} \end{aligned} $$
(g)(iii)
$$ \begin{aligned} &\text { nilai } c \text { apabila } a=3 \text { dan } b=3\\ &\begin{aligned} \sqrt{3 a} & =9 b-4 c \\ \sqrt{3(3)} & =9(3)-4 c \\ \sqrt{9} & =27-4 c \\ 3 & =27-4 c \\ 4 c & =24 \\ c & =6 \end{aligned} \end{aligned} $$
Soalan 3(h):
Selesaikan yang berikut.
$$ \text { Diberi } 1 \frac{1}{2} s=\frac{3}{5} t^2+\frac{1}{3} u^2 \text {, hitung } $$
$$ \text { (i) nilai } s \text { apabila } t=-5 \text { dan } u=3 $$
$$ \text { (ii) nilai } t \text { apabila } u=-6 \text { dan } s=28 $$
$$ \text { (iii) nilai } u \text { apabila } s=\frac{4}{6} \text { dan } t=\frac{5}{6} $$
Penyelesaian:
(h)(i)
$$ \begin{aligned} &\text { nilai } s \text { apabila } t=-5 \text { dan } u=3\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2} s & =\frac{3}{5}(-5)^2+\frac{1}{3}(3)^2 \\ \frac{3}{2} s & =\frac{3}{5}(25)+\frac{1}{3}(9) \\ \frac{3}{2} s & =15+3 \\ \frac{3}{2} s & =18 \\ s & =(18)\left(\frac{2}{3}\right) \\ s & =12 \end{aligned} \end{aligned} $$
(h)(ii)
$$ \begin{aligned} &\text { Nilai } t \text { apabila } u=-6 \text { dan } s=28\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2}(28) & =\frac{3}{5} t^2+\frac{1}{3}(-6)^2 \\ 42 & =\frac{3}{5} t^2+\frac{1}{3}(36) \\ 42-12 & =\frac{3}{5} t^2 \\ 30 & =\frac{3}{5} t^2 \\ (30)\left(\frac{5}{3}\right) & =t^2 \\ 50 & =t^2 \\ \sqrt{50} & =t \\ t & =\sqrt{50} \end{aligned} \end{aligned} $$
(h)(iii)
$$ \begin{aligned} &\text { Nilai } u \text { apabila } s=\frac{4}{6} \text { dan } t=\frac{5}{6}\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2}\left(\frac{4}{6}\right) & =\frac{3}{5}\left(\frac{5}{6}\right)^2+\frac{1}{3} u^2 \\ 1 & =\frac{3}{5}\left(\frac{25}{36}\right)+\frac{1}{3} u^2 \\ 1 & =\frac{5}{12}+\frac{1}{3} u^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} 1-\frac{5}{12} & =\frac{1}{3} u^2 \\ \frac{7}{12} & =\frac{1}{3} u^2 \\ \frac{7}{12} \times 3 & =u^2 \\ \frac{7}{4} & =u^2 \\ \sqrt{\frac{7}{4}} & =u \\ u & =\sqrt{\frac{7}{4}} \end{aligned} $$
Selesaikan yang berikut.
$$ \text { Diberi } 1 \frac{1}{2} s=\frac{3}{5} t^2+\frac{1}{3} u^2 \text {, hitung } $$
$$ \text { (i) nilai } s \text { apabila } t=-5 \text { dan } u=3 $$
$$ \text { (ii) nilai } t \text { apabila } u=-6 \text { dan } s=28 $$
$$ \text { (iii) nilai } u \text { apabila } s=\frac{4}{6} \text { dan } t=\frac{5}{6} $$
Penyelesaian:
(h)(i)
$$ \begin{aligned} &\text { nilai } s \text { apabila } t=-5 \text { dan } u=3\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2} s & =\frac{3}{5}(-5)^2+\frac{1}{3}(3)^2 \\ \frac{3}{2} s & =\frac{3}{5}(25)+\frac{1}{3}(9) \\ \frac{3}{2} s & =15+3 \\ \frac{3}{2} s & =18 \\ s & =(18)\left(\frac{2}{3}\right) \\ s & =12 \end{aligned} \end{aligned} $$
(h)(ii)
$$ \begin{aligned} &\text { Nilai } t \text { apabila } u=-6 \text { dan } s=28\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2}(28) & =\frac{3}{5} t^2+\frac{1}{3}(-6)^2 \\ 42 & =\frac{3}{5} t^2+\frac{1}{3}(36) \\ 42-12 & =\frac{3}{5} t^2 \\ 30 & =\frac{3}{5} t^2 \\ (30)\left(\frac{5}{3}\right) & =t^2 \\ 50 & =t^2 \\ \sqrt{50} & =t \\ t & =\sqrt{50} \end{aligned} \end{aligned} $$
(h)(iii)
$$ \begin{aligned} &\text { Nilai } u \text { apabila } s=\frac{4}{6} \text { dan } t=\frac{5}{6}\\ &\begin{aligned} 1 \frac{1}{2} s & =\frac{3}{5} t^2+\frac{1}{3} u^2 \\ \frac{3}{2}\left(\frac{4}{6}\right) & =\frac{3}{5}\left(\frac{5}{6}\right)^2+\frac{1}{3} u^2 \\ 1 & =\frac{3}{5}\left(\frac{25}{36}\right)+\frac{1}{3} u^2 \\ 1 & =\frac{5}{12}+\frac{1}{3} u^2 \end{aligned} \end{aligned} $$
$$ \begin{aligned} 1-\frac{5}{12} & =\frac{1}{3} u^2 \\ \frac{7}{12} & =\frac{1}{3} u^2 \\ \frac{7}{12} \times 3 & =u^2 \\ \frac{7}{4} & =u^2 \\ \sqrt{\frac{7}{4}} & =u \\ u & =\sqrt{\frac{7}{4}} \end{aligned} $$