Solving Multi-Step Equations: 21 = 6 - x - 4x

math87690

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Nov 14, 2006
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I have a question, here is the problem 21=6-x-4x. And then I need to know how to check the answer. I would like you to reply as soon as possible please. I thank you for your help to. And when you reply could you tell me how long it regularly takes to reply.
 
Hello, math87690!

So many variable I can't tell you how long it take to reply.

You can check your answer by filling what you found \(\displaystyle x\) to be and see if the equation is balanced (same number on both side of the equations).

So with that being said:

\(\displaystyle \L \;21\,=\,6\,-x\,-4x\)


Combine like terms: \(\displaystyle \L \;21\,=\,6\,-5x\)


Subtract 6: \(\displaystyle \L \;15\,=\,-5x\)


Divide by -5: \(\displaystyle \L \;-3\,=\,x\)


...........Transitive Property: \(\displaystyle \L \;x\,=\,-3\)

Now you fill \(\displaystyle x\) back in our original equation (\(\displaystyle 21\,=\,6\,-x\,-4x\)) and see if it is balanced.
 
math87690 said:
I have a question, here is the problem 21=6-x-4x.
Keep it simple :shock:
Add 5x to both sides:
5x + 21 = 6
Subtract 21 from each sides:
5x = -15
Divide each side by 5:
x = -3
 
here's the proof it worked

here's the proof it worked: I assigned -3 for x then plugged in the right hand side of the equation which, with x equivalent to -3, should equal 21 (the left side of the equation)
 
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