Solving Linear–Quadratic Systems - Algebra
Card 1 of 30
Find the intersection points of $x=0$ and $x^2+y^2=9$.
Find the intersection points of $x=0$ and $x^2+y^2=9$.
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$(0,-3)$ and $(0,3)$. Set $x=0$ in the circle equation and solve.
$(0,-3)$ and $(0,3)$. Set $x=0$ in the circle equation and solve.
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Find the intersection points of $y=x+1$ and $y=x^2$.
Find the intersection points of $y=x+1$ and $y=x^2$.
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$( -1,0)$ and $(2,3)$. Set $x+1=x^2$ and solve the quadratic.
$( -1,0)$ and $(2,3)$. Set $x+1=x^2$ and solve the quadratic.
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Find the intersection points of $y=x$ and $x^2+y^2=2$.
Find the intersection points of $y=x$ and $x^2+y^2=2$.
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$( -1,-1)$ and $(1,1)$. Substitute $y=x$ into $x^2+y^2=2$ and solve.
$( -1,-1)$ and $(1,1)$. Substitute $y=x$ into $x^2+y^2=2$ and solve.
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Find the intersection points of $y=-x$ and $x^2+y^2=2$.
Find the intersection points of $y=-x$ and $x^2+y^2=2$.
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$( -1,1)$ and $(1,-1)$. Substitute $y=-x$ into $x^2+y^2=2$ and solve.
$( -1,1)$ and $(1,-1)$. Substitute $y=-x$ into $x^2+y^2=2$ and solve.
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Find the intersection points of $y=1$ and $x^2+y^2=2$.
Find the intersection points of $y=1$ and $x^2+y^2=2$.
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$( -1,1)$ and $(1,1)$. Set $y=1$ in the circle equation and solve.
$( -1,1)$ and $(1,1)$. Set $y=1$ in the circle equation and solve.
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Find the intersection points of $y=-x+2$ and $y= -x^2+2$.
Find the intersection points of $y=-x+2$ and $y= -x^2+2$.
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$(0,2)$ and $(1,1)$. Set $-x+2=-x^2+2$ and solve for intersections.
$(0,2)$ and $(1,1)$. Set $-x+2=-x^2+2$ and solve for intersections.
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Find the intersection points of $y=x+2$ and $y= -x^2+2$.
Find the intersection points of $y=x+2$ and $y= -x^2+2$.
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$(0,2)$ and $(-1,1)$. Set $x+2=-x^2+2$ and solve the quadratic.
$(0,2)$ and $(-1,1)$. Set $x+2=-x^2+2$ and solve the quadratic.
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What does it mean if a line is tangent to a parabola in a linear-quadratic system?
What does it mean if a line is tangent to a parabola in a linear-quadratic system?
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The system has $1$ real solution (one intersection point). The line touches the parabola at exactly one point.
The system has $1$ real solution (one intersection point). The line touches the parabola at exactly one point.
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Find the intersection points of $y=4-x^2$ and $y=3$.
Find the intersection points of $y=4-x^2$ and $y=3$.
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$( -1,3)$ and $(1,3)$. Set $4-x^2=3$ and solve for $x$.
$( -1,3)$ and $(1,3)$. Set $4-x^2=3$ and solve for $x$.
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Find the intersection points of $y=-x+2$ and $y= -x^2+2$.
Find the intersection points of $y=-x+2$ and $y= -x^2+2$.
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$(0,2)$ and $(1,1)$. Set $-x+2=-x^2+2$ and solve for intersections.
$(0,2)$ and $(1,1)$. Set $-x+2=-x^2+2$ and solve for intersections.
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Find the intersection points of $y= -2x+1$ and $x^2+y^2=5$.
Find the intersection points of $y= -2x+1$ and $x^2+y^2=5$.
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$(0,1)$ and $(2,-3)$. Substitute $y=-2x+1$ into $x^2+y^2=5$ and solve.
$(0,1)$ and $(2,-3)$. Substitute $y=-2x+1$ into $x^2+y^2=5$ and solve.
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Find the intersection points of $y=2x+1$ and $x^2+y^2=5$.
Find the intersection points of $y=2x+1$ and $x^2+y^2=5$.
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$(0,1)$ and $( -2,-3)$. Substitute $y=2x+1$ into $x^2+y^2=5$ and solve.
$(0,1)$ and $( -2,-3)$. Substitute $y=2x+1$ into $x^2+y^2=5$ and solve.
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What must you always do after finding $x$-values from substitution in a system?
What must you always do after finding $x$-values from substitution in a system?
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Substitute back to find the matching $y$-value(s). Complete the solution by finding corresponding coordinates.
Substitute back to find the matching $y$-value(s). Complete the solution by finding corresponding coordinates.
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What form should solutions to a system in two variables be written in?
What form should solutions to a system in two variables be written in?
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As ordered pairs $(x,y)$. Solutions are coordinate pairs showing intersection points.
As ordered pairs $(x,y)$. Solutions are coordinate pairs showing intersection points.
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What is the standard form of a line used for substitution if it is already solved for $y$?
What is the standard form of a line used for substitution if it is already solved for $y$?
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$y=mx+b$. Slope-intercept form ready for direct substitution.
$y=mx+b$. Slope-intercept form ready for direct substitution.
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What is the standard form of a circle centered at the origin with radius $r$?
What is the standard form of a circle centered at the origin with radius $r$?
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$x^2+y^2=r^2$. Circle equation with center at origin.
$x^2+y^2=r^2$. Circle equation with center at origin.
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What is the standard form of a parabola that opens up or down with vertex at the origin?
What is the standard form of a parabola that opens up or down with vertex at the origin?
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$y=ax^2$. Basic parabola form opening vertically.
$y=ax^2$. Basic parabola form opening vertically.
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Identify the correct substitution for the system $y=-3x$ and $x^2+y^2=3$.
Identify the correct substitution for the system $y=-3x$ and $x^2+y^2=3$.
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Replace $y$ with $-3x$ in $x^2+y^2=3$. Substitute the linear expression into the circle equation.
Replace $y$ with $-3x$ in $x^2+y^2=3$. Substitute the linear expression into the circle equation.
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Find the intersection points of $y=2x$ and $y=x^2$.
Find the intersection points of $y=2x$ and $y=x^2$.
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$(0,0)$ and $(2,4)$. Set $2x=x^2$ and solve for intersection points.
$(0,0)$ and $(2,4)$. Set $2x=x^2$ and solve for intersection points.
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Find the intersection points of $y=-x$ and $y=x^2$.
Find the intersection points of $y=-x$ and $y=x^2$.
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$(0,0)$ and $(-1,1)$. Set $-x=x^2$ and solve the resulting quadratic.
$(0,0)$ and $(-1,1)$. Set $-x=x^2$ and solve the resulting quadratic.
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Find the intersection points of $y=x$ and $y=x^2$.
Find the intersection points of $y=x$ and $y=x^2$.
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$(0,0)$ and $(1,1)$. Set $x=x^2$ and solve for the intersection points.
$(0,0)$ and $(1,1)$. Set $x=x^2$ and solve for the intersection points.
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Find the intersection points of $y=-1$ and $y=x^2$.
Find the intersection points of $y=-1$ and $y=x^2$.
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No real solution. Parabola $y=x^2$ never reaches negative values.
No real solution. Parabola $y=x^2$ never reaches negative values.
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Find the intersection points of $y=0$ and $y=x^2-4$.
Find the intersection points of $y=0$ and $y=x^2-4$.
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($( -2,0)$ and $(2,0)$). Set $0=x^2-4$ and solve for $x$.
($( -2,0)$ and $(2,0)$). Set $0=x^2-4$ and solve for $x$.
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Find the intersection points of $y=2x+3$ and $y=x^2+3$.
Find the intersection points of $y=2x+3$ and $y=x^2+3$.
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$(0,3)$ and $(2,7)$. Set $2x+3=x^2+3$ and solve the quadratic.
$(0,3)$ and $(2,7)$. Set $2x+3=x^2+3$ and solve the quadratic.
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What is the key algebraic step to solve $y=2x+1$ and $y=x^2$ together?
What is the key algebraic step to solve $y=2x+1$ and $y=x^2$ together?
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Set $2x+1=x^2$. Set the expressions for $y$ equal to each other.
Set $2x+1=x^2$. Set the expressions for $y$ equal to each other.
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What does $b^2-4ac>0$ tell you about a linear-quadratic system after substitution?
What does $b^2-4ac>0$ tell you about a linear-quadratic system after substitution?
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There are $2$ distinct real solutions. Positive discriminant means two intersection points exist.
There are $2$ distinct real solutions. Positive discriminant means two intersection points exist.
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What does $b^2-4ac=0$ tell you about a linear-quadratic system after substitution?
What does $b^2-4ac=0$ tell you about a linear-quadratic system after substitution?
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There is exactly $1$ real solution (a tangent intersection). Zero discriminant means the line is tangent to the curve.
There is exactly $1$ real solution (a tangent intersection). Zero discriminant means the line is tangent to the curve.
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What does $b^2-4ac<0$ tell you about a linear-quadratic system after substitution?
What does $b^2-4ac<0$ tell you about a linear-quadratic system after substitution?
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There are $0$ real solutions (no real intersections). Negative discriminant means no real intersection points exist.
There are $0$ real solutions (no real intersections). Negative discriminant means no real intersection points exist.
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Find the intersection points of $y=2$ and $x^2+y^2=1$.
Find the intersection points of $y=2$ and $x^2+y^2=1$.
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No real solution. The line $y=2$ is outside the unit circle.
No real solution. The line $y=2$ is outside the unit circle.
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Identify the number of real solutions if substitution gives $5x^2+1=0$.
Identify the number of real solutions if substitution gives $5x^2+1=0$.
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$0$ real solutions. The equation $5x^2=-1$ has no real solutions.
$0$ real solutions. The equation $5x^2=-1$ has no real solutions.
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