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Consider the following system of equations: -x + y = 5 y = (x-3)2 + 2Asked by the p
Consider the following system of equations:
− x + y = 5
y = (x−3)^2 +2
− x + y = 5
y = (x−3)^2 +2
Answers
Answered by
the p
Is (1,6)
a solution to the system of equations?
a solution to the system of equations?
Answered by
the p
Solve the system of equations using substitution.
Answered by
the p
Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions? If so, how could you tell from the graph? If not, explain why not.
Answered by
the p
shorter
Answered by
GPT-4o mini
To solve the system of equations
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\),
we can substitute the expression for \(y\) from the second equation into the first equation.
Starting with equation 1:
\[
y = -x + 5
\]
Now we substitute this into equation 2:
\[
-x + 5 = (x - 3)^2 + 2
\]
Next, we will expand and simplify the right side:
\[
(x - 3)^2 = x^2 - 6x + 9
\]
So, the second equation becomes:
\[
-x + 5 = x^2 - 6x + 9 + 2
\]
\[
-x + 5 = x^2 - 6x + 11
\]
Now, we can set up the equation by bringing all terms to one side:
\[
0 = x^2 - 6x + 11 + x - 5
\]
Combining like terms:
\[
0 = x^2 - 5x + 6
\]
Now we factor the quadratic equation:
\[
0 = (x - 2)(x - 3)
\]
Setting each factor to zero gives us the possible values for \(x\):
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Next, we find the corresponding \(y\) values for each \(x\) value using \(y = -x + 5\):
1. For \(x = 2\):
\[
y = -2 + 5 = 3
\]
2. For \(x = 3\):
\[
y = -3 + 5 = 2
\]
Thus, the solutions to the system of equations are:
\[
(2, 3) \quad \text{and} \quad (3, 2)
\]
In conclusion, the solutions are:
\[
\boxed{(2, 3) \text{ and } (3, 2)}
\]
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\),
we can substitute the expression for \(y\) from the second equation into the first equation.
Starting with equation 1:
\[
y = -x + 5
\]
Now we substitute this into equation 2:
\[
-x + 5 = (x - 3)^2 + 2
\]
Next, we will expand and simplify the right side:
\[
(x - 3)^2 = x^2 - 6x + 9
\]
So, the second equation becomes:
\[
-x + 5 = x^2 - 6x + 9 + 2
\]
\[
-x + 5 = x^2 - 6x + 11
\]
Now, we can set up the equation by bringing all terms to one side:
\[
0 = x^2 - 6x + 11 + x - 5
\]
Combining like terms:
\[
0 = x^2 - 5x + 6
\]
Now we factor the quadratic equation:
\[
0 = (x - 2)(x - 3)
\]
Setting each factor to zero gives us the possible values for \(x\):
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
Next, we find the corresponding \(y\) values for each \(x\) value using \(y = -x + 5\):
1. For \(x = 2\):
\[
y = -2 + 5 = 3
\]
2. For \(x = 3\):
\[
y = -3 + 5 = 2
\]
Thus, the solutions to the system of equations are:
\[
(2, 3) \quad \text{and} \quad (3, 2)
\]
In conclusion, the solutions are:
\[
\boxed{(2, 3) \text{ and } (3, 2)}
\]
Answered by
GPT-4o mini
To determine if the point \((1, 6)\) is a solution to the system of equations, we need to check if it satisfies both equations:
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)
Let's substitute \(x = 1\) and \(y = 6\) into both equations.
### Checking the first equation:
\[
-x + y = 5
\]
Substituting in the values:
\[
-1 + 6 = 5 \quad \Rightarrow \quad 5 = 5 \quad \text{(True)}
\]
### Checking the second equation:
\[
y = (x - 3)^2 + 2
\]
Substituting in the values:
\[
6 = (1 - 3)^2 + 2
\]
Calculating the right side:
\[
(1 - 3)^2 = (-2)^2 = 4 \quad \Rightarrow \quad 6 = 4 + 2
\]
\[
6 = 6 \quad \text{(True)}
\]
Since \((1, 6)\) satisfies both equations, we conclude that:
\[
(1, 6) \text{ is a solution to the system of equations.}
\]
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)
Let's substitute \(x = 1\) and \(y = 6\) into both equations.
### Checking the first equation:
\[
-x + y = 5
\]
Substituting in the values:
\[
-1 + 6 = 5 \quad \Rightarrow \quad 5 = 5 \quad \text{(True)}
\]
### Checking the second equation:
\[
y = (x - 3)^2 + 2
\]
Substituting in the values:
\[
6 = (1 - 3)^2 + 2
\]
Calculating the right side:
\[
(1 - 3)^2 = (-2)^2 = 4 \quad \Rightarrow \quad 6 = 4 + 2
\]
\[
6 = 6 \quad \text{(True)}
\]
Since \((1, 6)\) satisfies both equations, we conclude that:
\[
(1, 6) \text{ is a solution to the system of equations.}
\]
Answered by
GPT-4o mini
To solve the system of equations using substitution, let's first rewrite the two equations we are working with:
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)
### Step 1: Solve one equation for one variable.
We can solve the first equation for \(y\):
\[
y = x + 5
\]
### Step 2: Substitute this expression into the second equation.
Now we substitute \(y\) in the second equation:
\[
x + 5 = (x - 3)^2 + 2
\]
### Step 3: Expand and simplify the equation.
Now, expand the right-hand side:
\[
x + 5 = (x^2 - 6x + 9) + 2
\]
\[
x + 5 = x^2 - 6x + 11
\]
### Step 4: Set the equation to zero.
Rearranging gives us:
\[
0 = x^2 - 6x - x + 11 - 5
\]
\[
0 = x^2 - 7x + 6
\]
### Step 5: Factor the quadratic equation.
Next, we factor the quadratic:
\[
0 = (x - 1)(x - 6)
\]
### Step 6: Solve for \(x\).
Setting each factor equal to zero gives:
1. \(x - 1 = 0 \quad \Rightarrow \quad x = 1\)
2. \(x - 6 = 0 \quad \Rightarrow \quad x = 6\)
### Step 7: Substitute back to find \(y\).
Now, we will substitute back to find the corresponding \(y\) values.
#### For \(x = 1\):
\[
y = 1 + 5 = 6
\]
The first solution is \((1, 6)\).
#### For \(x = 6\):
\[
y = 6 + 5 = 11
\]
The second solution is \((6, 11)\).
### Final Solutions:
The solutions to the system of equations are:
\[
\boxed{(1, 6) \text{ and } (6, 11)}
\]
1. \(-x + y = 5\)
2. \(y = (x - 3)^2 + 2\)
### Step 1: Solve one equation for one variable.
We can solve the first equation for \(y\):
\[
y = x + 5
\]
### Step 2: Substitute this expression into the second equation.
Now we substitute \(y\) in the second equation:
\[
x + 5 = (x - 3)^2 + 2
\]
### Step 3: Expand and simplify the equation.
Now, expand the right-hand side:
\[
x + 5 = (x^2 - 6x + 9) + 2
\]
\[
x + 5 = x^2 - 6x + 11
\]
### Step 4: Set the equation to zero.
Rearranging gives us:
\[
0 = x^2 - 6x - x + 11 - 5
\]
\[
0 = x^2 - 7x + 6
\]
### Step 5: Factor the quadratic equation.
Next, we factor the quadratic:
\[
0 = (x - 1)(x - 6)
\]
### Step 6: Solve for \(x\).
Setting each factor equal to zero gives:
1. \(x - 1 = 0 \quad \Rightarrow \quad x = 1\)
2. \(x - 6 = 0 \quad \Rightarrow \quad x = 6\)
### Step 7: Substitute back to find \(y\).
Now, we will substitute back to find the corresponding \(y\) values.
#### For \(x = 1\):
\[
y = 1 + 5 = 6
\]
The first solution is \((1, 6)\).
#### For \(x = 6\):
\[
y = 6 + 5 = 11
\]
The second solution is \((6, 11)\).
### Final Solutions:
The solutions to the system of equations are:
\[
\boxed{(1, 6) \text{ and } (6, 11)}
\]
Answered by
GPT-4o mini
A system of two equations, consisting of one linear equation and one quadratic equation, generally cannot have an infinite number of solutions. Here's why, along with a graphical explanation:
### Why Not Infinite Solutions?
1. **Nature of the Equations**:
- A linear equation represents a straight line in a two-dimensional space (e.g., \(y = mx + b\)).
- A quadratic equation represents a parabola (e.g., \(y = ax^2 + bx + c\)).
2. **Intersection Points**:
- The possible intersections between a straight line and a parabola can be:
- **No Intersection**: The line does not touch the parabola at all.
- **One Intersection**: The line is tangent to the parabola at exactly one point.
- **Two Intersections**: The line intersects the parabola at two distinct points.
Since the straight line is a linear function and the parabola is a quadratic function, they can share at most two intersection points in a 2D Cartesian coordinate system but cannot share an infinite number of points.
### Graphical Representation:
- If you graph a linear equation and a quadratic equation, it will look like this:
- **Linear Equation**: A straight line.
- **Quadratic Equation**: A U-shaped curve (upward or downward facing).
- **Possible Graph Scenarios**:
- **No Solutions**: The line might be completely above or below the parabola.
- **One Solution**: The line could touch the parabola at one point (tangent).
- **Two Solutions**: The line might cross the parabola at two distinct points.
These graphical scenarios illustrate that while a linear function and a quadratic function can intersect at various points, it's impossible for them to intersect an infinite number of times—they are fundamentally different in their shapes and equations.
### Conclusion:
In summary, a system of one linear equation and one quadratic equation cannot have an infinite number of solutions, and this can be viewed clearly in a graph where the behavior of lines versus parabolas indicates a maximum of two points of intersection or none at all.
### Why Not Infinite Solutions?
1. **Nature of the Equations**:
- A linear equation represents a straight line in a two-dimensional space (e.g., \(y = mx + b\)).
- A quadratic equation represents a parabola (e.g., \(y = ax^2 + bx + c\)).
2. **Intersection Points**:
- The possible intersections between a straight line and a parabola can be:
- **No Intersection**: The line does not touch the parabola at all.
- **One Intersection**: The line is tangent to the parabola at exactly one point.
- **Two Intersections**: The line intersects the parabola at two distinct points.
Since the straight line is a linear function and the parabola is a quadratic function, they can share at most two intersection points in a 2D Cartesian coordinate system but cannot share an infinite number of points.
### Graphical Representation:
- If you graph a linear equation and a quadratic equation, it will look like this:
- **Linear Equation**: A straight line.
- **Quadratic Equation**: A U-shaped curve (upward or downward facing).
- **Possible Graph Scenarios**:
- **No Solutions**: The line might be completely above or below the parabola.
- **One Solution**: The line could touch the parabola at one point (tangent).
- **Two Solutions**: The line might cross the parabola at two distinct points.
These graphical scenarios illustrate that while a linear function and a quadratic function can intersect at various points, it's impossible for them to intersect an infinite number of times—they are fundamentally different in their shapes and equations.
### Conclusion:
In summary, a system of one linear equation and one quadratic equation cannot have an infinite number of solutions, and this can be viewed clearly in a graph where the behavior of lines versus parabolas indicates a maximum of two points of intersection or none at all.
Answered by
GPT-4o mini
A system of two equations, consisting of one linear equation and one quadratic equation, cannot have an infinite number of solutions.
### Explanation:
1. **Linear Equation**: Represents a straight line.
2. **Quadratic Equation**: Represents a parabola (U-shaped curve).
### Intersection Scenarios:
- **No Solutions**: The line does not intersect the parabola.
- **One Solution**: The line is tangent to the parabola (touches at one point).
- **Two Solutions**: The line crosses the parabola at two points.
Since a line can intersect a parabola at most twice, it cannot produce an infinite number of intersection points.
### Graphically:
When graphed:
- The linear equation is a straight line.
- The quadratic equation is a curve.
This visual representation shows that they can intersect at a maximum of two points but never infinitely.
### Conclusion:
In summary, a linear equation and a quadratic equation can intersect at most twice, so they cannot have an infinite number of solutions.
### Explanation:
1. **Linear Equation**: Represents a straight line.
2. **Quadratic Equation**: Represents a parabola (U-shaped curve).
### Intersection Scenarios:
- **No Solutions**: The line does not intersect the parabola.
- **One Solution**: The line is tangent to the parabola (touches at one point).
- **Two Solutions**: The line crosses the parabola at two points.
Since a line can intersect a parabola at most twice, it cannot produce an infinite number of intersection points.
### Graphically:
When graphed:
- The linear equation is a straight line.
- The quadratic equation is a curve.
This visual representation shows that they can intersect at a maximum of two points but never infinitely.
### Conclusion:
In summary, a linear equation and a quadratic equation can intersect at most twice, so they cannot have an infinite number of solutions.
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