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Calista A
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this is a problem that i am working on -6(x+2)+10+3(x-3) in which i got the answer of 9x+-11 and apparently i am wrong, can someone help me understand this more.
i added -6x to 3x and got 9x so now i have 9x-12+10 - 9 is this correct?skeeter said:$-6(x+2) + 10 + 3(x-3) = -6x - 12 + 10 + 3x - 9$
now, combine like terms
Calista A said:i added -6x to 3x and got 9x so now i have 9x-12+10 - 9 is this correct?
Calista A said:i added -6x to 3x and got 9x so now i have 9x-12+10 - 9 is this correct?
why... this is what i was stuck on, i thought you add on both sides which is why i was getting 9xProve It said:-6x + 3x is not 9x, it's -3x.
How did you get -12 + 10 - 9 = -11 but not -6x + 3x = (-6 + 3)x = -3x? The addition of numbers works the same way!Calista A said:why... this is what i was stuck on, i thought you add on both sides which is why i was getting 9x
First "+-11" is not a number and makes no sense.Calista A said:this is a problem that i am working on -6(x+2)+10+3(x-3) in which i got the answer of 9x+-11 and apparently i am wrong, can someone help me understand this more.
The first step in solving this problem is to combine like terms. In this case, we can combine the 9x and -11 terms to get 9x-11.
To solve for x, we need to isolate it on one side of the equation. In this case, we can add 11 to both sides to get 9x = 11. Then, we can divide both sides by 9 to get the final solution of x = 11/9.
Yes, you can always check your answer by plugging it back into the original equation. In this case, when we substitute x = 11/9 into the equation, we get 9(11/9) - 11 = 11 - 11 = 0, which confirms our solution.
Yes, there are multiple methods for solving equations. Another common method is to use the distributive property to eliminate parentheses, then combine like terms and solve for x.
If you encounter a problem with multiple variables, the general approach is still the same. Start by combining like terms and isolating the variable you want to solve for. If there are still multiple variables present, you may need to use substitution or elimination methods to solve for the remaining variables.