Solving for unknown twice in equation. Example 8 in Reading 35.

I’m not very good at algebra and normally learn from examples how to do it and repeat the process because I never learned it properly. Side note if anyone knows a good online course for algebra help for solving questions like this please let me know.

In example 8 it has: s2/1.05 + s2/1.06^2 + 1/1.06^2 = 1 and it just has the solution as 5.97%. I can’t find anything similar on how to do it and can’t remember myself how to do it. I’ve had teachers help me in the past but since I’m out of school now I can’t for the life of me remember how to solve something like this.


instead of keep the /1.05

convert it to

s2*0.952381 + s2*0.889996=1-1/1.06^2 = 1-0.889996 = 0.110004

s2*1.842377 = 0.110004

s2 = 0.0597076

= 5.97%

I always find it easier to have the unknown multiplied by a coefficient, rather than divided by one, so here’s what I would do:

1 / 1.05 = 0.9524

1 / 1.06² = 0.8900

S2 / 1.05 + S2 / 1.06² + 1 / 1.06² = 1

(0.9524)S2 + (0.8900)S2 + 0.8900 = 1

(1.8424)S2 = 1 – 0.8900 = 0.1100

S2 = 0.1100 / 1.8424 = 0.059705 = 5.9705%

Honestly, thank you so much both for the step by step process there. I managed to get to the .11 but was still confused on the left hand side but now I see how it works. Makes me realize I definitely need to put some work in to learn algebra before the exam. I really appreciate the help, thanks.

My pleasure.

Get yourself a copy of Algebra for Dummies; it’ll help.

In two weeks I start teaching two first-semester calculus classes at Chapman University. I will guaranty you that I will have a number of students who will each, at some point in the semester, perpetrate what I call _ weird algebra _. I can but hope and pray that it occurs early, and that I can correct it once and for all.

Best of luck!