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# how do I solve for x in this... xy+5=x+7? I cant figure it out and I really need help..

0
779
7
+49

how do I solve for x in this... xy+5=x+7? I cant figure it out and I really need help.

Oct 12, 2014

#7
+101771
+5

Hi Braksess,

I really like your dogged determination to understand this!

I think you are having problems with this

$$xy-x=x(y-1)$$

so that is the bit I am going to look at.

Look at the rectangle below.

The area of the whole rectangle is      $$x\times y \quad units^2$$

I have cut a little rectangle off the end.  The area of the cut off bit is   $$1\times x=x \quad units^2$$

so the area that is left when you take away the litle bit at the end is

$$\\x\times y - x\\ or\; just\\ xy-x$$

The area that is left after you take away the bit at the end is displayed in yellow below.

The length of the top is y-1  (because 1 unit was cut off)

The side is still x  units

The area is   x(y-1)

SO          $$xy-x=x(y-1)$$

LETS look at it the other way around

you probably know that     3(x-1)=3*x-3*1 = 3x-3

It works with letters too

$$x(y-1)=x\times y - x\times 1 = xy-x$$

NOW lets look at how to factorise it going in the other direction.

$$\\xy-x=x\times y - x\times 1\\ x is a common factor so we can factor it out and we will be left with y-1 in the bracket.\\ =x(y-1)$$

.
Oct 14, 2014

#1
+17774
+5

To solve for x, get all the x-terms to one side and everything else to the other side:

xy + 5  =  x + 7

Subtract 5 from both sides:

xy  =  x + 2

Subtract x from both sides:

xy - x  =  2

Factor out the x:

x(y - 1)  =  2

Divide both sides by y - 1:

x  =  2 / (y - 1)

Oct 12, 2014
#2
+49
+5

how did you keep the x after you subtracted it from both sides?

Oct 12, 2014
#3
+101431
+5

I think you're getting a little confused here, Braksess....here's what geno did

xy  =  x + 2     ( subtract x from both sides )

-x      -x

------------

xy - x  = 2       (note that he didn't "divide away" x......I think this is what you were thinking......!)

Can you take it from here??

Oct 13, 2014
#4
+49
0

y=2? i know im wrong but im still really confused,  xy/-x=x-2/x im trying to find x. maybe its x-x=-y-2...so....0=y-2

Oct 13, 2014
#5
+101431
+5

Let's try this again.............

Note that we have......

xy + 5 = x + 7    now....just subtract 5 from both sides....

-5 =      -5

--------------------

xy        =  x + 2

Now...I want to get rid of the x on the right side.....so I can just subtract it from both sides...so we have

xy       =  x + 2

-x      -x

------------------

xy - x  =        2         Notice how the x "disappeared from the right and ended up on the left as a "negative??"

Now...I just want to "factor the x on the left out of both terms....so we have....

x(y - 1)  = 2       do you see that???

So this is like having

x * (y - 1)  = 2        and since I want the "x" by itself on the left....we can divide both sides by  (y - 1)

So

x* (y - 1)/(y-1)  = 2/(y-1)      and the (y -1)s on the left "cancel" leaving us with......

x   =   2/(y - 1)     !!!!

Does that help ???

(We can't get any "numerical" values for x or y, because we don't know the value of either  !!!!  )

Oct 13, 2014
#6
+49
+5

i still dont get how an x turned into a 1? i get it all up till there

Oct 13, 2014
#7
+101771
+5

Hi Braksess,

I really like your dogged determination to understand this!

I think you are having problems with this

$$xy-x=x(y-1)$$

so that is the bit I am going to look at.

Look at the rectangle below.

The area of the whole rectangle is      $$x\times y \quad units^2$$

I have cut a little rectangle off the end.  The area of the cut off bit is   $$1\times x=x \quad units^2$$

so the area that is left when you take away the litle bit at the end is

$$\\x\times y - x\\ or\; just\\ xy-x$$

The area that is left after you take away the bit at the end is displayed in yellow below.

The length of the top is y-1  (because 1 unit was cut off)

The side is still x  units

The area is   x(y-1)

SO          $$xy-x=x(y-1)$$

LETS look at it the other way around

you probably know that     3(x-1)=3*x-3*1 = 3x-3

It works with letters too

$$x(y-1)=x\times y - x\times 1 = xy-x$$

NOW lets look at how to factorise it going in the other direction.

$$\\xy-x=x\times y - x\times 1\\ x is a common factor so we can factor it out and we will be left with y-1 in the bracket.\\ =x(y-1)$$

Melody Oct 14, 2014