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Concept

A linear-quadratic system is a set of equations where at least one is linear and at least one is quadratic. They are expressed identical to linear systems. $\begin{cases}y=2x & \, \text {(I)}\\ y=\text{-} x^2+4x-1 & \text {(II)}\end{cases}$ Similar to systems of linear equations, linear-quadratic systems contain more than one unknown variable, and the solution is the set of coordinates that make all equations true simultaneously. In the example above, the solution is $x=1$ and $y=2.$ These coordinates make the sides equal in both equations. The solution is usually written as a point: $(1,2).$

They can be solved graphically, by substitution or by elimination.
Method

## Solving a Linear-Quadratic System Graphically

Similar to solving linear systems, a linear-quadratic system can be solved by graphing both functions in the system. The solution(s) can be found by identifying the coordinates of the point(s) of intersection.

However, since linear and quadratic graphs cannot coincide, it's impossible to have infinitely many solutions. Thus, a linear-quadratic system can have $0,$ $1$ or $2$ solutions.
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Exercise

Solve the linear-quadratic system graphically. $\begin{cases}y=2x^2-14x+26 \\ y=2x-7 \end{cases}$

Show Solution
Solution

Solving a linear-quadratic system graphically means finding the point or points where the line and the parabola intersect. Since the linear equation is given in slope-intercept form, we can see that the $y$-intercept of the line is $(0,\text{-} 7)$ and the slope is $2.$ Since the quadratic function is given in standard form, we can see the following characteristics of the parabola. \begin{aligned} \textbf{direction} &: \text{upward} \\ \textbf{vertex} &: (3.5,1.5), \ \text{minimum} \\ \mathbf{y}\textbf{-intercept} &: (0,26) \end{aligned} We'll use the characteristics to graph both functions on the same coordinate plane.

From the graph we can see that the parabola and the line do not intersect. Thus, there are no solutions to the system.

Method

## Solving a Linear-Quadratic System using Substitution

There are different ways to solve systems of equations. Recall that linear systems can be solved using the substitution method. Linear-quadratic systems can be solved in the same way. The main difference is that a quadratic equation — rather than a linear equation — will need to be solved. Consider the following system of equations. $\begin{cases}y=x^2+2x-19 \\ y=5x-1 \end{cases}$

### 1

Substitute one equation into the other

First, it is necessary to substitute one equation into the other. It does not matter which equation is chosen. Here, $y=x^2+2x-19$ will be substituted into $y=5x-1.$ $y=5x-1 \quad \Rightarrow \quad x^2+2x-19=5x-1$

### 2

Solve the resulting equation
Since the resulting equation has degree $2,$ it is a quadratic equation. Moving all terms to one side of the equation ensures it is set equal to $0.$
$x^2+2x-19=5x-1$
$x^2-3x-19=\text{-}1$
$x^2-3x-18=0$
A quadratic equation can be solved in different ways. Here, the quadratic formula will be used. Substitute $a=1,$ $b=\text{-}3$ and $c=\text{-}18$ into the formula and simplify.
$x^2-3x-18=0$
$x=\dfrac{\text{-}({\color{#009600}{\text{-}3}})\pm\sqrt{({\color{#009600}{\text{-}3}})^2-4\cdot{\color{#0000FF}{1}}({\color{#FF0000}{\text{-}18}})}}{2\cdot{\color{#0000FF}{1}}}$
$x=\dfrac{3\pm\sqrt{(\text{-}3)^2-4\cdot1(\text{-}18)}}{2\cdot1}$
$x=\dfrac{3\pm\sqrt{9-4\cdot1(\text{-}18)}}{2\cdot1}$
$x=\dfrac{3\pm\sqrt{9+72}}{2}$
$x=\dfrac{3\pm\sqrt{81}}{2}$
$x=\dfrac{3\pm9}{2}$
$\begin{array}{l}x=\dfrac{3+9}{2}\\[.1em] \\ x=\dfrac{3-9}{2} \end{array}$
$\begin{array}{l}x=\dfrac{12}{2}\\[.1em] \\ x=\dfrac{\text{-}6}{2} \end{array}$
$\begin{array}{l}x_1=6 \\ x_2=\text{-}3 \end{array}$
Since, $x=6$ and $x=\text{-}3,$ there are two solutions to the system.

### 3

Substitute found variable value(s) into either given equation

The $x$-coordinates of the points of intersections are $x=6$ and $x=\text{-}3.$ They can be used to find the corresponding $y$-values. To do this, substitute them into either equation and solve. Here, $y=5x-1$ will be used. \begin{aligned} y &= 5 \cdot 6 \phantom{\text{-}} -1 = \phantom{\text{-}}29\\ y &= 5(\text{-} 3) -1 =\text{-} 16 \end{aligned} Thus, the solutions to the system are $(6,29)\text{ and }(\text{-}3,\text{-}15).$

Method

## Solving a Linear-Quadratic System using Elimination

Systems of linear-quadratic equations can be solved algebraically by elimination. The goal of this method is to eliminate one of the variables, resulting in a one-variable equation. This is done by intentionally combining the two equations. For example, the following system can be solved in this way. $\begin{cases}\text{-} x^2 +2x +8=y \\ x+4 = y \end{cases}$

### 1

Analyze variables

Since the $x$-variable has a different degree in a linear and quadratic equation, it can not be eliminated. Thus, the $y$-variable should be eliminated instead.

### 2

Manipulate equations

To be able to eliminate $y,$ the coefficients of the chosen variable should be the same. $\begin{cases}\text{-} x^2 +2x +8={\color{#0000FF}{y}} \\ x+4 = {\color{#0000FF}{y}} \end{cases}$ In this system, they already are.

### 3

Combine equations

The equations can now be combined, by either adding or subtracting one of the equations from the other, whichever will lead to eliminating one of the variables. $\begin{cases}\text{-} x^2 +2x +8=y \\ x+4 = y \end{cases}$ In this example, the coefficients of $y$ have the same sign. Therefore, subtracting one equation from the other will eliminate $y.$ The second equation will now be subtracted from the first, though it is an arbitrary choice. This is done by subtracting the left-hand side of the second equation from the first, and doing the same for the right-hand side. $\begin{array}{c r c l}& \text{-} x^2+2x+8 & = & y \\ - & x+4 & = & y \\ \hline & \text{-} x^2+x+4 & = & 0\end{array}$

### 4

The resultant equation is quadratic and can be solved with the quadratic formula.
$\text{-} x^2 + x + 4$
$x=\dfrac{\text{-}{\color{#009600}{1}}\pm\sqrt{{\color{#009600}{1}}^2-4({\color{#0000FF}{\text{-}1}})\cdot{\color{#FF0000}{4}}}}{2({\color{#0000FF}{\text{-}1)}}}$
$x=\dfrac{\text{-}1\pm\sqrt{1^2+4\cdot1\cdot4}}{2(\text{-}1)}$
$x=\dfrac{\text{-}1\pm\sqrt{1+16}}{\text{-}2}$
$x=\dfrac{\text{-}1\pm\sqrt{17}}{\text{-}2}$
The square root of $17$ is an irrational number. Therefore, $\sqrt{17}$ should not be calculated to get an exact answer. $x=\dfrac{\text{-}1\pm\sqrt{17}}{\text{-}2} \Leftrightarrow \begin{array}{l}x_1=\frac{1}{2} + \frac{\sqrt{17}}{2} \\ \ \\ x_2 = \frac{1}{2} - \frac{\sqrt{17}}{2} \end{array}$

### 5

Solve the linear equation for the other variable
The solutions from the quadratic equation is now used to solve the linear equation for $y.$ Since there are two solutions, one is used at a time.
$x+4=y$
${\color{#0000FF}{\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}}}+4=y$
Simplify left-hand side
$\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}+\dfrac{8}{2}=y$
$\dfrac{9}{2}+\dfrac{\sqrt{17}}{2}=y$
$y=\dfrac{9}{2}+\dfrac{\sqrt{17}}{2}$
The second solution is now substituted into the linear equation.
$x+4=y$
${\color{#0000FF}{\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}}}+4=y$
Simplify left-hand side
$\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}+\dfrac{8}{2}=y$
$\dfrac{9}{2}-\dfrac{\sqrt{17}}{2}=y$
$y=\dfrac{9}{2}-\dfrac{\sqrt{17}}{2}$
The solutions to the system of the linear-quadratic system are $\left(\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}, \dfrac{9}{2}+\dfrac{\sqrt{17}}{2}\right) \quad \text{ and } \quad \left(\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}, \dfrac{9}{2}-\dfrac{\sqrt{17}}{2}\right).$