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.