SOLUTION: Solve the system and verify by replacing the values in the original equation 3x+2y = 5 x-y=1 So what I have gotten so far is: y= -x-1 and 3x+2y=5 2y=-3x+5

Algebra ->  Linear-equations -> SOLUTION: Solve the system and verify by replacing the values in the original equation 3x+2y = 5 x-y=1 So what I have gotten so far is: y= -x-1 and 3x+2y=5 2y=-3x+5       Log On


   



Question 1106695: Solve the system and verify by replacing the values in the original equation
3x+2y = 5
x-y=1
So what I have gotten so far is:
y= -x-1
and

3x+2y=5
2y=-3x+5
2/2y =3/2x+5/2
y=-1.5x + 2.5
1.5x +2.5 = -x -1
1.5x +x =-1 -2.5
2.5x = 1.5

where am I going wrong?


Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52890) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve the system and verify by replacing the values in the original equation
3x+2y = 5
x-y=1

So what I have gotten so far is:

y= -x-1     <<<---=== This line is WRONG.    The correct version is  THIS:  y = x-1.   Try again !!

and


3x+2y=5

2y=-3x+5

2/2y =3/2x+5/2
y=-1.5x + 2.5

1.5x +2.5  = -x -1

1.5x +x =-1 -2.5

2.5x = 1.5

                       where am I going wrong?

-----------
I like very much that you started your own trials !

Find my corrections inside your text.



Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Thanks for taking the time to show your work. That makes it easy for us to help you learn to solve problems like this.

The method you chose for solving the pair of equation was to solve each given equation for y and set the two expressions equal to each other. That requires far more work than either of the basic methods for solving pairs of linear equations.

And in the work you show, you are very careless with you signs; in both cases where you solved the given equations for y you have an error in signs.

You don't show the work you did to solve the first equation x-y=1 for y; however you did it, you ended up with the wrong sign. You should get y+=+x-1, not y+=+-x-1.

In the work you show for solving the second equation for y, you go back and forth between -3/2 and 3/2 for the coefficient of x. In the end, you end up with the correct coefficient, which is -3/2.

But then when you set the two expressions for y equal to each other, you show 3/2 instead of -3/2 for the coefficient of x.

So my first suggestion is for you to pay close attention to your signs....

My second suggestion is to use a different method for solving the pair of equations.

One basic method for solving a pair of linear equations is to solve one equation for one variable or the other and then substitute that expression for that variable in the other equation.

In this problem, with the given equations, you can solve the second equation easily for either x or y. And with the given equation x-y=1 there is less work involved solving for x, so I would change this equation to x+=+y%2B1.

Then I can replace x with y+1 in the first equation:
3%28y%2B1%29%2B2y+=+5 [x is replaced with (y+1)]
3y%2B3%2B2y+=+5
5y=2
y+=+2%2F5

Then I can use this value in either of the original equations to solve for x:
x-2%2F5+=+1
x+=+1%2B2%2F5+=+7%2F5

The solution is (x,y) = (7/5,2/5)

That is still a lot of steps; but less work than what you tried to do.

The other basic method for solving pairs of linear equations is elimination. With this method, you multiply one or both equations by appropriate constants so that when you add the resulting equations one of the variables is eliminated.

With the given equations we can just multiply the second equation by 2; then when we add the two equations y is eliminated, giving us an equation we can solve for x:

3x%2B2y=5 [original first equation]
2x-2y=2 [original second equation, multiplied by 2]
5x=7 [the two equations added to each other]
x+=+7%2F5

Then, as before, substitute this value in one of the original equations to solve for y:
7%2F5-y+=+1
7%2F5-1+=+y
2%2F5+=+y

And of course by this method we find the same solution: (x,y) = (7/5,2/5).