Giving 40 Points...need help as soon as possible

The correct answers are:

The first system matches with (-4, 5);
The second system matches with (2, 2); 
The third system matches with (-1, -3);
The fourth system matches with (3, 4);
The fifth system matches with (2, -1); and 
The sixth system matches with (-5, 6).

Explanation:

The sixth system is the easiest one to find.  We are given the value of x and the value of y in the equations: x=-5 and y=6.

For all other systems, we must solve the system.  For the first system:
[tex] \left \{ {{2x-y=-13} \atop {y=x+9}} \right. [/tex]

Since we have the y-variable isolated in the second equation, we will use substitution.  We substitute this value in place of y in the first equation:
2x-y=-13
2x-(x+9)=-13

Distributing the subtraction sign,
2x-x-9=-13

Combining like terms:
x-9=-13

Add 9 to each side:
x-9+9=-13+9
x=-4

Substitute this into the second equation:
y=x+9
y=-4+9
y=5

The coordinates (-4, 5) represent the solution point.

For the second system:
[tex] \left \{ {{3x+2y=10} \atop {6x-y=10}} \right. [/tex]

We can make the coefficients of x the same and use elimination.  To do this, we will multiply the first equation by 2:
[tex] \left \{ {{2(3x+2y=10)} \atop {6x-y=10}} \right. \\ \\ \left \{ {{6x+4y=20} 
\atop {6x-y=10}} \right. [/tex]

Since the coefficients of x are now the same, we can cancel it.  They are both positive, so we subtract:
[tex] \left \{ {{6x+4y=20} \atop {-(6x-y=10)}} \right. \\ \\5y=10[/tex]

Divide both sides by 5:
5y/5=10/5
y=2

Substitute this into the second equation
6x-y=10
6x-2=10

Add 2 to each side:
6x-2+2=10+2
6x=12

Divide both sides by 6:
6x/6 = 12/6
x=2

The coordinates (2, 2) represent the solution to this system.

For the third system:
[tex] \left \{ {{4x-3y=5} \atop {3x+2y=-9}} \right. [/tex]

We can make the coefficients of x the same by multiplying the first equation by 3 and the second by 4:
[tex] \left \{ {{3(4x-3y=5)} \atop {4(3x+2y=-9)}} \right. \\ \\ \left \{ {{12x-9y=15} \atop {12x+8y=-36}} \right. [/tex]

Since the coefficients of x are the same, we can cancel them.  Since they are both positive, we will subtract:
[tex] \left \{ {{12x-9y=15} \atop {-(12x+8y=-36)}} \right. \\ \\-17y=51[/tex]

Divide both sides by -17:
-17y/-17 = 51/-17
y=-3

Substitute this into the first equation:
4x-3y=5
4x-3(-3)=5
4x--9=5
4x+9=5

Subtract 9 from each side:
4x+9-9=5-9
4x=-4

Divide each side by 4:
4x/4 = -4/4
x=-1

The coordinates (-1, -3) represent the solution of the third system..

For the fourth system:
[tex] \left \{ {{x+y=7} \atop {x-y=-1}} \right. [/tex]

Since the coordinates of x are the same and both are positive, we can cancel x by subtracting:
[tex] \left \{ {{x+y=7} \atop {-(x-y=-1)}} \right. \\ \\2y=8[/tex]

Divide both sides by 2:
2y/2 = 8/2
y=4

Substitute this into the first equation
x+y=7
x+4=7

Subtract 4 from each side:
x+4-4=7-4
x=3

The coordinates (3, 4) represent the solution to the fourth system.

For the fifth system:
[tex] \left \{ {{y=3x-7} \atop {y=2x-5}} \right. [/tex]

Since y is isolated in each equation, we can set them equation to each other:
3x-7=2x-5

Subtract 2x from each side:
3x-7-2x=2x-5-2x
x-7=-5

Add 7 to each side:
x-7+7=-5+7
x=2

Substitute this into the first equation:
y=3x-7
y=3(2)-7
y=6-7
y=-1

The coordinates (2, -1) represent the solution to the fifth system.