Solve by any method. Assume $a$ and $b$ represent nonzero constants.
-$a x+b y=1$
Ax - by $=14$

A) $\left\{\left(\frac{15}{2 a},-\frac{13}{2 b}\right)\right\}$

B) $\left\{\left\{\frac{13}{a},-\frac{15}{2 b}\right)\right\}$

C) $\left\{\left(\frac{15}{2 a}, \frac{13}{2 b}\right)\right\}$

D) $\left\{\left(-\frac{15}{2 a},-\frac{13}{2 b}\right)\right\}$

Question

Solve by any method. Assume $a$ and $b$ represent nonzero constants.
-$a x+b y=1$
Ax - by $=14$

A) $\left\{\left(\frac{15}{2 a},-\frac{13}{2 b}\right)\right\}$

B) $\left\{\left\{\frac{13}{a},-\frac{15}{2 b}\right)\right\}$

C) $\left\{\left(\frac{15}{2 a}, \frac{13}{2 b}\right)\right\}$

D) $\left\{\left(-\frac{15}{2 a},-\frac{13}{2 b}\right)\right\}$

Quizplus · Accepted Answer

To solve the system of equations $ax + by = 1$ and $Ax - by = 14$, we can add the two equations to eliminate $y$, resulting in $a x + A x = 15$, which simplifies to $x = \frac{15}{a + A}$. Since $A = a$, $x = \frac{15}{2a}$. Subtracting the second equation from the first gives $2by = -13$, so $y = -\frac{13}{2b}$. Therefore, the solution is $\left(\frac{15}{2a}, -\frac{13}{2b}\right)$.

Solve by Any Method aaa And bbb Represent Nonzero Constants ax+by=1a x+b y=1ax+by=1

Solve by Any Method $a$ And $b$ Represent Nonzero Constants $a x+b y=1$