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image of a coordinate plane with two lines
what system does it represent, and what is the solution of this system?
- ( y + 4 = x ) and ( y - x = -6 ); solution ( = (-1, -5) )
- ( y + 4 = x ) and ( y - x = -6 ); solution ( = (-5, -1) )
- ( y + 4 = x ) and ( x + y = -6 ); solution ( = (-5, -1) )
- ( y + 4 = x ) and ( x + y = -6 ); solution ( = (-1, -5) )
question id 69523
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# Brief Explanations:
1. First, analyze the equations. The line $ y + 4=x $ can be rewritten as $ y=x - 4 $. For the other line, if we consider the intersection and the slope - intercept form, let's check the equations. The equation $ x + y=-6 $ can be rewritten as $ y=-x - 6 $.
   - To find the solution of the system of equations, we can solve the system $ \begin{cases}y=x - 4\x + y=-6\end{cases} $. Substitute $ y=x - 4 $ into $ x + y=-6 $, we get $ x+(x - 4)=-6 $.
   - Simplify the left - hand side: $ x+x - 4=-6 $, which is $ 2x-4=-6 $. Add 4 to both sides: $ 2x=-6 + 4=-2 $, then $ x=-1 $.
   - Substitute $ x = - 1 $ into $ y=x - 4 $, we get $ y=-1-4=-5 $. Wait, no, let's check the third and fourth options. Wait, the two lines in the graph: one has a positive slope (from the equation $ y + 4=x $ or $ y=x - 4 $) and the other has a negative slope. The equation $ y - x=-6 $ is the same as $ y=x - 6 $, which has the same slope as $ y=x - 4 $, so they are parallel and have no solution. But the equation $ x + y=-6 $ (or $ y=-x - 6 $) has a slope of - 1, which is different from the slope of $ y=x - 4 $ (slope = 1).
   - Now, solve the system $ \begin{cases}y=x - 4\x + y=-6\end{cases} $. Substitute $ y=x - 4 $ into $ x + y=-6 $: $ x+(x - 4)=-6\Rightarrow2x-4=-6\Rightarrow2x=-2\Rightarrow x=-1 $, then $ y=-1 - 4=-5 $? No, wait, if we substitute $ x=-5 $ into $ y=x - 4 $, $ y=-5 - 4=-9 $, which is wrong. Wait, let's solve $ \begin{cases}y=x - 4\x + y=-6\end{cases} $ again. $ x+(x - 4)=-6\Rightarrow2x=-2\Rightarrow x=-1 $, $ y=-1-4=-5 $? No, that's not correct. Wait, let's check the intersection point of the two lines in the graph. The intersection point of the two lines (the solution of the system) should satisfy both equations. Let's check the fourth option: equations $ y + 4=x $ (or $ y=x - 4 $) and $ x + y=-6 $ (or $ y=-x - 6 $).
     - Solve $ \begin{cases}y=x - 4\y=-x - 6\end{cases} $. Set $ x - 4=-x - 6 $. Add $ x $ to both sides: $ 2x-4=-6 $. Add 4 to both sides: $ 2x=-2 $, so $ x=-1 $. Then $ y=-1 - 4=-5 $? No, wait, if we solve $ \begin{cases}y=x - 4\x + y=-6\end{cases} $, when $ x=-5 $, $ y=-5 - 4=-9 $, no. Wait, let's check the third option: equations $ y + 4=x $ ( $ y=x - 4 $) and $ x + y=-6 $ ( $ y=-x - 6 $). Solve $ x - 4=-x - 6\Rightarrow2x=-2\Rightarrow x=-1 $, $ y=-1 - 4=-5 $? No, I made a mistake. Wait, let's check the intersection point of the two lines in the graph. The two lines intersect at $ (-5,-1) $? Wait, no, let's substitute $ x=-5 $ and $ y=-1 $ into $ y + 4=x $: $ -1 + 4=3
eq - 5 $. Substitute $ x=-5 $ and $ y=-1 $ into $ x + y=-6 $: $ -5+( - 1)=-6 $, which is correct. Substitute $ x=-5 $ and $ y=-1 $ into $ y + 4=x $: $ -1 + 4 = 3
eq - 5 $. Wait, no, $ y + 4=x $ when $ x=-5 $, $ y=x - 4=-5 - 4=-9 $, which is wrong. Wait, I think I messed up. Let's start over.
     - The equation $ y + 4=x $ can be written as $ y=x - 4 $. Let's find two points on this line: when $ x = 0 $, $ y=-4 $; when $ x = 4 $, $ y=0 $. The other line: let's assume it's $ x + y=-6 $, when $ x = 0 $, $ y=-6 $; when $ x=-6 $, $ y = 0 $. The intersection of $ y=x - 4 $ and $ x + y=-6 $:
       - Substitute $ y=x - 4 $ into $ x + y=-6 $: $ x+(x - 4)=-6\Rightarrow2x-4=-6\Rightarrow2x=-2\Rightarrow x=-1$, $ y=-1 - 4=-5 $. But the graph shows the intersection at $ (-5,-1) $? Wait, no, maybe I misread the graph. Wait, the two lines: one goes from $ (-6,0) $ to $ (4,0) $? No, the first line $ y + 4=x $ passes through $ (4,0) $ (when $ y = 0 $, $ x = 4 $) and $ (0,-4) $. The second line: let's take two points. If the intersection point is $ (-5,-1) $, substitute into $ y + 4=x $: $ -1+4 = 3
eq - 5 $. Substitute into $ x + y=-6 $: $ -5+( - 1)=-6 $, which is correct. Substitute $ (-5,-1) $ into $ y + 4=x $: $ -1 + 4=3
eq - 5 $. Wait, the correct system should be $ y + 4=x $ ( $ y=x - 4 $) and $ x + y=-6 $ ( $ y=-x - 6 $). Let's solve them:
         - $ x - 4=-x - 6\Rightarrow2x=-2\Rightarrow x=-1$, $ y=-1 - 4=-5 $. But the graph's intersection seems to be at $ (-5,-1) $. Wait, maybe the first line is $ y+4 = x $ ( $ y=x - 4 $) and the second line is $ y - x=-6 $ ( $ y=x - 6 $)? But these two lines are parallel (same slope $ m = 1 $) and have no solution. So the only non - parallel lines are when the second equation is $ x + y=-6 $.
         - Now, check the options:
           - Option 1: Equations $ y + 4=x $ and $ y - x=-6 $ (same as $ y=x - 6 $) are parallel, no solution.
           - Option 2: Equations $ y + 4=x $ and $ y - x=-6 $ are parallel, no solution.
           - Option 3: Equations $ y + 4=x $ ( $ y=x - 4 $) and $ x + y=-6 $ ( $ y=-x - 6 $). Solve them: $ x - 4=-x - 6\Rightarrow2x=-2\Rightarrow x=-1$, $ y=-5 $. But the solution given is $ (-5,-1) $. Wait, substitute $ x=-5 $ and $ y=-1 $ into $ y + 4=x $: $ -1 + 4 = 3
eq - 5 $. Substitute into $ x + y=-6 $: $ -5+( - 1)=-6 $, correct. Substitute $ x=-5 $ and $ y=-1 $ into $ y + 4=x $: no. Wait, maybe the first equation is $ y+4=x $ ( $ y=x - 4 $) and the second equation is $ x + y=-6 $ ( $ y=-x - 6 $). Let's check the intersection point of the two lines. The line $ y=x - 4 $ has a slope of 1, $ y=-x - 6 $ has a slope of - 1. The intersection is at $ x=-1,y=-5 $. But the graph in the picture: let's count the grid. If the intersection is at $ (-5,-1) $, maybe I made a mistake in the equation of the first line. Wait, the first line: when $ x = 4 $, $ y = 0 $; when $ x=0 $, $ y=-4 $. The second line: when $ x=-6 $, $ y = 0 $; when $ x = 0 $, $ y=-6 $. The intersection of these two lines:
             - $ x - 4=-x - 6\Rightarrow2x=-2\Rightarrow x=-1$, $ y=-5 $. But the option 3 has solution $ (-5,-1) $. Wait, substitute $ (-5,-1) $ into $ y + 4=x $: $ -1 + 4=3
eq - 5 $. Substitute into $ x + y=-6 $: $ -5+( - 1)=-6 $, correct. Substitute $ (-5,-1) $ into $ y + 4=x $: no. Substitute $ (-1,-5) $ into $ x + y=-6 $: $ -1+( - 5)=-6 $, correct. Substitute $ (-1,-5) $ into $ y + 4=x $: $ -5 + 4=-1$, correct. Oh! I made a mistake earlier. $ y+4=x $ when $ y=-5 $, $ x=-5 + 4=-1 $, which is correct. And $ x + y=-6 $ when $ x=-1 $ and $ y=-5 $, $ -1+( - 5)=-6 $, correct. And the two lines: $ y=x - 4 $ (slope 1) and $ y=-x - 6 $ (slope - 1) intersect at $ (-1,-5) $. The first two options have parallel lines (same slope) and no solution. So the correct option is the fourth one: $ y + 4=x $ and $ x + y=-6 $; solution $ =(-1,-5) $.
# Answer:
$ y + 4 = x $ and $ x + y=-6 $; solution $ =(-1,-5) $ (the fourth option: $ y + 4 = x $ and $ x + y=-6 $; solution $ =(-1,-5) $)

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