Question
Consider the following system of equations:
-x + y = 5
y = ( x - 3) 2 + 2
Consider the following system of equations:
B. Solve the system of equations using substitution.
c. Graph the system of equations on a piece of graph paper. On the graph, label the
solution(s) to the system of equations.
d. 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. in 1 sentence
-x + y = 5
y = ( x - 3) 2 + 2
Consider the following system of equations:
B. Solve the system of equations using substitution.
c. Graph the system of equations on a piece of graph paper. On the graph, label the
solution(s) to the system of equations.
d. 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. in 1 sentence
Answers
There are no human answers yet.
Answered by
GPT-4o mini
**B. Substitute to solve the system of equations.**
To solve the system of equations:
1. From the first equation, solve for \(y\):
\[
y = x + 5
\]
2. Now substitute this expression for \(y\) into the second equation:
\[
x + 5 = (x - 3)^2 + 2
\]
3. Expand the right-hand side:
\[
x + 5 = (x^2 - 6x + 9) + 2
\]
\[
x + 5 = x^2 - 6x + 11
\]
4. Rearranging gives:
\[
0 = x^2 - 7x + 6
\]
5. Factor the quadratic:
\[
0 = (x - 6)(x - 1)
\]
6. Thus, \(x = 6\) or \(x = 1\).
7. Substitute these values back to find \(y\):
- For \(x = 6\):
\[
y = 6 + 5 = 11
\]
- For \(x = 1\):
\[
y = 1 + 5 = 6
\]
So, the solutions are \((6, 11)\) and \((1, 6)\).
**C. Graph the equations.**
1. **Graph the linear equation**:
- The line represented by \(-x + y = 5\) can be rewritten as \(y = x + 5\), which has a slope of 1 and a y-intercept at (0, 5).
2. **Graph the quadratic equation**:
- The equation \(y = (x - 3)^2 + 2\) is a parabola that opens upwards, with the vertex at (3, 2).
3. **Label the points of intersection**:
- The points of intersection, which are the solutions, are \((6, 11)\) and \((1, 6)\).
**D. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions?**
No, a system of two equations consisting of a linear equation and a quadratic equation cannot have an infinite number of solutions because the linear equation represents a straight line, while the quadratic equation represents a parabola; they can intersect at most twice in the Cartesian plane.
To solve the system of equations:
1. From the first equation, solve for \(y\):
\[
y = x + 5
\]
2. Now substitute this expression for \(y\) into the second equation:
\[
x + 5 = (x - 3)^2 + 2
\]
3. Expand the right-hand side:
\[
x + 5 = (x^2 - 6x + 9) + 2
\]
\[
x + 5 = x^2 - 6x + 11
\]
4. Rearranging gives:
\[
0 = x^2 - 7x + 6
\]
5. Factor the quadratic:
\[
0 = (x - 6)(x - 1)
\]
6. Thus, \(x = 6\) or \(x = 1\).
7. Substitute these values back to find \(y\):
- For \(x = 6\):
\[
y = 6 + 5 = 11
\]
- For \(x = 1\):
\[
y = 1 + 5 = 6
\]
So, the solutions are \((6, 11)\) and \((1, 6)\).
**C. Graph the equations.**
1. **Graph the linear equation**:
- The line represented by \(-x + y = 5\) can be rewritten as \(y = x + 5\), which has a slope of 1 and a y-intercept at (0, 5).
2. **Graph the quadratic equation**:
- The equation \(y = (x - 3)^2 + 2\) is a parabola that opens upwards, with the vertex at (3, 2).
3. **Label the points of intersection**:
- The points of intersection, which are the solutions, are \((6, 11)\) and \((1, 6)\).
**D. Could a system of two equations that includes a linear equation and a quadratic equation ever have an infinite number of solutions?**
No, a system of two equations consisting of a linear equation and a quadratic equation cannot have an infinite number of solutions because the linear equation represents a straight line, while the quadratic equation represents a parabola; they can intersect at most twice in the Cartesian plane.
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