Question

which systems of equations do not have any solutions? select all that apply. click or tap the correct systems of equations.
$x + y = 5$
$x + y = 6$
$x + y = 7$
$x - y = 2$
$3x + y = 6$
$2x + y = 8$
$6x + 2y = 12$
$2x + y = 4$

Explanation:

Step1: Analyze first system

Check $x+y=5$ and $x+y=6$. Rearrange to slope-intercept form: $y=-x+5$, $y=-x+6$. Same slope ($-1$), different y-intercepts. Parallel lines, no solution.

Step2: Analyze second system

Check $x+y=7$ and $x-y=2$. Rearrange: $y=-x+7$, $y=x-2$. Different slopes, lines intersect. Has one solution.

Step3: Analyze third system

Check $3x+y=6$ and $2x+y=8$. Rearrange: $y=-3x+6$, $y=-2x+8$. Different slopes, lines intersect. Has one solution.

Step4: Analyze fourth system

Check $6x+2y=12$ and $2x+y=4$. Simplify first equation: divide by 2, get $3x+y=6$? No, divide $6x+2y=12$ by 2: $3x+y=6$? Wait, no: $\frac{6x}{2}+\frac{2y}{2}=\frac{12}{2}$ → $3x+y=6$. Wait, no, the second equation is $2x+y=4$. Wait, correct simplification: $6x+2y=12$ can be simplified to $3x+y=6$, which has slope $-3$, y-intercept 6. $2x+y=4$ has slope $-2$, y-intercept 4. Different slopes, intersect? Wait no, wait another way: subtract equations. Multiply $2x+y=4$ by 3: $6x+3y=12$. Subtract $6x+2y=12$: $(6x+3y)-(6x+2y)=12-12$ → $y=0$. Then $2x+0=4$ → $x=2$. So this system has solution $(2,0)$.
Wait, correction for first system: $x+y=5$ and $x+y=6$: subtract equations: $(x+y)-(x+y)=5-6$ → $0=-1$, which is false, no solution.
Wait another system: $3x+y=6$ and $2x+y=8$: subtract: $(3x+y)-(2x+y)=6-8$ → $x=-2$, then $3(-2)+y=6$ → $y=12$, so solution $(-2,12)$.
Wait, did I miss a system? No, the four systems are:

1. $x+y=5$; $x+y=6$
2. $x+y=7$; $x-y=2$
3. $3x+y=6$; $2x+y=8$
4. $6x+2y=12$; $2x+y=4$

Wait, recheck system 4: $6x+2y=12$ divide by 2 is $3x+y=6$, and $2x+y=4$. These are not identical, so they intersect. The only system with no solution is the first one? Wait no, wait system 1: $x+y=5$ and $x+y=6$: same slope, different intercepts, parallel, no solution.
Wait, is there another? Let's check system 4 again: $6x+2y=12$ and $2x+y=4$. If we multiply $2x+y=4$ by 3, we get $6x+3y=12$. Subtract $6x+2y=12$: $y=0$, then $x=2$. So that's a valid solution.
Wait, the second system: $x+y=7$, $x-y=2$. Add equations: $2x=9$ → $x=4.5$, $y=2.5$. Solution exists.
Third system: $3x+y=6$, $2x+y=8$. Subtract: $x=-2$, $y=12$. Solution exists.
Only the first system has no solution? Wait no, wait the first system is $x+y=5$ and $x+y=6$. Yes, that's the only one with no solution.

Wait, wait, maybe I misread system 4: $6x+2y=12$ and $2x+y=4$. Wait $2x+y=4$ multiplied by 2 is $4x+2y=8$. Subtract from $6x+2y=12$: $2x=4$ → $x=2$, $y=0$. Correct, solution exists.

Answer:

The system with no solution is:

1. $x + y = 5$; $x + y = 6$