Solve the system for x and y using Cramer's rule. Assume a and b are nonzero constants.
-$$\begin{array} { l }
a ^ { 2 } x + b ^ { 2 } y = a ^ { 2 } \
a x + b y = b
\end{array}$$
A) $\left\{ \left\{ \frac { a + b } { a } , \frac { a } { b } \right) \right\}$

B) $\left\{ \left( \frac { a - b } { a } , - \frac { a } { b } \right) \right\}$

C) $\left\{ \left( \frac { a + b } { a } , - \frac { a } { b } \right) \right\}$

D) $\left\{ \left\{ \frac { a + b } { a } , \frac { b - a } { b } \right) \right\}$

Question

Solve the system for x and y using Cramer's rule. Assume a and b are nonzero constants.
-$$\begin{array} { l } 
a ^ { 2 } x + b ^ { 2 } y = a ^ { 2 } \
a x + b y = b
\end{array}$$
A) $\left\{ \left\{ \frac { a + b } { a } , \frac { a } { b } \right) \right\}$

B) $\left\{ \left( \frac { a - b } { a } , - \frac { a } { b } \right) \right\}$

C) $\left\{ \left( \frac { a + b } { a } , - \frac { a } { b } \right) \right\}$

D) $\left\{ \left\{ \frac { a + b } { a } , \frac { b - a } { b } \right) \right\}$

Quizplus · Accepted Answer

Cramer's rule involves calculating determinants. The determinant of the coefficient matrix is $D = ab^2 - a^2b = ab(b-a)$. For $x$, replace the first column with the constants to get $D_x = b^2b - a^2b = b^3 - a^2b$, so $x = D_x / D = (b^3 - a^2b) / ab(b-a) = (b^2 - a^2) / a(b-a) = (b+a)(b-a) / a(b-a) = (b+a)/a$. For $y$, replace the second column with the constants to get $D_y = a^2a - a^2b = a^3 - a^2b$, so $y = D_y / D = (a^3 - a^2b) / ab(b-a) = a(a^2 - ab) / ab(b-a) = a(a-b) / b(b-a) = -a/b$. Thus, the solution is $\left( \frac { a + b } { a } , - \frac { a } { b } \right)$.

Solve the System for X and Y Using Cramer's Rule a2x+b2y=a2ax+by=b\begin{array} { l } a ^ { 2 } x + b ^ { 2 } y = a ^ { 2 } \\a x + b y = b\end{array}a2x+b2y=a2ax+by=b​

Solve the System for X and Y Using Cramer's Rule $\begin{array} { l } a ^ { 2 } x + b ^ { 2 } y = a ^ { 2 } \\a x + b y = b\end{array}$