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Topic: Can you help solve this equation 6(-4) = 2/3 + 8?**Question:**
been out of school for the last 15 years so i am trying to study for a test to go to community college

July 18, 2019 / By Derick

method 1: multiply both sides of your equation by 3 (which is the number in the denominator), this way you're not working with fractions which makes things much easier. (3)6(-4) = 3[2/3 + 8] => 18(-4) = 2 + 24 => -72 = 26 as you can see we have a problem because -72 is not equal to 26. I'm assuming there's a variable somewhere that you need to solve for. Hopefully you got the idea on how to approach this type of problem. You can also turn the 2/3 into decimal which gives you 1.667 and work with that.

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Did you like the answer? 2x + 3(11/3) = 12 3/1 * 11/3 = 33/3 When multiplying, just multiply across, numerator to numerator and denominator to denominator. (I know I put a 1 under the 3, all whole numbers have a 1 under them) (example : 3/1 = 3) or ( lets say 5 in fraction form is 5/1 ) (Another fraction example 2/5 * 3/4 mult across and you get 2 * 3 = 6 and 5 * 4 = 20 answer 6/20 then, simplify and get 3/10) Now simplify 33/3 = 11 2x + 11 = 12 Now subtract 11 from both sides. and you get 2x = 1 2x = 1 Now divide 2 on both sides and you get x = 1/2 ordered pair is (1/2 , 11/3)

2x + 3(11/3) = 12 look at 3(11/3) = (3 * 11) / 3 = 11 ( 3/3) = 11 that gives you 2x + 11 = 12 2x = 1 x = 1/2 the point (1/2 , 11/3) is a solution to the equation remember, when you multiply fractions, you multiply the tops and multiply the bottoms if there isn't a bottom for one of the fractions, treat it like 1 3(11/3) = (3/1)(11/3) = (3*11) / (1 * 3) = 33 / 3 = 11

2x + 3 (11/3) = 12 first you multiply the 3 into 11/3 and you'll now have 2x + 11 = 12 next subtract 11 from both sides 2x = 1 finally divide by 2 in both sides x = 1/2 or .5

The left is -24. The right is 8 2/3. This is not an equation since there are different numbers on each side of the equals sign.

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With substitution it becomes a quadratic. Let u = w² and rewrite as: u² + 7u - 18 = 0 (u+9)(u-2) = 0 u = -9 and 2 = w² w = ±3i and ±√2

Quartic equations have quadratic factors. w⁴ + 7w² - 18 = 0 can be factored as (w² - 2)(w² + 9) = 0. This means that w² - 2 = 0 or w² + 9 = 0. w² - 2 = 0 means that w² = 2, so w = ±√2. w² + 9 = 0 means that w² = -9, so w = ±√(-9) = ±3i. The solution set is {√2, -√2, 3i, -3i}.

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