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Systems of Equations

edited Sun Jul 30, 2017 3:40 pm in Lesson 2
fontainedfontained Joined: 07/22/2017Posts: 24
Section Lesson 2, Systems of Equations Problem set---question #3

If cd = 6 and 2c + 2d = 4, then 2c²d + 2cd² = 

A. 2
B. 10
C. 24
D. 36
E. 48

I tried working it out---2*c*c*d +2*c*d*d =... but couldn't get it.


Posts

  • edited Sun Jul 30, 2017 3:39 pm
    Jonathan EvansJonathan Evans PowerScore Staff Joined: 10/31/2016Posts: 109
    Hi, Fontained, 

    As with many algebra problems involving binomials or quadratics, the name of the game is to get rid of as many variables as possible.

    Start by using cd = 6 to get rid of every cd in 2c²d + 2cd².

    Now we have 2c × 6 + 2d × 6 =

    Now factor out 6.

    You get 6(2c + 2d) =

    Notice that we know 2c + 2d = 4

    So 6 × 4 = 24.

    What are the keys here?
    1. Record What You Know. 
    2. Keep Your Pencil Moving. 
    When you're dealing with binomials, roots, exponents, or quadratics, the key will often be factoring.

    I've played around with this problem a bit, and this approach is pretty clearly the most direct solution. Reverse foil is frustrating, and the quadratic formula (which you do not need for the GRE) gives complex (imaginary) roots (which again you do not need for the GRE). Thus, the factoring and substitution approach outlined above is far and away the best approach to this problem.

    If you find yourself stuck in a complicated mess of algebra, do a gut check and remind yourself that there's always an easier approach. 

    I hope this helps!


  • fontainedfontained Joined: 07/22/2017Posts: 24
    That makes perfect sense, thank you! 
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