If r and s are positive integers, each greater than 1, and if 11(s-1) =13(r-1), what is the least possible value of (r+s)?

The least value of (r+s) is (12+14)=26

Step-by-step explanation:

Well, first let us solve equation of 11(s-1)=13(r-1), which results in 11s-11=13r-13. Hence, 11s+2=13r. It is stated that r and s both are integers and greater than 1.

To make sure that r and s are integers, the least value of s must be equal to 14 (s=14) then the least value of r becomes 12 (r=12).

Finally, the least value of (r+s) is (12+14)=26.

arrange the equations with like terms in columns.

analyze the coefficients of x or y.

add the equations and solve for the remaining variable.

substitute the value into either equation and solve.

check the solution.

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.