Systems of Equations

[susieq00]

I can't figure out these problems i keep getting different answers everytime i solve them.

Identify the type of system of equations for the system below:

2y=4-6x
3y+9x=6


Identify the system of equations which has no solution.
a. 5x+2y=4
2x-2y=10
b. -x=3y+1
x=3y-1
c. x-2y=-6
2x=4y-12
d. x=2y-1
2x=4y

 

[Gene]

2y=4-6x
3y+9x=6
Divide the first by 2 and the second by 3 to get
y=2-3x
y+3x=2
Move the x's to the right
y=2-3x
y=2-3x
The two linear equations are the same so there is no solution.

Do the same with the second part. Move the y terms to the left and the x terms and constants to the right. Two of them will be the same

 

[Mrspi]

Hmmmm......

I've always taught that if the two equations in a system represent the same line then the system has infinitely many solutions (or is consistent and dependent). Each ordered pair (x, y) which is a solution for one of the equations is also a solution for the other.

If the two equations represent lines with the same slope but different y-intercepts, then the system has no solutions (or is inconsistent). The lines are parallel and there is no point whose coordinates satisfy both equations.

If the two equations represent lines with different slopes, regardless of the y-intercepts, the lines will intersect in exactly one point and the system has just one solution (is consistent and independent).

So, my suggestion is that you take the equations in each system and solve each for y so that you can examine the slopes (and if they are the same, then examine the y-intercepts). You should then be able to identify the type of system you are dealing with.

 

[Gene]

Technically correct but looking at the second question THIS book says there is no solution in this situation.
---------------
Gene

 

[Mrspi]

I
Identify the system of equations which has no solution.
a. 5x+2y=4
2x-2y=10
b. -x=3y+1
x=3y-1
c. x-2y=-6
2x=4y-12
d. x=2y-1
2x=4y

Ok...I am perfectly content to look at the parts of problem 2:

a) 5x + 2y = 4
and 2x - 2y = 10

Solve each equation for y:
2y = -5x + 4
y = (-5/2)x + 2

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

Different slopes, so the lines intersect in exactly one point. One solution, and the system is consistent and independent.

b) -x = 3y + 1
x = 3y - 1

Multiply both sides of the first equation by -1:
x = 3y - 1
and the second equation is
x = 3y - 1

Since the equations are clearly the same, the lines represented by these equations are the same line. Each point which is a solution for the first equation is also a solution for the second. There are infinitely many solutions for this system.

c) x - 2y = -6
2x = 4y - 12

Solve the first equation for y:
-2y = -x - 6
y = (1/2)x + 3

Solve the second equation for y:
2x + 12 = 4y
(2/4)x + (12/4) = y
(1/2)x + 3 = y

Again, the two equations are identical, and any solution for the first is also a solution for the second. This system has infinitely many solutions.

d) x = 2y - 1
2x = 4y

Again, solve each equation for y:
x + 1 = 2y
(1/2)x + (1/2) = y

2x = 4y
(1/2)x = y

The two equations,
y = (1/2)x + (1/2)
and
y = (1/2)x

have the same slope, but different y-intercepts. So, the lines are parallel and have no point of intersection. So, there will be no point whose coordinates satisfy both equations, and this system has no solution.

I think previous answer is both technically and actually correct.

For problem 2, only part d represents a system which has no solution.

 

[Gene]

I stand corrected. I stopped at b and considered that to be the one they were refering to.
-------------------
Gene