Solve a System of Linear Equations in Two Variables Using Cramer's Rule
-$$\begin{array} { l }
7 x + 8 y = 4 \
2 x + 3 y = 2
\end{array}$$
A) $\left\{ \left( - \frac { 4 } { 5 } , \frac { 6 } { 5 } \right) \right\}$
B) $\left\{ \left( \frac { 6 } { 5 } , - \frac { 4 } { 5 } \right) \right\}$
C) $\left\{ \left( - \frac { 5 } { 4 } , \frac { 5 } { 6 } \right) \right\}$
D) $\left\{ \left( \frac { 28 } { 37 } , \frac { 22 } { 37 } \right) \right\}$

Question

Solve a System of Linear Equations in Two Variables Using Cramer's Rule
-$$\begin{array} { l } 
7 x + 8 y = 4 \
2 x + 3 y = 2
\end{array}$$
A) $\left\{ \left( - \frac { 4 } { 5 } , \frac { 6 } { 5 } \right) \right\}$
B) $\left\{ \left( \frac { 6 } { 5 } , - \frac { 4 } { 5 } \right) \right\}$
C) $\left\{ \left( - \frac { 5 } { 4 } , \frac { 5 } { 6 } \right) \right\}$
D) $\left\{ \left( \frac { 28 } { 37 } , \frac { 22 } { 37 } \right) \right\}$

Quizplus · Accepted Answer

Cramer's Rule uses determinants to solve a system of linear equations. For the given system:$$D = \begin{vmatrix} 7 & 8 \ 2 & 3 \end{vmatrix} = (7*3) - (8*2) = 21 - 16 = 5$$$$D_x = \begin{vmatrix} 4 & 8 \ 2 & 3 \end{vmatrix} = (4*3) - (8*2) = 12 - 16 = -4$$$$D_y = \begin{vmatrix} 7 & 4 \ 2 & 2 \end{vmatrix} = (7*2) - (4*2) = 14 - 8 = 6$$$$x = \frac{D_x}{D} = \frac{-4}{5}$$$$y = \frac{D_y}{D} = \frac{6}{5}$$Thus, the solution is $\left( - \frac { 4 } { 5 } , \frac { 6 } { 5 } \right)$.

Solve a System of Linear Equations in Two Variables Using 7x+8y=42x+3y=2\begin{array} { l } 7 x + 8 y = 4 \\2 x + 3 y = 2\end{array}7x+8y=42x+3y=2​

Solve a System of Linear Equations in Two Variables Using $\begin{array} { l } 7 x + 8 y = 4 \\2 x + 3 y = 2\end{array}$