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Sagot :
To determine which point is a solution to the given system of linear equations:
[tex]\[ \begin{cases} y = -x + 1 \\ 3x - y = 3 \end{cases} \][/tex]
we need to check each point to see if it satisfies both equations.
Let's consider each point one by one.
1. Point [tex]\(\left(\frac{3}{2}, -\frac{1}{2}\right)\)[/tex]:
- Substitute [tex]\(x = \frac{3}{2}\)[/tex] and [tex]\(y = -\frac{1}{2}\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\left(\frac{3}{2}\right) + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\frac{3}{2} + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\frac{1}{2} \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3\left(\frac{3}{2}\right) - \left(-\frac{1}{2}\right) = 3 \quad \Rightarrow \quad \frac{9}{2} + \frac{1}{2} = 3 \quad \Rightarrow \quad 5 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
2. Point [tex]\((1, 0)\)[/tex]:
- Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad 0 = -1 + 1 \quad \Rightarrow \quad 0 = 0 \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3(1) - 0 = 3 \quad \Rightarrow \quad 3 = 3 \quad \text{(True)} \][/tex]
- This point satisfies both equations.
3. Point [tex]\(\left(-\frac{1}{2}, \frac{3}{2}\right)\)[/tex]:
- Substitute [tex]\(x = -\frac{1}{2}\)[/tex] and [tex]\(y = \frac{3}{2}\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad \frac{3}{2} = -\left(-\frac{1}{2}\right) + 1 \quad \Rightarrow \quad \frac{3}{2} = \frac{1}{2} + 1 \quad \Rightarrow \quad \frac{3}{2} = \frac{3}{2} \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3\left(-\frac{1}{2}\right) - \frac{3}{2} = 3 \quad \Rightarrow \quad -\frac{3}{2} - \frac{3}{2} = 3 \quad \Rightarrow \quad -3 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
4. Point [tex]\((0, 1)\)[/tex]:
- Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 1\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad 1 = -0 + 1 \quad \Rightarrow \quad 1 = 1 \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3(0) - 1 = 3 \quad \Rightarrow \quad -1 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
Therefore, the point [tex]\((1, 0)\)[/tex] is the solution to the system of linear equations.
[tex]\[ \begin{cases} y = -x + 1 \\ 3x - y = 3 \end{cases} \][/tex]
we need to check each point to see if it satisfies both equations.
Let's consider each point one by one.
1. Point [tex]\(\left(\frac{3}{2}, -\frac{1}{2}\right)\)[/tex]:
- Substitute [tex]\(x = \frac{3}{2}\)[/tex] and [tex]\(y = -\frac{1}{2}\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\left(\frac{3}{2}\right) + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\frac{3}{2} + 1 \quad \Rightarrow \quad -\frac{1}{2} = -\frac{1}{2} \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3\left(\frac{3}{2}\right) - \left(-\frac{1}{2}\right) = 3 \quad \Rightarrow \quad \frac{9}{2} + \frac{1}{2} = 3 \quad \Rightarrow \quad 5 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
2. Point [tex]\((1, 0)\)[/tex]:
- Substitute [tex]\(x = 1\)[/tex] and [tex]\(y = 0\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad 0 = -1 + 1 \quad \Rightarrow \quad 0 = 0 \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3(1) - 0 = 3 \quad \Rightarrow \quad 3 = 3 \quad \text{(True)} \][/tex]
- This point satisfies both equations.
3. Point [tex]\(\left(-\frac{1}{2}, \frac{3}{2}\right)\)[/tex]:
- Substitute [tex]\(x = -\frac{1}{2}\)[/tex] and [tex]\(y = \frac{3}{2}\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad \frac{3}{2} = -\left(-\frac{1}{2}\right) + 1 \quad \Rightarrow \quad \frac{3}{2} = \frac{1}{2} + 1 \quad \Rightarrow \quad \frac{3}{2} = \frac{3}{2} \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3\left(-\frac{1}{2}\right) - \frac{3}{2} = 3 \quad \Rightarrow \quad -\frac{3}{2} - \frac{3}{2} = 3 \quad \Rightarrow \quad -3 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
4. Point [tex]\((0, 1)\)[/tex]:
- Substitute [tex]\(x = 0\)[/tex] and [tex]\(y = 1\)[/tex]:
[tex]\[ y = -x + 1 \quad \Rightarrow \quad 1 = -0 + 1 \quad \Rightarrow \quad 1 = 1 \quad \text{(True)} \][/tex]
[tex]\[ 3x - y = 3 \quad \Rightarrow \quad 3(0) - 1 = 3 \quad \Rightarrow \quad -1 \neq 3 \quad \text{(False)} \][/tex]
- This point does not satisfy the second equation.
Therefore, the point [tex]\((1, 0)\)[/tex] is the solution to the system of linear equations.
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