# Help with linear equations.

• Feb 23rd 2013, 12:41 PM
rickyricardo
Help with linear equations.
Hello All,

I'm having trouble with linear equations.

2=x+7

The answer is -5, but I can't arrive at the answer systematically.

I've tried adding or subtracting 7 from both sides.

Also, I'm not sure why you would add or subtract in a given situation.

Any help would be appreciated.

Thanks!
• Feb 23rd 2013, 03:18 PM
Bashyboy
Re: Help with linear equations.
To find what x is equal to, simply isolate it, that is, put any x terms on one side of the equation, and the remaning things you'll put on the opposite side of the equation.

The reason why you would subtract 7 from both sides is because it leads to isolating the x-variable. The reason why you DON'T add seven to both sides of the equation is because you won't isolate the x-variable. By adding 7 to both sides, you'll get $7+2=x+7+7 \biggr\rightarrow 9 = x +14$. We want to know what x is, not what $x+14$ is.

EDIT: As a side note, equations are alive! They say things in natural language. If you have the equation $x=-1$, this is saying that $x$ IS $-1$, and that $-1$ IS $x$. Similarly, if you have the equation $8=x+1$, this is saying that $x+1$ IS $8$, and that $8$ IS $x+1$
• Feb 23rd 2013, 03:18 PM
Paze
Re: Help with linear equations.
Hi rickyricardo and welcome to MHF!

You are correct when you subtract 7 from both sides of your equation. However, you might be miscalculating the outcome.

$2=x+7\\2-7=x+7-7\\-5=x$

Do you see? :)
• Feb 24th 2013, 07:15 AM
rickyricardo
Re: Help with linear equations.
Thanks for the replies guys!

Here is what I was doing.

2=x+7

2=7-x+7-7

Looks like I was putting the 7 in the wrong places?

Paze, you subtracted the 7 from the 2 on the left of the = sign. I had it on the right.

Can you clarify why the position is important?

And how can you tell when you should add or subtract? I thought it didn't matter as long as you did the same to both sides.

Thanks!
• Feb 24th 2013, 08:59 AM
Paze
Re: Help with linear equations.
Quote:

Originally Posted by rickyricardo
Thanks for the replies guys!

Here is what I was doing.

2=x+7

2=7-x+7-7

Looks like I was putting the 7 in the wrong places?

Paze, you subtracted the 7 from the 2 on the left of the = sign. I had it on the right.

Can you clarify why the position is important?

And how can you tell when you should add or subtract? I thought it didn't matter as long as you did the same to both sides.

Thanks!

Hi again rickyricardo.

You can apply any function (division, subtraction, addition, multiplication etc.) to both sides of an equation at all times.
Let me clarify:

Imagine the equation is a even scale:
http://s14.postimage.org/sq7pfbsap/scale.png

Now imagine that we add some balls which are all the exact same weight. If we add 3 to one side and 2 to the other, the scale will tip and be uneven:
http://s13.postimage.org/89qo9hbjr/scale2.png

However, when we add the same amount of balls to each side, the weight stays even!
http://s17.postimage.org/75hu6mlwf/scale3.png

This works with any function you can think of. Adding, subtracting, multiplying, dividing, etc. seeing as you will do the exact same thing to both sides of the scale (equation).

So now you see why I subtracted 7 from BOTH sides of the equation. I need to keep the scale (equation) even for it to add up. The reason I subtracted 7, was so that at one side $2-7=x+7-7$ the 7's would cancel out at the 'x side' leaving us with $-5=x$ which means we have successfully isolated the x, which is the goal. You want to add, subtract, multiply, divide etc. until you are left with an equation looking somewhat like $x=...$. That means you isolated for x and found out what x equals. Note that $-5=x$ is the same as writing $x=-5$

Putting this knowledge to practice, imagine you had the equation $x+2=5$ we would express the equation on our scale like this:
http://s14.postimage.org/s77o3677l/scalez1.png
We know that every ball not marked with 'x' is 1kg. We don't know how heavy the ball marked with 'x' is. It could be made out of steel, led or even gold. It could vary in weight. How do we find out how heavy the ball marked with x is? We subtract every ball on the 'x side' of the scale but to keep the scale even, we subtract the SAME amount from the other side. So we remove 2 balls from each side of the equation:
http://s15.postimage.org/vicxov4vv/scalez2.png
We see the the ball marked with 'x' weighs the same as 3 balls. 'x' must therefore equal 3kg.
$x+2=5\\x+2-2=5-2\\x=3$
• Feb 24th 2013, 09:05 AM
Bashyboy
Re: Help with linear equations.
Well, rickyricardo, I think you should re-read my post. The primary goal of solving an equation is isolating the variable, so that you may determine what x is. It appears what you did was added AND subtracted 7 from the right-side of the equation, which has the effect of just adding zero ( $+7 -7=0$); this is a valid move, there is no mathematical law that says you can not do it, but it isnt very helpful. As I said, the whole point of solving an equation is determining JUST what $x$ is.

What an equation says is that whatever is on the left-side is the exact same thing as whatever is on the right-side. So, if you subtract 7 just from one side of the equation, you ruin the equality--the left-hand side is no longer equal to the right-hand side. In order to keep that "equalness" that an equation has, you use that old adage: "whatever you do to one side of an equation, you do to the other side of the side of the equation." In our case, we would subtract 7 from BOTH sides of the equation.

Does this help?
• Mar 4th 2013, 04:52 PM
rickyricardo
Re: Help with linear equations.
I guess what I don't understand, is what number from the example would you choose to add to each side? How do you know which one to use? If I understand correctly, it doesn't matter as long as it's the same on both sides. Does it even have to be from the example? If I can add, subtract, multiply, or divide ANY number on each side, I should get the correct answer.

I also don't understand how you choose the placement of the numbers within the equation. Would you place the numbers on the far left and right, or to the direct left and right of the = sign?

• Mar 4th 2013, 06:26 PM
Paze
Re: Help with linear equations.
Quote:

Originally Posted by rickyricardo
I guess what I don't understand, is what number from the example would you choose to add to each side? How do you know which one to use? If I understand correctly, it doesn't matter as long as it's the same on both sides. Does it even have to be from the example? If I can add, subtract, multiply, or divide ANY number on each side, I should get the correct answer.

I also don't understand how you choose the placement of the numbers within the equation. Would you place the numbers on the far left and right, or to the direct left and right of the = sign?

Correct, you can add or subtract ANY number from the equation. Your goal is to get the x on either side. Let's just say the left side. So you want the equation to look something like this in the end $x=50$.

In your example $2=x+7$ we found that it was very convenient for us to subtract 7 from both sides, because by doing that, we isolated our x, which is our goal. The equation became $-5=x$ which is the same thing as writing $x=-5$. We have reached our goal of isolating x to the left side of the equation.

Do you see how $2-7=x+7-7$ becomes $-5=x$?
• Mar 9th 2013, 08:37 AM
rickyricardo
Re: Help with linear equations.
I do see how you arrive at the correct answer.

But I get different answers depending on where I place the 7's.

Do you automatically know when you see an equation, where the placement of the numbers will be? Or do you have to experiment a few times to find the right one?
• Mar 9th 2013, 12:39 PM
fanna1119
Re: Help with linear equations.
Think of it this way: If lets say Joe has 7 apples and John stole a few apples(x), after John stole the apples Joe only had 2 left.
Meaning
2=x+7
2-7=x
-5=x

Always test your answer that's how you figure out how the equation works and if you done it right
2=-5+7
2=2
:)
• Mar 9th 2013, 01:31 PM
Paze
Re: Help with linear equations.
Quote:

Originally Posted by rickyricardo
I do see how you arrive at the correct answer.

But I get different answers depending on where I place the 7's.

Do you automatically know when you see an equation, where the placement of the numbers will be? Or do you have to experiment a few times to find the right one?

I don't follow. What do you mean by 'placement'?
• Mar 10th 2013, 06:18 PM
rickyricardo
Re: Help with linear equations.
For instance, in the original example; 2=x+7, this is what I was doing: 2=7-x+7-7.

You did it like this: 2-7=x+7-7

How can you always know where to place the numbers that you are, in this case, subtracting?
• Mar 10th 2013, 07:14 PM
Paze
Re: Help with linear equations.
Quote:

Originally Posted by rickyricardo
For instance, in the original example; 2=x+7, this is what I was doing: 2=7-x+7-7.

You did it like this: 2-7=x+7-7

How can you always know where to place the numbers that you are, in this case, subtracting?

I explained why we apply the function on both sides of the equation in depth in my previous replies. I don't think I can explain it any better. I suggest you read my previous replies again and try with various equations.

Perhaps you can attempt to think of it as moving the 7 over the equation sign, but when you move numbers, you must switch the numbers from positive to negative or from negative to positive as so:

$2=x+7$ we swap the 7 to the other side, but like I said, the positive number becomes a negative number once we swap it over.

$2-7=x\\-5=x$.

This method will yield you correct answers but you will not understand WHY we apply the method unless you understand my previous replies.