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Sagot :
To find the solution of the system of equations
[tex]\[ \left\{ \begin{array}{r} x^2 + y^2 = 4 \\ x - y = 1 \end{array} \right. \][/tex]
we will analyze and solve these equations step-by-step.
### Step 1: Understand the Equations
1. The first equation [tex]\(x^2 + y^2 = 4\)[/tex] represents a circle with a radius of 2 centered at the origin [tex]\((0,0)\)[/tex].
2. The second equation [tex]\(x - y = 1\)[/tex] represents a straight line with a slope of 1 and a y-intercept of -1.
### Step 2: Solve the System of Equations Algebraically
To find the points where these two graphs intersect, we will solve the system of equations.
1. From the second equation [tex]\(x - y = 1\)[/tex], express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = y + 1 \][/tex]
2. Substitute [tex]\(x = y + 1\)[/tex] into the first equation [tex]\(x^2 + y^2 = 4\)[/tex]:
[tex]\[ (y + 1)^2 + y^2 = 4 \][/tex]
3. Expand and simplify this equation:
[tex]\[ y^2 + 2y + 1 + y^2 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 = 4 \][/tex]
4. Subtract 4 from both sides to set the equation to zero:
[tex]\[ 2y^2 + 2y + 1 - 4 = 0 \][/tex]
[tex]\[ 2y^2 + 2y - 3 = 0 \][/tex]
5. Divide the entire equation by 2:
[tex]\[ y^2 + y - \frac{3}{2} = 0 \][/tex]
6. Factor this quadratic equation (or use the quadratic formula):
[tex]\[ y = \frac{-1 \pm \sqrt{1 + 6}}{2} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{7}}{2} \][/tex]
7. Hence, we get two solutions for [tex]\(y\)[/tex]:
[tex]\[ y_1 = \frac{-1 - \sqrt{7}}{2}, \quad y_2 = \frac{-1 + \sqrt{7}}{2} \][/tex]
8. Substitute these [tex]\(y\)[/tex] values back into the equation [tex]\(x = y + 1\)[/tex] to find the corresponding [tex]\(x\)[/tex] values:
[tex]\[ x_1 = \frac{-1 - \sqrt{7}}{2} + 1 = \frac{2 - 1 - \sqrt{7}}{2} = \frac{1 - \sqrt{7}}{2} \][/tex]
[tex]\[ x_2 = \frac{-1 + \sqrt{7}}{2} + 1 = \frac{2 - 1 + \sqrt{7}}{2} = \frac{1 + \sqrt{7}}{2} \][/tex]
### Step 3: Write the Solution Pairs
The solutions for the system of equations are the points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right), \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right) \][/tex]
So the graph representing the solution of the system [tex]\(\left\{\begin{array}{r} x^2 + y^2 = 4 \\ x - y = 1 \end{array}\right.\)[/tex] intersects at the points:
[tex]\[ \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right), \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right). \][/tex]
Thus, when you plot these points on the graph, the circle [tex]\(x^2 + y^2 = 4\)[/tex] intersects with the line [tex]\(x - y = 1\)[/tex] at these two points.
[tex]\[ \left\{ \begin{array}{r} x^2 + y^2 = 4 \\ x - y = 1 \end{array} \right. \][/tex]
we will analyze and solve these equations step-by-step.
### Step 1: Understand the Equations
1. The first equation [tex]\(x^2 + y^2 = 4\)[/tex] represents a circle with a radius of 2 centered at the origin [tex]\((0,0)\)[/tex].
2. The second equation [tex]\(x - y = 1\)[/tex] represents a straight line with a slope of 1 and a y-intercept of -1.
### Step 2: Solve the System of Equations Algebraically
To find the points where these two graphs intersect, we will solve the system of equations.
1. From the second equation [tex]\(x - y = 1\)[/tex], express [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]:
[tex]\[ x = y + 1 \][/tex]
2. Substitute [tex]\(x = y + 1\)[/tex] into the first equation [tex]\(x^2 + y^2 = 4\)[/tex]:
[tex]\[ (y + 1)^2 + y^2 = 4 \][/tex]
3. Expand and simplify this equation:
[tex]\[ y^2 + 2y + 1 + y^2 = 4 \][/tex]
[tex]\[ 2y^2 + 2y + 1 = 4 \][/tex]
4. Subtract 4 from both sides to set the equation to zero:
[tex]\[ 2y^2 + 2y + 1 - 4 = 0 \][/tex]
[tex]\[ 2y^2 + 2y - 3 = 0 \][/tex]
5. Divide the entire equation by 2:
[tex]\[ y^2 + y - \frac{3}{2} = 0 \][/tex]
6. Factor this quadratic equation (or use the quadratic formula):
[tex]\[ y = \frac{-1 \pm \sqrt{1 + 6}}{2} \][/tex]
[tex]\[ y = \frac{-1 \pm \sqrt{7}}{2} \][/tex]
7. Hence, we get two solutions for [tex]\(y\)[/tex]:
[tex]\[ y_1 = \frac{-1 - \sqrt{7}}{2}, \quad y_2 = \frac{-1 + \sqrt{7}}{2} \][/tex]
8. Substitute these [tex]\(y\)[/tex] values back into the equation [tex]\(x = y + 1\)[/tex] to find the corresponding [tex]\(x\)[/tex] values:
[tex]\[ x_1 = \frac{-1 - \sqrt{7}}{2} + 1 = \frac{2 - 1 - \sqrt{7}}{2} = \frac{1 - \sqrt{7}}{2} \][/tex]
[tex]\[ x_2 = \frac{-1 + \sqrt{7}}{2} + 1 = \frac{2 - 1 + \sqrt{7}}{2} = \frac{1 + \sqrt{7}}{2} \][/tex]
### Step 3: Write the Solution Pairs
The solutions for the system of equations are the points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex]:
[tex]\[ \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right), \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right) \][/tex]
So the graph representing the solution of the system [tex]\(\left\{\begin{array}{r} x^2 + y^2 = 4 \\ x - y = 1 \end{array}\right.\)[/tex] intersects at the points:
[tex]\[ \left( \frac{1 - \sqrt{7}}{2}, \frac{-1 - \sqrt{7}}{2} \right), \quad \left( \frac{1 + \sqrt{7}}{2}, \frac{-1 + \sqrt{7}}{2} \right). \][/tex]
Thus, when you plot these points on the graph, the circle [tex]\(x^2 + y^2 = 4\)[/tex] intersects with the line [tex]\(x - y = 1\)[/tex] at these two points.
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