Question
Solve the following equations by Cramer’s method.
6x – 3y = –10; 3x + 5y – 8 = 0
6x – 3y = –10; 3x + 5y – 8 = 0
✓
Answer
$6x-3y = -10$
$3 x+5 y=8 $
$D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39$
$ D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 $
$ D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \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 $
$ \therefore(x, y)=\left(-\frac{2}{3}, 2\right)$
$3 x+5 y=8 $
$D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39$
$ D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 $
$ D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \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 $
$ \therefore(x, y)=\left(-\frac{2}{3}, 2\right)$
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