Solve the system for x and y using Cramerʹs rule. Assume a and b are nonzero constants.
-$$\begin{array} { l }
\frac { 1 } { a } x + \frac { 1 } { b } y = a b \
x + y = a
\end{array}$$
A) $\left. \left\{ \frac { a ( b - a ) \left( b ^ { 2 } - 1 \right) } { a b ^ { 2 } } , \frac { ( b - a ) ( 1 - a b ) } { a b } \right) \right\}$
B) $\left\{ \left( a b ( 1 - a b ) , a ^ { 2 } \left( b ^ { 2 } - 1 \right) \right) \right\}$
C) $\left\{ \left( \frac { a ^ { 2 } \left( b ^ { 2 } - 1 \right) } { b - a } , \frac { a b ( 1 - a b ) } { b - a } \right) \right\}$
D) $\left\{ \left( \frac { a ^ { 2 } b \left( b ^ { 2 } - 1 \right) } { b - a } , \frac { a b ( 1 - a b ) } { b - a } \right) \right\}$

Question

Solve the system for x and y using Cramerʹs rule. Assume a and b are nonzero constants.
-$$\begin{array} { l } 
\frac { 1 } { a } x + \frac { 1 } { b } y = a b \
x + y = a
\end{array}$$
A) $\left. \left\{ \frac { a ( b - a ) \left( b ^ { 2 } - 1 \right) } { a b ^ { 2 } } , \frac { ( b - a ) ( 1 - a b ) } { a b } \right) \right\}$
B) $\left\{ \left( a b ( 1 - a b ) , a ^ { 2 } \left( b ^ { 2 } - 1 \right) \right) \right\}$
C) $\left\{ \left( \frac { a ^ { 2 } \left( b ^ { 2 } - 1 \right) } { b - a } , \frac { a b ( 1 - a b ) } { b - a } \right) \right\}$
D) $\left\{ \left( \frac { a ^ { 2 } b \left( b ^ { 2 } - 1 \right) } { b - a } , \frac { a b ( 1 - a b ) } { b - a } \right) \right\}$

Quizplus · Accepted Answer

Cramer's rule involves finding determinants. The determinant of the coefficient matrix is $\frac{1}{a} \cdot 1 - \frac{1}{b} \cdot 1 = \frac{b-a}{ab}$. Solving for x and y using Cramer's rule gives the solutions in option C.

Solve the System for X and Y Using Cramerʹs Rule 1ax+1by=abx+y=a\begin{array} { l } \frac { 1 } { a } x + \frac { 1 } { b } y = a b \\x + y = a\end{array}a1​x+b1​y=abx+y=a​

Solve the System for X and Y Using Cramerʹs Rule $\begin{array} { l } \frac { 1 } { a } x + \frac { 1 } { b } y = a b \\x + y = a\end{array}$