solve the system of equations: first: x+3y=5, second: 3x-y=5?

Question

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Answer 1

1) x + 3y = 5
2) 3x - y = 5

There are various ways to solve these systems. One approach here is to solve equation 1 for x

x = 5 - 3y

and substitute for x in equation 2

3(5 - 3y) - y = 5
15 - 9y - y = 5
- 10 y = - 10
y = 1

Substitute this into equation 1

x + 3(1) = 5
x + 3 = 5
x = 2

Check using equation 2

3(2) - 1 = 5
6 - 1 = 5
5 = 5

Answer 2

The solution is x = 2, y = 1

Here's how to solve such equations. I will CAREFULLY explain EVERY step, assuming that that is what you really want to see. (My answer will then look long --- but that's because I'm NOT assuming that any step is obvious, as some other responders will probably do.) Then YOU will be equipped to solve such equations in future.

First write the equations one below the other :

x + 3 y = 5 ........(1)
3 x - y = 5 ........(2)

Looking at this, if you multiply equation (2) by 3, you'll get "- 3 y" in it; you'll see why that is good, almost immediately:

9 x - 3 y = 15 ........(2 ')

NOW add eqns (1) and (2 ') : then y WILL DISAPPEAR! This is called "eliminating one of the variables" --- in this case y. (And THAT'S WHY we did that multiplication by 3, so that that could happen.) Then what we get with the addition of eqns (1) and (2 ') is:

10 x = 20, so that x = 2.

Putting x = 2 back into eqn (1) we have 2 + 3 y = 5, or 3 y = 3, so y = 1.

(That's it! If you ignore the explanatory writing, you'll see how very few, and how quick, the mathematical steps really are.)

We now have x = 3 and y = 1, but one IMPORTANT step is still needed for a really complete answer: CHECK IT! Here's how we do that.

We got the value for y, the last variable we solved for, from eqn (1). That's all very well, but : do BOTH variables satisfy the other original equation, eqn (2)? So, one writes:

CHECK : With the values we deduced, 3 x - y = (3 x 2) - 1 = 6 - 1 = 5. That's correct, so it checks out.

What was the point of that last step? It was to confirm that we didn't make an algebraic mistake in our work. Without doing that, we could be just fooling ourselves into thinking that we had the right answer. (I've seen all too many "answers" recently that were simply wrong, their writers not having performed such a straightforward check on their results.)

Live long and prosper.

Answer 3

3x - y = 5 and x + 3y = 5 i began with the small numbers so i attempted x = 2 and y =a million. That labored so I went to the subsequent one. 2x - 3y = 3 and x + 4y = 7 I regarded on the 2d of those sums and realised that y might desire to be a million because of the fact if it grew to become into any bigger, than 7 grew to become into out of attain. (except negatives are in this, yet i've got no longer executed algebra with negatives till now.) So y = a million lower back and x + (4x1) = 7 could make x 3. (2x3) - (3x1) = 3 additionally so the respond grew to become into shown.

Answer 4

x + 3y = 5- - - - Eqution 1
3x - y = 5- - -- -Equation 2
- - - - - - - -

Multiply equation 2 by 3

3x - y = 5

3(3x) - 3(y) = 3(5)

9x - 3y = 15. . . .New equation 2

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

Elimination of y

x + 3y = 5
9x - 3y = 15
- - - - - - - -

10x = 20

10x / 10 = 20 / 10

x = 2

The answer is x = 2

Insert the x value into equation 1

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

x + 3y = 5

2 + 3y = 5

2 +3y - 2 = 5 - 3

3y = 3

3y/3 = 3/3

y = 3/3

y = 1

The answer is y = 1

Insert the y answer into equation 1

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

Check for equation 1

x + 3y = 5

2 + 3(1) = 5

2 + 3 = 5

5 = 5

- - - - - - - - -

Check for equation 2

3x - y = 5

3(2) - 1 = 5

6 - 1 = 5

5 = 5

- - - - - - - - - -

The solution set is { 2, 1 }

- - - - - - - - - - -s-

Answer 5

x+3y=5
3y=5-x
y=(5-x)/3substitute this into the equation 3x-y=5

3x-[(5-x)/3]=5
multiply 3x by 3
9x/3 - (5-x)/3 =5
(9x-5+x)/3 =5
(10x-5)/3 =5
multiply denominator to 5
10x-5=15
10x=20
x=2
or
3x-y=5
3x-5=y substitute this into the equation x+3y=5
x+3(3x-5)=5
x+9x-15=5
10x=20
x=2

Answer 6

x=5-3y
substitue x in 3x-y=5
the equation becomes
(15-9y)-y=5;
which becomes 10y=10;
so y=1 and x=2

Answer 7

x+3y=5
x-3y=5-3y
x=5-3y

x+3y=5
3y=5-x
divide both sides by 3
y=5-xdivided by y



3x-y=5
3x+y=5+y
3x=5+y
again divide both sides by 3
x=5+ydivided by 3


3x-y=5
subtract both 3x
get -y=5-3x
multiply by -1 to get y=-5+3x
I on't think you can switch it but if you can it would be y=3x-5

good luck.

Answer 8

A similar question was asked a few minutes ago:

http://answers.yahoo.com/question/index;_ylt=Av1jHguSiG091M3cXYxivpwCxgt.?qid=20070201103624AAgpXnS

just look at the working for that...

Answer 9

see http://answers.yahoo.com/question/index;_ylt=AqfNWAoAgdfAtYRvwIPVFVsCxgt.?qid=20070201103624AAgpXnS

This gives a good detailed explanation of how to do these problems.

Answer 10

-3x-9y=-15
3x-y=5
adding
-10y=-10
y=1
substituting
x=2