If 2x^(2a − b − 1) − 3y^(3a + 2b − 16) = 10 is a linear equation in two variables, find a and b.

Step-by-step solution

1. 1

   Translate "linear in two variables" into conditions on the exponents. By definition such an equation has both $x$ and $y$ appearing to the first power with nonzero coefficients. Since the coefficients $2$ and $-3$ are already nonzero, the only requirement is

   $2a-b-1=1\qquad\text{and}\qquad 3a+2b-16=1$

2. 2

   Tidy the two conditions into a system.

   $\begin{cases}2a-b=2\\3a+2b=17\end{cases}$

3. 3

   Solve by substitution. From the first equation, $b=2a-2$. Substituting into the second:

   $3a+2(2a-2)=17\quad\Longrightarrow\quad 3a+4a-4=17\quad\Longrightarrow\quad 7a=21\quad\Longrightarrow\quad a=3$

4. 4

   Back-substitute for b.

   $b=2(3)-2=4$

   $\boxed{a=3,\ b=4}$

5. 5

   Verify both exponents really become 1. $2a-b-1=6-4-1=1$ ✓ and $3a+2b-16=9+8-16=1$ ✓, so the equation becomes $2x-3y=10$ — visibly linear. Checking both exponents (not just one) matters, because a pair satisfying only one condition would leave a stray square or cube behind.