 XP Math - Forums - View Single Post - What is the process in Algebra if there are two unknown numbers?
 View Single Post 07-30-2007 #3 kael Guest   Posts: n/a hi there...^,^if there are two unknowns then most probably you'll have 2 equations to solve that problem.so here goes...lets take the first no. as x and then the second one as y...for simplicitythe first eq isx + y = 45 (eq1)the second eq would bex/y + y/x = 2.05 (eq2)from eq1 we have x = 45 - yand then substituting this to eq2 we have(45-y)/y +y(45-y) = 2.05that eliminates one unknown leading to the other...^,^that equation is equivalent to:[(45-y)(45-y) + y]/(45-y)y = 2.05next step, is to transpose the denominator to the other side...so we have...(45-y)Â²+ y = 2.05(45-y)yexpanding it...45Â² - 90y + yÂ² = 2.05(45y) - 2.05yÂ²simplifyin...2025 - 90y - 92.25y + yÂ² + 2.05yÂ² = 0voila! we have a quadratic equation...3.05yÂ² - 182.25y + 2025 = 0 _______y = [bÂ±âˆš(bÂ² - 4ac)]/2a ___________________y = [182.25Â±âˆš(182.25Â² - 4(3.05)(2025)]/2(3.05)y = (182.25Â±92.25)/6.1y = 45 or y = 14.7545 would be an improbable answer since the sum of the two numbers is 45! get it? it's like this...let's look at eq2 which is -----> x/y + y/x = 2.05if y is equal to 45 then x would be 0 which means that it'll not satisfy the second eq...ok?therefore, y = 14.75 since, x = 45 - ythen, x = 30.25So, basically you are just goin to identify first the two availble equations...then combine it to eliminate one of the two unknowns...and voila! you'll get the answer...hope this will you help you...^-^â€*Â¬Ù¼ 