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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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:
$(1,2).$

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.

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

Show 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.

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

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

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 $0.$
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.
Since, $x=6$ and $x=\text{-}3,$ there are two solutions to the system.
### 3

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 $(6,29)\text{ and }(\text{-}3,\text{-}15).$

Substitute one equation into the other

Solve the resulting equation

$x^2+2x-19=5x-1$

SubEqn$\text{LHS}-5x=\text{RHS}-5x$

$x^2-3x-19=\text{-}1$

AddEqn$\text{LHS}+1=\text{RHS}+1$

$x^2-3x-18=0$

$x^2-3x-18=0$

UseQuadFormUse the Quadratic Formula: $a = {\color{#0000FF}{1}}, \, b={\color{#009600}{\text{-}3}}, \, c={\color{#FF0000}{\text{-}18}}$

$x=\dfrac{\text{-}({\color{#009600}{\text{-}3}})\pm\sqrt{({\color{#009600}{\text{-}3}})^2-4\cdot{\color{#0000FF}{1}}({\color{#FF0000}{\text{-}18}})}}{2\cdot{\color{#0000FF}{1}}}$

NegNeg$\text{-} (\text{-} a)=a$

$x=\dfrac{3\pm\sqrt{(\text{-}3)^2-4\cdot1(\text{-}18)}}{2\cdot1}$

CalcPowCalculate power

$x=\dfrac{3\pm\sqrt{9-4\cdot1(\text{-}18)}}{2\cdot1}$

MultiplyMultiply

$x=\dfrac{3\pm\sqrt{9+72}}{2}$

AddTermsAdd terms

$x=\dfrac{3\pm\sqrt{81}}{2}$

CalcRootCalculate root

$x=\dfrac{3\pm9}{2}$

StateSolState solutions

$\begin{array}{l}x=\dfrac{3+9}{2}\\[.1em] \\ x=\dfrac{3-9}{2} \end{array}$

AddSubTermsAdd and subtract terms

$\begin{array}{l}x=\dfrac{12}{2}\\[.1em] \\ x=\dfrac{\text{-}6}{2} \end{array}$

SimpQuotSimplify quotient

$\begin{array}{l}x_1=6 \\ x_2=\text{-}3 \end{array}$

Substitute found variable value(s) into either given equation

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

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

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

### 4

The resultant equation is quadratic and can be solved with the quadratic formula.
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

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.
The second solution is now substituted into the linear equation.
The solutions to the system of the linear-quadratic system are
$\left(\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}, \dfrac{9}{2}+\dfrac{\sqrt{17}}{2}\right) \quad \text{ and } \quad \left(\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}, \dfrac{9}{2}-\dfrac{\sqrt{17}}{2}\right).$

Analyze variables

Manipulate equations

Combine equations

**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.
$\begin{array}{c r c l}& \text{-} x^2+2x+8 & = & y \\ - & x+4 & = & y \\ \hline & \text{-} x^2+x+4 & = & 0\end{array}$

Solve the resulting quadratic equation

$\text{-} x^2 + x + 4$

UseQuadFormUse the Quadratic Formula: $a = {\color{#0000FF}{\text{-}1}}, \, b={\color{#009600}{1}}, \, c={\color{#FF0000}{4}}$

$x=\dfrac{\text{-}{\color{#009600}{1}}\pm\sqrt{{\color{#009600}{1}}^2-4({\color{#0000FF}{\text{-}1}})\cdot{\color{#FF0000}{4}}}}{2({\color{#0000FF}{\text{-}1)}}}$

Solve using the quadratic formula

MultNegNegOnePar$\text{-} a(\text{-} b)=a\cdot b$

$x=\dfrac{\text{-}1\pm\sqrt{1^2+4\cdot1\cdot4}}{2(\text{-}1)}$

CalcPowProdCalculate power and product

$x=\dfrac{\text{-}1\pm\sqrt{1+16}}{\text{-}2}$

AddTermsAdd terms

$x=\dfrac{\text{-}1\pm\sqrt{17}}{\text{-}2}$

Solve the linear equation for the other variable

$x+4=y$

Substitute$x={\color{#0000FF}{\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}}}$

${\color{#0000FF}{\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}}}+4=y$

Simplify left-hand side

NumberToFrac$a = \dfrac{2\cdot a}{2}$

$\dfrac{1}{2}+\dfrac{\sqrt{17}}{2}+\dfrac{8}{2}=y$

AddFracAdd fractions

$\dfrac{9}{2}+\dfrac{\sqrt{17}}{2}=y$

RearrangeEqnRearrange equation

$y=\dfrac{9}{2}+\dfrac{\sqrt{17}}{2}$

$x+4=y$

Substitute$x={\color{#0000FF}{\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}}}$

${\color{#0000FF}{\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}}}+4=y$

Simplify left-hand side

NumberToFrac$a = \dfrac{2\cdot a}{2}$

$\dfrac{1}{2}-\dfrac{\sqrt{17}}{2}+\dfrac{8}{2}=y$

AddFracAdd fractions

$\dfrac{9}{2}-\dfrac{\sqrt{17}}{2}=y$

RearrangeEqnRearrange equation

$y=\dfrac{9}{2}-\dfrac{\sqrt{17}}{2}$

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