Thread: Solving the system of two equations in two unknowns

I'm going through these tutorials for Calc III, and I came upon this slide:

main-22.pdf

source:MATH 311 - Calculus III Resources

How did the author solve for s and t?

Originally Posted by Lexy

I'm going through these tutorials for Calc III, and I came upon this slide:

main-22.pdf

source:MATH 311 - Calculus III Resources

How did the author solve for s and t?

first equation ...

1 + t = 2s , solve for t

t = 2s - 1

substitute (2s - 1) for t in the second equation ... solve for s, then determine the value of t

That's one good method. Another is this: the two equations are 1+ t= 2s and -2+ 3t= 3+s. Multiply the first equation by 3 to get 3+ 3t= 6s. Now subtract the second equation from that: (3+ 3t)- (-2+ 3t)= 6s- (3+ s). The "t" terms cancel giving 5= 5s- 3. Solve that for s then put that value into either of the original equations to get an equation to solve for t.