Question

Solve the following equations by Cramer’s method.
(1) 6x – 3y = –10 ; 3x + 5y – 8 = 0
(2) 4m – 2n = –4 ; 4m + 3n = 16
(3) 3x – 2y = $\frac{5}{2}$ ; $\frac{1}{3}x+3y=-\frac{4}{3}$
(4) 7x + 3y = 15 ; 12y – 5x = 39
(5) $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$

Answer 1

(1) 6x – 3y = –10 ; 3x + 5y – 8 = 0
$$\begin{gathered} D=\left|\begin{array}{cc}6& -3\\ 3& 5\end{array}\right|=6\times 5-\left(-3\right)\times 3=30+9=39 \\ {D}_x=\left|\begin{array}{cc}-10& -3\\ 8& 5\end{array}\right|=-10\times 5-\left(-3\right)\times 8=-50+24=-26 \\ {D}_y=\left|\begin{array}{cc}6& -10\\ 3& 8\end{array}\right|=6\times 8-\left(-10\right)\times 3=48+30=78 \\ x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} \\ y=\frac{D_y}{D}=\frac{78}{39}=2 \\ \left(x,y\right)=\left(\frac{-2}{3},2\right) \end{gathered}$$

(2) 4m – 2n = –4 ; 4m + 3n = 16
$$\begin{gathered} D=\left|\begin{array}{cc}4& -2\\ 4& 3\end{array}\right|=4\times 3-\left(-2\right)\times 4=12+8=20 \\ {D}_x=\left|\begin{array}{cc}-4& -2\\ 16& 3\end{array}\right|=-4\times 3-\left(-2\right)\times 16=-12+32=20 \\ {D}_y=\left|\begin{array}{cc}4& -4\\ 4& 16\end{array}\right|=4\times 16-\left(-4\right)\times 4=64+16=80 \\ x=\frac{D_x}{D}=\frac{20}{20}=1 \\ y=\frac{D_y}{D}=\frac{80}{20}=4 \\ \left(x,y\right)=\left(1,4\right) \end{gathered}$$
(3) 3x – 2y = $\frac{5}{2}$ ; $\frac{1}{3}x+3y=-\frac{4}{3}$
$$\begin{gathered} D=\left|\begin{array}{cc}3& -2\\ \frac{1}{3}& 3\end{array}\right|=9+\frac{2}{3}=\frac{29}{3} \\ {D}_x=\left|\begin{array}{cc}\frac{5}{2}& -2\\ \frac{-4}{3}& 3\end{array}\right|=\frac{15}{2}-\frac{8}{3}=\frac{29}{6} \\ {D}_y=\left|\begin{array}{cc}3& \frac{5}{2}\\ \frac{1}{3}& \frac{-4}{3}\end{array}\right|=-4-\frac{5}{6}=\frac{-29}{6} \\ x=\frac{D_x}{D}=\frac{{\displaystyle \frac{29}{6}}}{{\displaystyle \frac{29}{3}}}=\frac{1}{2} \\ y=\frac{D_y}{D}=\frac{\frac{-29}{6}}{\frac{29}{3}}=\frac{-1}{2} \\ \left(x,y\right)=\left(\frac{1}{2},\frac{-1}{2}\right) \end{gathered}$$

(4) 7x + 3y = 15 ; 12y – 5x = 39
$$\begin{gathered} D=\left|\begin{array}{cc}7& 3\\ -5& 12\end{array}\right|=7\times 12-\left(-5\right)\times 3=84+15=99 \\ {D}_x=\left|\begin{array}{cc}15& 3\\ 39& 12\end{array}\right|=15\times 12-39\times 3=180-117=63 \\ {D}_y=\left|\begin{array}{cc}7& 15\\ -5& 39\end{array}\right|=7\times 39-\left(-5\right)\times 15=273+75=348 \\ x=\frac{D_x}{D}=\frac{{\displaystyle 63}}{{\displaystyle 99}}=\frac{7}{11} \\ y=\frac{D_y}{D}=\frac{348}{99}=\frac{116}{33} \\ \left(x,y\right)=\left(\frac{7}{11},\frac{116}{33}\right) \end{gathered}$$

(5) $\frac{x+y-8}{2}=\frac{x+2y-14}{3}=\frac{3x-y}{4}$
$$\begin{gathered} \frac{x+y-8}{2}=\frac{x+2y-14}{3} \\ \Rightarrow 3x+3y-24=2x+4y-28 \\ \Rightarrow x-y=-4.....\left(\mathrm{I}\right) \\ \mathrm{and}\frac{x+2y-14}{3}=\frac{3x-y}{4} \\ \Rightarrow 4x+8y-56=9x-3y \\ \Rightarrow 5x-11y=-56.....\left(\mathrm{II}\right) \end{gathered}$$

From (I) and (II)
$$\begin{gathered} D=\left|\begin{array}{cc}1& -1\\ 5& -11\end{array}\right|=-11\times 1-\left(-1\right)\times 5=-11+5=-6 \\ {D}_x=\left|\begin{array}{cc}-4& -1\\ -56& -11\end{array}\right|=-11\times \left(-4\right)-\left(-1\right)\times \left(-56\right)=44-56=-12 \\ {D}_y=\left|\begin{array}{cc}1& -4\\ 5& -56\end{array}\right|=-56\times 1-\left(-4\right)\times 5=-56+20=-36 \\ x=\frac{D_x}{D}=\frac{-12}{-6}=2 \\ y=\frac{D_y}{D}=\frac{-36}{-6}=6 \\ \left(x,y\right)=\left(2,6\right) \end{gathered}$$