Solution of a system of equations - one of them is linear and the other quadratic

πŸ†Practice solution of a system of equations, one of which is linear and the other quadratic.

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 XX or YY 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.

Start practice

Test yourself on solution of a system of equations, one of which is linear and the other quadratic.!

einstein

Solve the following system of equations:

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

Practice more now

Let's look at an example

Given the following system of equations: y=x2+2x+7y=x^2+2x+7,2x+y=42x+y=4

Solution:
Let's isolate one of the variables from one of the equations:
2x+y=42x+y=4
y=βˆ’2x+4y=-2x+4

Now, let's substitute the value we isolated into the second equation. We will obtain:

βˆ’2x+4=x2+2x+7-2x+4=x^2+2x+7

Let's combine like terms and transpose members. We will obtain:
βˆ’X2βˆ’4xβˆ’3=0-X^2-4x-3=0
We will solve it with the help of the quadratic formula and will obtain:
X=βˆ’1,βˆ’3X=-1 ,-3
Now we will gradually place an XX we found into one of the equations to find the complete points: Let's start with X=βˆ’1X=-1

2Γ—(βˆ’1)+y=4 2\times(-1)+y=4
βˆ’2+y=4-2+y=4
y=6y=6
Let's write this solution (βˆ’1,6)(-1,6)
let's move to the second XX we found X=βˆ’3X=-3 and we will obtain:
2βˆ—(βˆ’3)+y=42*(-3)+y=4
βˆ’6+y=4-6+y=4
y=10y=10 let's write this solution (βˆ’3,10)(-3,10)

These solutions are the points of intersection of the parabola with the axis.


If you are interested in this article, you might also be interested in the following articles:

The functions y=xΒ²

Family of parabolas y=xΒ²+c: Vertical shift

Family of parabolas y=(x-p)Β²

Family of parabolas y=(x-p)Β²+k (combination of horizontal and vertical shift)

Vertex form of the quadratic function

Factored form of the quadratic function

Completing the square in a quadratic equation

Standard form of the quadratic function

System of quadratic equations - Algebraic and graphical solution

In the blog of Tutorela you will find a variety of articles about mathematics.


Examples and exercises with solutions of linear and quadratic equation systems

Exercise #1

Solve the following system of equations:

{xβˆ’y=61βˆ’6xy=9 \begin{cases} \sqrt{x}-\sqrt{y}=\sqrt{\sqrt{61}-6} \\ xy=9 \end{cases}

Video Solution

Answer

x=612βˆ’2.5 x=\frac{\sqrt{61}}{2}-2.5

y=612+2.5 y=\frac{\sqrt{61}}{2}+2.5

or

x=612+2.5 x=\frac{\sqrt{61}}{2}+2.5

y=612βˆ’2.5 y=\frac{\sqrt{61}}{2}-2.5

Exercise #2

Solve the following system of equations:

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

Video Solution

Answer

x=612βˆ’2.5 x=\frac{\sqrt{61}}{2}-2.5

y=612+2.5 y=\frac{\sqrt{61}}{2}+2.5

or

x=612+2.5 x=\frac{\sqrt{61}}{2}+2.5

y=612βˆ’2.5 y=\frac{\sqrt{61}}{2}-2.5

Join Over 30,000 Students Excelling in Math!
Endless Practice, Expert Guidance - Elevate Your Math Skills Today
Test your knowledge
Start practice