Solving Systems Consisting of Linear and Quadratic Equations Practice

A system solution is an ordered pair, so recover the output for every quadratic root.

Set the two expressions for \(y\) equal, solve the resulting quadratic to get \(x=-2\) and \(x=3\), then substitute each into the line. This yields \((-2,0)\text{ and }(3,5)\), and both pairs satisfy both equations.

A system solution is an ordered pair, so recover the output for every quadratic root.

Set the two expressions for \(y\) equal, solve the resulting quadratic to get \(x=-1\) and \(x=4\), then substitute each into the line. This yields \((-1,5)\text{ and }(4,0)\), and both pairs satisfy both equations.

A system solution is an ordered pair, so recover the output for every quadratic root.

Set the two expressions for \(y\) equal, solve the resulting quadratic to get \(x=0\) and \(x=5\), then substitute each into the line. This yields \((0,-1)\text{ and }(5,9)\), and both pairs satisfy both equations.

A system solution is an ordered pair, so recover the output for every quadratic root.

Set the two expressions for \(y\) equal, solve the resulting quadratic to get \(x=-3\) and \(x=2\), then substitute each into the line. This yields \((-3,9)\text{ and }(2,-1)\), and both pairs satisfy both equations.

A system solution is an ordered pair, so recover the output for every quadratic root.

Set the two expressions for \(y\) equal, solve the resulting quadratic to get \(x=1\) and \(x=6\), then substitute each into the line. This yields \((1,4)\text{ and }(6,19)\), and both pairs satisfy both equations.

A missing coordinate is found by substitution, not by reading unrelated coefficients.

Substitute \(x=-1\) into the quadratic: \(k=1(-1)^2+-3(-1)+-4=0\). A shared point must use this output coordinate.

A missing coordinate is found by substitution, not by reading unrelated coefficients.

Substitute \(x=2\) into the quadratic: \(k=1(2)^2+2(2)+-8=0\). A shared point must use this output coordinate.

A missing coordinate is found by substitution, not by reading unrelated coefficients.

Substitute \(x=5\) into the quadratic: \(k=-1(5)^2+4(5)+5=0\). A shared point must use this output coordinate.

A missing coordinate is found by substitution, not by reading unrelated coefficients.

Substitute \(x=1\) into the quadratic: \(k=2(1)^2+-6(1)+1=-3\). A shared point must use this output coordinate.

A missing coordinate is found by substitution, not by reading unrelated coefficients.

Substitute \(x=-3\) into the quadratic: \(k=-2(-3)^2+-4(-3)+6=0\). A shared point must use this output coordinate.

Count graph intersections and confirm the count with the reduced quadratic's discriminant.

Substitution produces a quadratic whose real roots correspond to the plotted intersection points. It has 0 real roots, matching the 0 graph intersections.

Count graph intersections and confirm the count with the reduced quadratic's discriminant.

Substitution produces a quadratic whose real roots correspond to the plotted intersection points. It has 1 real root, matching the 1 graph intersection.

Count graph intersections and confirm the count with the reduced quadratic's discriminant.

Substitution produces a quadratic whose real roots correspond to the plotted intersection points. It has 2 real roots, matching the 2 graph intersections.

Count graph intersections and confirm the count with the reduced quadratic's discriminant.

Substitution produces a quadratic whose real roots correspond to the plotted intersection points. It has 2 real roots, matching the 2 graph intersections.

Count graph intersections and confirm the count with the reduced quadratic's discriminant.

Substitution produces a quadratic whose real roots correspond to the plotted intersection points. It has 0 real roots, matching the 0 graph intersections.

Read both coordinates at each intersection and verify them algebraically.

The marked intersections are \((-1,0)\text{ and }(2,3)\). Substitution confirms that each ordered pair satisfies both the line and parabola equations.

Read both coordinates at each intersection and verify them algebraically.

The marked intersections are \((-2,4)\text{ and }(3,-1)\). Substitution confirms that each ordered pair satisfies both the line and parabola equations.

Read both coordinates at each intersection and verify them algebraically.

The marked intersections are \((0,0)\text{ and }(4,8)\). Substitution confirms that each ordered pair satisfies both the line and parabola equations.

Read both coordinates at each intersection and verify them algebraically.

The marked intersections are \((-1,7)\text{ and }(5,-5)\). Substitution confirms that each ordered pair satisfies both the line and parabola equations.

Read both coordinates at each intersection and verify them algebraically.

The marked intersections are \((1,1)\text{ and }(3,7)\). Substitution confirms that each ordered pair satisfies both the line and parabola equations.

For a horizontal tangent, find the parabola's extreme output value.

A horizontal tangent occurs at the vertex. The vertex is \((2,3)\), so the tangent line is \(y=3\). Equivalently, substitution makes the reduced discriminant zero.

For a horizontal tangent, find the parabola's extreme output value.

A horizontal tangent occurs at the vertex. The vertex is \((-2,-5)\), so the tangent line is \(y=-5\). Equivalently, substitution makes the reduced discriminant zero.

For a horizontal tangent, find the parabola's extreme output value.

A horizontal tangent occurs at the vertex. The vertex is \((-3,2)\), so the tangent line is \(y=2\). Equivalently, substitution makes the reduced discriminant zero.

For a horizontal tangent, find the parabola's extreme output value.

A horizontal tangent occurs at the vertex. The vertex is \((2,-7)\), so the tangent line is \(y=-7\). Equivalently, substitution makes the reduced discriminant zero.

For a horizontal tangent, find the parabola's extreme output value.

A horizontal tangent occurs at the vertex. The vertex is \((5,-9)\), so the tangent line is \(y=-9\). Equivalently, substitution makes the reduced discriminant zero.

Set the two expressions equal, solve for inputs, then recover and verify each output.

Set the functions equal and simplify. The resulting quadratic has roots \(-2\) and \(2\). Substitution gives \((-2,3)\text{ and }(2,3)\), and each pair satisfies both quadratics.

Set the two expressions equal, solve for inputs, then recover and verify each output.

Set the functions equal and simplify. The resulting quadratic has roots \(-3\) and \(1\). Substitution gives \((-3,3)\text{ and }(1,3)\), and each pair satisfies both quadratics.

Set the two expressions equal, solve for inputs, then recover and verify each output.

Set the functions equal and simplify. The resulting quadratic has roots \(-1\) and \(4\). Substitution gives \((-1,6)\text{ and }(4,11)\), and each pair satisfies both quadratics.

Set the two expressions equal, solve for inputs, then recover and verify each output.

Set the functions equal and simplify. The resulting quadratic has roots \(-4\) and \(1\). Substitution gives \((-4,18)\text{ and }(1,3)\), and each pair satisfies both quadratics.

Set the two expressions equal, solve for inputs, then recover and verify each output.

Set the functions equal and simplify. The resulting quadratic has roots \(0\) and \(5\). Substitution gives \((0,1)\text{ and }(5,6)\), and each pair satisfies both quadratics.

Reduce the system to one equation before using factoring or the quadratic formula.

Because both expressions equal \(y\), set them equal and move every term to one side. Combining like terms gives \(x^2+x-8=0\).

Reduce the system to one equation before using factoring or the quadratic formula.

Because both expressions equal \(y\), set them equal and move every term to one side. Combining like terms gives \(2x^2-6x+6=0\).

Reduce the system to one equation before using factoring or the quadratic formula.

Because both expressions equal \(y\), set them equal and move every term to one side. Combining like terms gives \(x^2-3x-6=0\).

Reduce the system to one equation before using factoring or the quadratic formula.

Because both expressions equal \(y\), set them equal and move every term to one side. Combining like terms gives \(3x^2+2x-6=0\).

Reduce the system to one equation before using factoring or the quadratic formula.

Because both expressions equal \(y\), set them equal and move every term to one side. Combining like terms gives \(2x^2-10x+2=0\).

When only the number of intersections is requested, the reduced discriminant is sufficient.

Substitution gives a one-variable quadratic. Its discriminant has the sign corresponding to 1 real root, so the graphs have 1 real intersection.

When only the number of intersections is requested, the reduced discriminant is sufficient.

Substitution gives a one-variable quadratic. Its discriminant has the sign corresponding to 0 real roots, so the graphs have 0 real intersections.

When only the number of intersections is requested, the reduced discriminant is sufficient.

Substitution gives a one-variable quadratic. Its discriminant has the sign corresponding to 2 real roots, so the graphs have 2 real intersections.

When only the number of intersections is requested, the reduced discriminant is sufficient.

Substitution gives a one-variable quadratic. Its discriminant has the sign corresponding to 1 real root, so the graphs have 1 real intersection.

When only the number of intersections is requested, the reduced discriminant is sufficient.

Substitution gives a one-variable quadratic. Its discriminant has the sign corresponding to 2 real roots, so the graphs have 2 real intersections.

For graph-based systems, verify each marked intersection in both equations.

Setting the two quadratics equal produces the marked input values; substitution gives \((-2,0)\text{ and }(2,0)\). Every displayed point satisfies both equations, so all 2 ordered pairs are retained.

For graph-based systems, verify each marked intersection in both equations.

Setting the two quadratics equal produces the marked input values; substitution gives \((-3.236,1)\text{ and }(1.236,1)\). Every displayed point satisfies both equations, so all 2 ordered pairs are retained.

For graph-based systems, verify each marked intersection in both equations.

Setting the two quadratics equal produces the marked input values; substitution gives \((-1.732,4)\text{ and }(1.732,4)\). Every displayed point satisfies both equations, so all 2 ordered pairs are retained.

For graph-based systems, verify each marked intersection in both equations.

Setting the two quadratics equal produces the marked input values; substitution gives \((-1.236,3)\text{ and }(3.236,3)\). Every displayed point satisfies both equations, so all 2 ordered pairs are retained.

For graph-based systems, verify each marked intersection in both equations.

Setting the two quadratics equal produces the marked input values; substitution gives \((-4.828,5)\text{ and }(0.828,5)\). Every displayed point satisfies both equations, so all 2 ordered pairs are retained.

Track inputs and outputs together, and verify both original equations.

System solutions require ordered pairs, so each corresponding output must be found. Substitution into both original equations is the final validity test for every proposed ordered pair.

Track inputs and outputs together, and verify both original equations.

Each real root must be substituted back because each can produce an intersection. Substitution into both original equations is the final validity test for every proposed ordered pair.

Track inputs and outputs together, and verify both original equations.

A system solution must satisfy both equations, so the point is invalid. Substitution into both original equations is the final validity test for every proposed ordered pair.

Track inputs and outputs together, and verify both original equations.

First reduce the system to a one-variable quadratic, then compute its discriminant. Substitution into both original equations is the final validity test for every proposed ordered pair.

Track inputs and outputs together, and verify both original equations.

After algebraic verification, apply the stated contextual domain restriction. Substitution into both original equations is the final validity test for every proposed ordered pair.