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whpalmer4Best ResponseYou've already chosen the best response.0
First, use the distributive property: a(bc) = ab  ac Then add and subtract the same thing from both sides to end up with all the y terms on the left side, and the leftover numbers on the right side. If y still has a number in front of it (coefficient), divide both sides by that number to give you y = <the answer> For example: 0.6(y2) + 0.5y = 0.3y  .2(10) 0.6y  1.2 + 0.5y = 0.3y  2 1.1y  1.2 = 0.3y 2 Add 1.2 to both sides 1.1y  1.2 + 1.2 = 0.3y  2 + 1.2 1.1y = 0.3y  0.8 now subtract 0.3y from both sides 1.1y  0.3y = 0.3y  0.3y  0.8 0.8y = 0.8 Now divide both sides by 0.8 0.8y / 0.8 = 0.8 / 0.8 y = 1 That's *not* your problem, so the answer probably isn't the same! I'm just illustrating typical steps to solve such a problem.
 one year ago

ZeHanzBest ResponseYou've already chosen the best response.0
Don't know about you, but I *hate* decimals in equations! So I always try to get rid of them. Everything gets much clearer then. If you multiply everything with 100, you get: 4(y4)+2y=14y3*30. See: looks much better! Now do the next steps yourself, maybe with a little help from @whpalmer4's scheme...
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.0
Yep, I agree with @ZeHand, eliminating fractions and decimals first often makes the work easier.
 one year ago

whpalmer4Best ResponseYou've already chosen the best response.0
But it isn't *necessary* to solve this problem, so I left it out. It is good to do some practice with decimals, as you'll likely encounter many problems later where you can't eliminate them,, and being proficient will make life easier. My philosophy is that you should definitely seek to find the easy route to solve something, but practice the hard routes too, so when you are forced to use them, you may grumble and curse a bit, but you still get the right answer!
 one year ago
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