anonymous
  • anonymous
What is the value of x? x/5 + 6 =36 x=?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathstudent55
  • mathstudent55
Is this the equation? \(\dfrac{x}{5} + 6 = 36\)
anonymous
  • anonymous
yes
mathstudent55
  • mathstudent55
We need to isolate x, In order to do that, we need to undo the operations being done to x. x is being divided by 5. Then a 6 is being added. We start by undoing the addition of 6. Since we need to do the same to both sides of an equation, we subtract 6 from both sides.

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More answers

mathstudent55
  • mathstudent55
|dw:1443651036765:dw|
mathstudent55
  • mathstudent55
What is 6 - 6 ?
anonymous
  • anonymous
0
mathstudent55
  • mathstudent55
Good. What is 36 - 6 ?
anonymous
  • anonymous
30
mathstudent55
  • mathstudent55
Good. Since there is no operation yet being done to x/5, we just copy it down. We get: |dw:1443651137106:dw|
mathstudent55
  • mathstudent55
We now have \(\dfrac{x}{5} = 30\) x is being divided by 5. We need x by itself, so we do the opposite of dividing by 5. The opposite operation to division is multiplication. We need to multiply x/5 by 5 to end up with just x. The rule is that we need to do the operation to both sides of the equation, so we multiply both sides of the equation by 5.
mathstudent55
  • mathstudent55
|dw:1443651276216:dw|
mathstudent55
  • mathstudent55
What is 30 * 5 ?
anonymous
  • anonymous
150?
mathstudent55
  • mathstudent55
Yes. Since 5/5 = 1, we end up with just x on the left side, so we have our solution: |dw:1443651347959:dw|
anonymous
  • anonymous
so 150 is the answer
mathstudent55
  • mathstudent55
Yes. Now we can check. If we plug in 150 into x in the original equation, we must come up with a true statement. Let's check.
mathstudent55
  • mathstudent55
|dw:1443651437740:dw|
anonymous
  • anonymous
Thank You so much! could you help me with 3 more?
mathstudent55
  • mathstudent55
Since replacing x with 150 makes the equation true, 150 is the correct solution to the equation.
mathstudent55
  • mathstudent55
You're welcome. I'll help, but pls start a new post.
anonymous
  • anonymous
ok :)

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