Question

Which system has (-2, -2) as the solution?
Which system has no solutions?
Which system has infinite solutions?


A. x + y = -1
4x + 4y = -4
B. x + y = -4
-3x + 2y = 2
C. 4x + 6y = -12
2x + 3y = 9
D. x - y = -2
3x - y = 0
E. x - 4y = -8
2x - 3y = -16

Answer 1

A) x +y =-1
4x + 4y = -4
The second equation is just 4 times the first equation.
So there are inifinty solutions.
i.e multiply the first by 4

=> 4x +4y =-4 ( first equation)
4x + 4y =-4
you cannot isolate a single variable either x or y



B) x + y = -4
-3x + 2y = 2

Multiply the first equation by 3 and add both equation to eliminate x and solve for y:

i.e.
3x + 3y = -12
-3x + 2y = 2
-----------------
=> 5y = -10
y = -2
substitute the value of y back into any one of the original equations to solve for y
e.g. substitute into x +y = -4
=> x -2 = -4
add 2 on both sides
=> x = -2

Hence the solution is ( -2, -2)


C)

C. 4x + 6y = -12
2x + 3y = 9

The left hand side of the equations are multiples of each other.

e.g Multiply the second equation by 2 yields:

3x + 6y = 18

i.e the system becomes:

4x + 6y = -12
4x + 6y = 18
This system has no solution
for example subtract the second from the first yields:
0 + 0 = -30



D.x - y = -2
3x - y = 0

Reqwrite the system as:

-3x +3y = -6 ( after multiplying the first equation by -3 )
3x -y = 0

Add the two to yied

2y = 6
y = 3

D has a solution (1, 3)


E.x - 4y = -8
2x - 3y = -16


Multiply the first equation by -2

-2x +8y = 16
2x -3y = -16
Add the two equations:
5y = 0
y =0
=> x = -8 ( by plugging the value of y=0 in any one of the original equations)

E has solution:

( -8,0)

Answer 2

Therefore:

A has ininifinity solutions
B has solution ( -2, -2)
C has no solution