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Sagot :
Sure, let's solve this step-by-step to find the ordered pair that satisfies both equations [tex]\( y = x \)[/tex] and [tex]\( y = 2x \)[/tex].
To determine the intersection [tex]\( A \cap B \)[/tex], we need to find points [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously.
1. Equation [tex]\( y = x \)[/tex]
- This means that for any point [tex]\((x, y)\)[/tex] on this line, the [tex]\(y\)[/tex]-coordinate is equal to the [tex]\(x\)[/tex]-coordinate.
2. Equation [tex]\( y = 2x \)[/tex]
- Here, the [tex]\(y\)[/tex]-coordinate is double the [tex]\(x\)[/tex]-coordinate.
To find the intersection of these two sets, we need [tex]\((x, y)\)[/tex] such that:
[tex]\[ y = x \][/tex]
and
[tex]\[ y = 2x \][/tex]
This implies:
[tex]\[ x = 2x \][/tex]
Solving for [tex]\(x\)[/tex], we get:
[tex]\[ x - 2x = 0 \][/tex]
[tex]\[ -x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Substitute [tex]\( x = 0 \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
[tex]\[ y = x \][/tex]
[tex]\[ y = 0 \][/tex]
So, the point [tex]\((0, 0)\)[/tex] satisfies both equations.
Let's verify the other given options:
- Option [tex]\((1, 1)\)[/tex]:
- [tex]\( y = x \Rightarrow 1 = 1 \)[/tex] (True)
- [tex]\( y = 2x \Rightarrow 1 = 2 \cdot 1 = 2 \)[/tex] (False)
- Option [tex]\((1, 2)\)[/tex]:
- [tex]\( y = x \Rightarrow 2 = 1 \)[/tex] (False)
- [tex]\( y = 2x \Rightarrow 2 = 2 \cdot 1 = 2 \)[/tex] (True)
- Option [tex]\((2, 1)\)[/tex]:
- [tex]\( y = x \Rightarrow 1 = 2 \)[/tex] (False)
- [tex]\( y = 2x \Rightarrow 1 = 2 \cdot 2 = 4 \)[/tex] (False)
Therefore, the only option that satisfies both [tex]\( y = x \)[/tex] and [tex]\( y = 2x \)[/tex] is:
[tex]\[ (0, 0) \][/tex]
So, the ordered pair that satisfies [tex]\( A \cap B \)[/tex] is [tex]\((0, 0)\)[/tex].
To determine the intersection [tex]\( A \cap B \)[/tex], we need to find points [tex]\((x, y)\)[/tex] that satisfy both equations simultaneously.
1. Equation [tex]\( y = x \)[/tex]
- This means that for any point [tex]\((x, y)\)[/tex] on this line, the [tex]\(y\)[/tex]-coordinate is equal to the [tex]\(x\)[/tex]-coordinate.
2. Equation [tex]\( y = 2x \)[/tex]
- Here, the [tex]\(y\)[/tex]-coordinate is double the [tex]\(x\)[/tex]-coordinate.
To find the intersection of these two sets, we need [tex]\((x, y)\)[/tex] such that:
[tex]\[ y = x \][/tex]
and
[tex]\[ y = 2x \][/tex]
This implies:
[tex]\[ x = 2x \][/tex]
Solving for [tex]\(x\)[/tex], we get:
[tex]\[ x - 2x = 0 \][/tex]
[tex]\[ -x = 0 \][/tex]
[tex]\[ x = 0 \][/tex]
Substitute [tex]\( x = 0 \)[/tex] back into either equation to find [tex]\( y \)[/tex]:
[tex]\[ y = x \][/tex]
[tex]\[ y = 0 \][/tex]
So, the point [tex]\((0, 0)\)[/tex] satisfies both equations.
Let's verify the other given options:
- Option [tex]\((1, 1)\)[/tex]:
- [tex]\( y = x \Rightarrow 1 = 1 \)[/tex] (True)
- [tex]\( y = 2x \Rightarrow 1 = 2 \cdot 1 = 2 \)[/tex] (False)
- Option [tex]\((1, 2)\)[/tex]:
- [tex]\( y = x \Rightarrow 2 = 1 \)[/tex] (False)
- [tex]\( y = 2x \Rightarrow 2 = 2 \cdot 1 = 2 \)[/tex] (True)
- Option [tex]\((2, 1)\)[/tex]:
- [tex]\( y = x \Rightarrow 1 = 2 \)[/tex] (False)
- [tex]\( y = 2x \Rightarrow 1 = 2 \cdot 2 = 4 \)[/tex] (False)
Therefore, the only option that satisfies both [tex]\( y = x \)[/tex] and [tex]\( y = 2x \)[/tex] is:
[tex]\[ (0, 0) \][/tex]
So, the ordered pair that satisfies [tex]\( A \cap B \)[/tex] is [tex]\((0, 0)\)[/tex].
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