Jump to content

Subracting Equations


Kieran Morrison

Recommended Posts

So if there is 9x+21=5x-4, you would minus 5x from each side, then minus 4 of each giving you 4x = 17 then divide them both by 4 giving you the answer- X = 14.25?

As an alternative to what Jason wrote, and in line with the direction you're going for:

9x+ 21 = 5x -4

Subtract 5x from both sides:

4x + 21 = -4

Add 4 to each side (if you subtract 4 from each side you wouldn't be helping to simplify the problem):

4x + 25 = 0

This can then be rearranged by subtracting 25 from each side:

4x = -25

Which gets you back to x = -6.25.

Edit: Also, how did you manage to end up with 14.25 when you divide 17 by 4?!

Link to comment
Share on other sites

Edit: Also, how did you manage to end up with 14.25 when you divide 17 by 4?!

My calculator did it! :lol: I must have typed it wrong or something.

So am i right though in saying if there is 2 minus's in an equation, it means you add them, the same with 2 + signs, and if there is a minus and a plus you minus them?

Link to comment
Share on other sites

So am i right though in saying if there is 2 minus's in an equation, it means you add them, the same with 2 + signs, and if there is a minus and a plus you minus them?

Pretty sure the answer is no... The only time when two minus signs make a positive is when you have the two signs together.

For example 4 - 1 = 3. But if you have 4 - (-1) ( read as four subtract minus one) then the two minuses combine to become an addition.

So 4 - (-1) = 4 + 1 = 5

Going back to equations lets make everything negative to show how it works.

Start with -4x - 6 = -14

We want all the x's on one side and all the numbers on the other so to get rid of the -6 on the left we add 6 to both sides:

-4x - 6 + 6 = -14 + 6

-4x = -8

We can now find x by dividing both sides by -4, to leave just one positive x on the left:

-4x/-4 = -8/-4

x = 2

Does that make sense?

Link to comment
Share on other sites

Still not sure on how to do equations like: if x is 9, and y is 10 whats xy2

When my teacher went over it (wasnt listening, dont remember what i was doing) he turned it into an equation somehow, like the kind of equations written like: 5x + 3 = 9x + 21 and i dont know how he did it :S

So how would you turn the x is 9, and y is 10 whats xy2into an equation like above?

Link to comment
Share on other sites

So how would you turn the x is 9, and y is 10 whats xy2into an equation like above?

The type you describe are some of the easiest you can come across as you're basically given the answer. All you need to do is substitute the numbers given (x=9, y=10) into the formula given (xy2). The only thing you need to remember is what to square. In this case if you were to write the formula out slightly differently you'd have

x * y2

Noting that only the y term is being squared.

Substituting x=9 and y=10 gives:

9 * 102

Which becomes:

9 * 100

Which = 900.

Edit: Not entirely sure how your teacher will have turned it into a proper equation without introducing another unknown (lets say z).

In this case we can write z = xy2

Because we know x and y we have one equation and one unknown (z) which means we can solve it.

So following what we did before:

z = 9 * 102

z = 900.

You can always check these things by going backwards. If we pretend we don't know what y is we can start with the same equation as before:

xy2 = z

So if we first rearrange for y:

y2 = z/x

y = squareroot(z/x)

And now substituting z = 900 and x = 9 into that we get:

y = squareroot(900/9)

y = squareroot(100)

y = 10

Is that taking it too far?

Link to comment
Share on other sites

Still not sure on how to do equations like: if x is 9, and y is 10 whats xy2

When my teacher went over it (wasnt listening, dont remember what i was doing) he turned it into an equation somehow, like the kind of equations written like: 5x + 3 = 9x + 21 and i dont know how he did it :S

So how would you turn the x is 9, and y is 10 whats xy2into an equation like above?

Well, it already is an equation.

xy2 you can turn into x*y*y or xyy but that's pretty much about it.

If you had to solve it, it's simply 9*102 which is 900.

There's not really much more that can be done with it.

Link to comment
Share on other sites

No, it isn't. Trust me. Did you actually read my post? If you have xy2, ONLY the y is being squared. As JT says, that means it can be rewritten as x * y * y. It's really not difficult...

So, could it be re-written as 9x + 81 = 10x + 100?

So would you get x squared, then y squared, then square the number you get?

I dont understand this :S

Link to comment
Share on other sites

So, could it be re-written as 9x + 81 = 10x + 100?

So would you get x squared, then y squared, then square the number you get?

Where did you get that equation from? Work it through:

9x + 81 = 10x + 100

x = -19

Which obviously isn't right and you've lost the y's...

9x + 81 = 10y +100

Substituting for x and y gives:

81 + 81 = 100 + 100

162 = 200

Which also obviously isn't right.

Link to comment
Share on other sites

No, it's squaring both the x and the y.

Well thats how we did it in school.

"BODMAS"

Brackets, orders (squared/cubed), division, multiplication addition subtraction.

That's the order of operations. Orders take priority over everything else other than brackets.

Link to comment
Share on other sites

"BODMAS"

Brackets, orders (squared/cubed), division, multiplication addition subtraction.

That's the order of operations. Orders take priority over everything else other than brackets.

BIDMAS, you mean.

I for indeces!

Edited by Revolver
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...