Can someone please help me with these algebra problems?
The directions are: Solve the equation or formula for the indicated variable.
m=ax+bx for x
8=mx+ax2 for x
ax+bx3=4 for x
Anything is helpful! Thanks.
For m=ax+bx for x you would use the distributive property to get
m=(a+b)x now divide both sides by (a+b) with the requirement that (a+b) is not 0.
For 8=mx+ax2 for x, now add 2 to both sides, use the distributive property as above, and find x. Be sure to test the answer in the original equation.
For ax+bx3=4 for x add 3 to both sides and use the distributive property as above. Be sure to show your work too.
Thankyou very much. Do you know if there are any restrictions/requirements in the two last answers? I think there are.
Yes, the same as the first. Anytime we have an equation and divide both sides by some quantity we should mention that that quantitiy must be nonzero. We should say this at least once in our solution, but sometimes our equation is such that all the values cannot be zero. In that case we can omit this statement.
Thanks again :)
I got
m/a+b=X
10/a+m=x
7/a+b+x all with the restrictions
Does that seem right?
m=ax+bx for x
8=mx+ax2 for x
ax+bx3=4 for x
Yes, that looks right. About the only thing I might suggest is that you use parentheses like this (a+b) for the denominators so it doesn't cause confusion. The requirement that a+b or a+m be nonzero could be written to the right of the equation. As I mentioned, if all of the variables were positive numbers say, then these restrictions could be ignored.
How do you combine like terms like in the problem (3m+5k)(2m+10k)?
Use the associative and distributive properties to get
(3m+5k)(2m+10k) = 3m+5k  2m 10k =
(3m2m) + (5k10k) = m5k
You should post this as a new question too. This is easy to miss after it's off page 1.
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what is r +12 help me you are no help
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