subsititution method questions?

Question

please help w/2 questions:

Use substitution method to solve this system

3x - y = 5
-6x + 2y = -10

I came up with x = 0
y int form for both eq's: y=3x-5
how do you write the answer?

plus i have one more where the first eq was x=0 ...i'm stuck on the other eq in the system


use sub.method
x + 5y = 5
3x + 15y = 11


Thanks in advance

Answer 1

3x - y = 5- - - - - - - -Equation 1
- 6x + 2y = - 10- - - Equation 2
- - - - - - - - - - - -

Substitute method equation 1

3x - y = 5

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

- y = - 3x + 5

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

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

y = 3x - 5

Substitute the y value into equation 2

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

- 6x + 2y = - 10

- 6x + 2(3x - 5)

- 6x + 6x - 10 = - 10

The coefficients and variables cancell. no solution

- - - - - - - - -s-

Answer 2

Answer 1:
3x - y = 5.....(1)
-6x + 2y = -10.....(2)

Divide (2) by -2
(-6x + 2y)/-2 = -10/-2
3x - y = 5....(3)

(3) is a direct result of (2).
It means that (2) and (1) are the same and the system has infinite solutions.

Answer 2:
x + 5y = 5.....(1)
3x + 15y = 11.....(2)

Simplify (2)
3x + 15y = 11
3(x + 5y) = 11
x + 5y = 11/3.....(3)

LHS of (3) and (1) are equal. So, RHS of both equations must be equal, which gives:

5 = 11/3 which is never possible. So this system of equations has no real solution.


There is an extremely simple way of checking how many solutions a system of linear equations in two variables has by using the coefficients of the variables and the constants. Here is the method

There are two equations to solve. They are:

ax + by = c.....(1)
px + qy = r.....(2)

The varibles are x and y.

If a/p = b/q but does not equal to c/r, the system has no solution.

If a/p does not equal b/q, the system is satisfied only for one value of x and y (has a unique solution).

If a/p = b/q = c/r, the system has infinite solutions. That is, the graph of both equations is the same line.


By using this checking method, let us verify if our result in the second question you asked is true.

x + 5y = 5
3x + 15y = 11

1/3 = 5/15 but does not equal 5/11. So, the system has no solution.

Check for the first question.

3x - y = 5
-6x + 2y = -10

3/-6 = -1/2 = 5/-10
So, the system has infinite solutions.

One can verify the long way, by solving the eqations. In the end, if you get LHS = RHS (for example, everything cancels to give 0 = 0), the system has infinite solutions. 0=0 doesn't imply that the system has no solutions. I'll make that absolutely clear to the guys who said that. The system has no solution only if you get an equality which is impossible like:
120 = 0
or 11/3 = 5

Answer 3

On the first one Jami did the work right but the answer wrong.

She got 0=0 which means "all solutions" not "no solutions".

0=0 is a true statement no matter what so you can put any x or y in the equation and it will still be true. Try it ...plug in any number for x and y and it will work.


Eturnal messed up here:
-y=5-3x
(-1)-y=(-1)(5-3x)
y=3x+5

it should be:
-y=5-3x
(-1)-y=(-1)(5-3x)
y=3x-5

Answer 4

3x-y=5 => x= (5+y)/3

substitute

-6((5+y)/3) +2y =-10
-2 (5+y) +2y =-10
-10+-2y +2y =-10
0=0

no solution

x +5y =5 ==> x=5-5y
3 (5-5y) +15y =11
15-15y+15y =11
0=-4

no solution..

:))