# Linear-Quadraric Systems of Equations - Examples, Exercises and Solutions

## Solving a system of equations when one of them is linear and the other is quadratic

When we have a system of equations where one of the equations is linear and the other quadratic
we will use the substitution method:

We will isolate one variable from an equation, place in the second equation the value of the expression of the variable we have isolated, and in this way, we will obtain an equation with one variable. We will solve for $X$ or $Y$ and then place it in one of the original equations to find the complete point. The point we discover will be the point of intersection of the line with the parabola, and it will also be the solution of the system of equations.

## examples with solutions for linear-quadraric systems of equations

### Exercise #1

Which formula represents line 2 shown in the graph below?

### Video Solution

$y=-2x+5$

### Exercise #2

The following functions are graphed below:

$f(x)=x^2-6x+8$

$g(x)=4x-17$

For which values of x is
f(x)<0 true?

2 < x < 4

### Exercise #3

The following functions are graphed below:

$f(x)=x^2-6x+8$

$g(x)=4x-17$

For which values of x is

f(x)>0 true?

x < 2, 4 < x

### Exercise #4

Solve the following system of equations:

$\begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases}$

### Video Solution

$x=\frac{\sqrt{61}}{2}-2.5$

$y=\frac{\sqrt{61}}{2}+2.5$

or

$x=\frac{\sqrt{61}}{2}+2.5$

$y=\frac{\sqrt{61}}{2}-2.5$

### Exercise #5

Solve the following system of equations:

$\begin{cases} \sqrt{x}+\sqrt{y}=\sqrt{\sqrt{61}+6} \\ xy=9 \end{cases}$

### Video Solution

$x=\frac{\sqrt{61}}{2}-2.5$
$y=\frac{\sqrt{61}}{2}+2.5$
$x=\frac{\sqrt{61}}{2}+2.5$
$y=\frac{\sqrt{61}}{2}-2.5$