3943 Shares

How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y?

How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y? Topic: How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y?
June 26, 2019 / By Beatrix
Question: I need to know the steps and the answer for these problems: 20x - 40xy = ?? and 28x^2 y^2 - 49x^2 y = ?? I know it might be a dumb question but I have forgotten algebra!! :-D I am helping my sister with her homework and I have not done this kind of stuff in years and I have completely forgotten how to do these types of problems. If someone could please help me out without being rude that would be great! THANKS!! :) Also, ^2 means squared. Like the little 2 in the upper corner. Hehe thanks! I messed up the first equation :( it is supposed to be 24x - 40xy! I'm sorry. I don't know how I missed that. So it is 24x not 20x. Thanks again! :)
Best Answer

Best Answers: How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y?

Adena Adena | 6 days ago
What is this, is this a factoring, well let assume that it is a factoring: Using the distributive property, we can solve this problem. Find the common factor of each term and put it outside. Common factor: 20x 20x - 40xy = 20x(1 - 2y) Common factor: 7x^2 y 28x^2 y^2 - 49x^2 y = 7x^2 y(4y - 7)
👍 220 | 👎 6
Did you like the answer? How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y? Share with your friends
Adena Originally Answered: Does anyone know how to solve these problems? Please solve for each variable. show work if possible.?
1) Subtract 12x from each side -3x = 6 Divide each side by -3 x = -2 2) Simplify the parentheses part first. -2y - 6y + 3 = 6y - 4 Combine like terms -8y + 3 = 6y - 4 Subtract 6y from each side and subtract 3 from each side -14y = -7 Divide both sides by -14 and reduce. y = -7/-14 y = 1/2 Now try the others on your own. Remember you can always check your answer by replacing the variable with the answer you just got to see if you get a true equation. The best way to learn math is to do the problems yourself.

Tel Tel
Hi, 20x - 40xy factors by dividing out a GCF of 20x from each term: 20x - 40xy = 20x(1 - 2y) <==ANSWER 28x²y² - 49x²y factors by dividing out a GCF of 7x²y from each term: 7x²y(4y - 7) <==ANSWER I hope that helps!! :-)
👍 90 | 👎 -1

Philander Philander
These are clearly simultaneous equations. I could solve them in time but I'm a bit too sleepy right now sorry. Basically, you isolate a variable (say, x) in the first equation (e.g. leaving it in the form x = ???). You then substitute this value of x into the second equation and simplify. I think that in this case it might require long division of polynomials: if I'm right, google it and you'll get the method of how to do it. You should be left at most with a quadratic equation in one variable to solve.
👍 85 | 👎 -8

Macey Macey
i know how to solve the first one but not the second sorry. 20x(1-2y)=0 x=0 1-2y=0 1=2y y=1/2 i hope it's not wrong but truthfully i am not sure but this is how i remember solving these kinda problems
👍 80 | 👎 -15

Macey Originally Answered: Could someone please solve these statistics problems for me I tried but cannot solve them Thanks ?
10. D If someone else wins, that means Jeff does not win. Thus, P(Jeff does not win) = 1-P(Jeff wins) = 1-0.2 = 0.8 20. I do not know what sample the problem is referring to. 12. C (40+30+20)/(15+25+40+30+20) = 0.69 13. C (15 + 40) / (130) = .42

If you have your own answer to the question How do you solve these problems : 20x - 40xy and 28x^2 y^2 - 49x^2 y?, then you can write your own version, using the form below for an extended answer.