susieq00 said:
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.