Having trouble with fractions used in substitutuion problems

alexyankeefan

New member
Joined
Jul 11, 2007
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3
I have been trying to solve the following problems with no avail. I'm hoping someone can end my misery!!!!

My problem is I dont know what operations to do when I encounter the fractions. Can anyone bring me back to the basics??????

The problems are:

1- x+2y=10
4x-y=13


2- 9x+10y=90
2x-8y=16


3 1/2x - 2/3y=5
x+1/5=10
 

jonboy

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Jun 8, 2006
Messages
547
Sometimes you can't avoid fractions but there are some in these problems when using substitution.

Like on 1, solve for x in the first equation: \(\displaystyle x\,=\,-10\,+\,2y\)

Then substitute that in for x in the second equation: \(\displaystyle 4(-10\,+\,2y)\,-\,y\,=\,13\)

Solve for y and then sub that in for x in the first equation. Also sometimes you can multiply an equation by a number to get rid of fractions.

Can you show me specifically in one of these problems what is the difficulty?
 

alexyankeefan

New member
Joined
Jul 11, 2007
Messages
3
I'm sorry. Problem 1 should look like this;

x+2y=10
4x-y=13

What I cant understand is the steps to take to deal with the fractions, specially on problems 2 and 3. What denominator do you use for problem 3 and can it be used for both the x and y fractions at the same time?

On number 3 1/2X-2/3Y=5
x+1/5=10

This is where I am so far before getting stumped.

x+1/5=10
-1/5 -1/5

x=10-1/5

Substitute for X 1/2 (10-1/5)-2/3Y=5

And here is where I stump*********

Can you help?
 

Subhotosh Khan

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Jun 18, 2007
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Code:
On number 3     1/2X-2/3Y=5
             x+1/5=10...........Are you sure this is not X + 1/5 Y = 10
 

alexyankeefan

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Jul 11, 2007
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3
It should look like this...

1/2 X - 2/3 Y = 5

X + 1/5 = 10
 

Subhotosh Khan

Super Moderator
Staff member
Joined
Jun 18, 2007
Messages
18,134
alexyankeefan said:
I'm sorry. Problem 1 should look like this;

x+2y=10
4x-y=13

What I cant understand is the steps to take to deal with the fractions, specially on problems 2 and 3. What denominator do you use for problem 3 and can it be used for both the x and y fractions at the same time?

On number 3 1/2X-2/3Y=5
x+1/5=10

This is where I am so far before getting stumped.

x+1/5=10
-1/5 -1/5

x=10-1/5 = 49/5

Substitute for X

1/2 * (49/5) - 2/3 * Y = 5

-2/3 * Y = 5 - (49/10)

Y = - 3/2 * 1/10 = - 3/20
......................DONE

1/2 (10-1/5)-2/3Y=5

And here is where I stump*********

Can you help?
 
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