Topics
2. Quadratic Functions & Equations
topic badge
United States of AmericaFL
Grade 912

2.05 Linear-quadratic systems

Lesson
Practice
Lesson

Concept summary

A system of equations is a set of equations that have the same variables.

The solution to a system of equations is any ordered pair that makes all of the equations in the system true. For graphs this will be the point(s) of intersection. Solutions can be found algebraically or graphically.

x
y
The line and parabola have no points of intersection so the system has no solution.
x
y
The parabolas do not have any points of intersection so the system has no solution.
x
y
The line and parabola have one point of intersection so the system has one solution.
x
y
The parabolas have one point of intersection so the system has one solution.
x
y
The line and parabola have two points of intersection so the system has two solutions.
x
y
The parabolas have two points of intersection so the system has two solutions.

The solution to a system of equations in a given context is viable if the solution makes sense in the context, and is non-viable if it does not make sense.

Worked examples

Example 1

Consider the following systems of equations:

\begin{cases} y= x^{2} - 2 x - 3 \\ y= - x + 3 \end{cases}

a

Graph the equations on the same coordinate plane.

Approach

The solution(s) to a system of equations can be represented graphically as their point(s) of intersection. We can use technology to graph the two equations, or, if drawing them by hand, it will be useful to first fill out a table of values for both equations.

x-3-2-101234
y=x^{2} - 2 x - 31250-3-4-305
x-3-2-101234
y=-x+36543210-1

Solution

-5
-4
-3
-2
-1
1
2
3
4
5
x
-5
-4
-3
-2
-1
1
2
3
4
5
y

Reflection

Notice that from the table of values both function have the columns with the points \left(-2,5\right) and \left(3,0\right). We want these points of intersection to be visible on our graph. We also want the vertex of the parabola, \left(1,-4\right), to be visible.

We want to ensure that the x-values on our graph cover at least the interval -3 \leq x \leq 4 and the y-values on our graph cover at least the interval -5 \leq y \leq 6.

b

Identify the coordinates of the solution(s) to the system of equations.

Solution

The points of intersection occur at \left(-2, -5\right) and \left(0, 3\right), so these coordinate pairs will be the solutions to the system of equations.

Reflection

We can check our work by solving the system of equations algebraically, by equating both equations and solving for x

\displaystyle x^2-2x-3 \displaystyle =\displaystyle -x+3Use that y=y, reflexive property
\displaystyle x^2-x-3\displaystyle =\displaystyle 3Add x to both sides
\displaystyle x^2-x-6\displaystyle =\displaystyle 0Subtract 3 from both sides
\displaystyle \left(x+2\right)\left(x-3\right)\displaystyle =\displaystyle 0Factor the quadratic

Using the zero product law we can see the two solutions to this new quadratic are x=-2 and x=3

We can now substitute these into one of the given equations to find the corresponding y-value.

Example 2

Find the solution(s) for the following linear-quadratic system of equations.\begin{cases} y = 3 x + 1 \\ y = x^{2} - 5x \end{cases}

Approach

We can approach this graphically or algebraically.

-4
-3
-2
-1
1
2
3
4
5
6
7
8
9
x
-5
5
10
15
20
25
y

But we can see from the graph that the points of intersection are not clearly identifiable. In cases like this, an algebraic approach is preferable. As both equations are already in terms of y we can use the substitution method to solve.

Solution

\displaystyle 3x+1 \displaystyle =\displaystyle x^2-5xUse that y=y, reflexive property
\displaystyle 1\displaystyle =\displaystyle x^2-8xSubtract 3x from both sides
\displaystyle 0\displaystyle =\displaystyle x^2-8x-1Subtract 1 from both sides

This equation is not easily factorable, and so we will use the quadratic formula to find the solutions.

\displaystyle x\displaystyle =\displaystyle \frac{-b\pm \sqrt{b^2-4ac}}{2a}State the quadratic formula
\displaystyle =\displaystyle \frac{-\left(-8 \right) \pm \sqrt{\left(-8\right)^2-4\left(1 \right)\left(-1\right)}}{2\left( 1\right)}Substitute a=1, b=-8, c=-1
\displaystyle =\displaystyle \frac{8\pm \sqrt{64+4}}{2}Evaluate the square and products
\displaystyle =\displaystyle \frac{8\pm \sqrt{68}}{2}Evaluate the difference in the radicand
\displaystyle =\displaystyle \frac{8\pm 2\sqrt{17}}{2}Simplify the radical
\displaystyle =\displaystyle \frac{8}{2}\pm \frac{2\sqrt{17}}{2}Rewrite as two fractions
\displaystyle =\displaystyle 4\pm \sqrt{17}Simplify the quotients

We have found the x-coordinates of the points of intersection. We can substitute these into either equation to find the corresponding y-coordinate.

\displaystyle y\displaystyle =\displaystyle 3x+1
\displaystyle y\displaystyle =\displaystyle 3\left(4-\sqrt{17}\right)+1
\displaystyle y\displaystyle =\displaystyle 13+3\sqrt{17}

So one solution is \left(4-\sqrt{17}, 13-3\sqrt{17}\right), and using the same method we find the other solution is \left(4+\sqrt{17}, 13+3\sqrt{17}\right)

Example 3

Forrest and his child Gustavo are driving remote control cars and are practicing making them turn around which follows a parabolic curve.

They are driving on the local basketball court before anyone gets there. Using one corner as the origin, the long side of the court as the x-axis and the short side of the court as the y-axis.

A first quadrant coordinate plane with no scales on its axes. A diagram of a basketball court is plotted as a rectangle using a part of the x axis as the length and y axis as the width. Two semi circles are drawn, one using a shorter segment in the middle of the y axis and another in the opposite side of the rectangle. A smaller circle is drawn at the center of the rectangle and a segment from the horizontal side to the x axis divides the circle in half.

Forrest and Gustavo are standing on opposite sides of the court. Forrest's car follows the curve \\y=\left(x-4\right)^2+3 and Gustavo's car follows the curve y=-\left(x-12 \right)^2+7.

a

Graph the two paths on the same coordinate plane.

Approach

Both of these equations are in vertex form, so we can see the direction of opening from the coefficient, then plot the vertex, and substitute in another value for x to get the shape.

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
x
1
2
3
4
5
6
7
8
9
y
b

Determine using the graph, or otherwise, if the two cars paths will ever cross.

Solution

These paths of the two cars will never cross. We can confirm this by noticing there are no points of intersection between the two graphs.

Reflection

We can confirm this algebraically as well:

\displaystyle \left(x-4\right)^2+3\displaystyle =\displaystyle -\left(x-12 \right)^2+7Use that y=y, reflexive property
\displaystyle x^2-8x+16+3\displaystyle =\displaystyle -\left(x^2-24x+144\right)+7Square the two binomials
\displaystyle x^2-8x+19\displaystyle =\displaystyle -x^2+24x-137Combine like terms
\displaystyle 2x^2-32x+156\displaystyle =\displaystyle 0Move all terms to one side
\displaystyle x^2-16x+78\displaystyle =\displaystyle 0Divide by 2

The discriminant of this would be \left(-16\right)^2-4 \cdot 1 \cdot 78=-56 which means there are no real solutions.

Outcomes

MA.912.AR.9.2

Given a mathematical or real-world context, solve a system consisting of a two-variable linear equation and a non-linear equation algebraically or graphically.

MA.912.AR.9.3

Given a mathematical or real-world context, solve a system consisting of two-variable linear or non-linear equations algebraically or graphically.

What is Mathspace

About Mathspace