SAT Exam > SAT Videos > Mathematics for Digital SAT > Non-linear systems of equations 2
Non-linear systems of equations 2 Video Lecture | Mathematics for Digital SAT
|
185 videos|124 docs|75 tests
|
FAQs on Non-linear systems of equations 2 Video Lecture - Mathematics for Digital SAT
| 1. What are non-linear systems of equations? |  |
Ans. Non-linear systems of equations are mathematical equations that involve variables with exponents, square roots, or other non-linear functions. Unlike linear systems, where the equations form straight lines, non-linear systems can have curves and more complex graphical representations.
| 2. How are non-linear systems of equations solved? | |
Ans. Solving non-linear systems of equations can be more challenging than solving linear systems. The most common methods include graphing, substitution, and elimination. However, in some cases, it may be necessary to use numerical methods or advanced techniques such as Newton's method or the method of Lagrange multipliers.
| 3. What is the importance of understanding non-linear systems of equations? | |
Ans. Non-linear systems of equations are prevalent in various fields, including physics, engineering, economics, and computer science. Understanding them is crucial for solving real-world problems that involve non-linear relationships between variables. It enables accurate modeling, optimization, and analysis of complex systems.
| 4. Are there any limitations or challenges in solving non-linear systems of equations? | |
Ans. Yes, solving non-linear systems can be challenging due to their complexity. Unlike linear systems, non-linear systems often lack a straightforward algebraic solution. They may have multiple solutions, no solution, or even an infinite number of solutions. In some cases, numerical approximation methods or software tools are required to obtain an accurate solution.
| 5. Can non-linear systems of equations have real-world applications? | |
Ans. Absolutely! Non-linear systems of equations have numerous real-world applications. For example, they can be used to model the growth of populations, predict the behavior of electrical circuits, analyze chemical reactions, optimize production processes, or simulate the motion of celestial bodies. Understanding and solving non-linear systems are essential for advancing in these fields and making informed decisions.
Text Transcript from Video
Solve the system of equations
by substitution.
Check your solution by graphing
the equation.
Let's do it by substitution.
We know that y is equal to this
thing over here, and we
also know that y is equal
to this thing over here.
So they're going to intersect
when they both equal the same
y, or when this thing is equal
to this thing, or when
negative x squared is equal to
2x squared plus 3x minus 6.
The x value, which these two are
equal, is going to be the
x values where these y's are
equal, so that's going to be
their point of intersection.
That will be an x and y
pair that satisfies
both of these equations.
Let's solve for x here.
A good starting point-- let's
just add x squared to both
sides, and we end up with
0 is equal to 3x squared
plus 3x minus 6.
Then, just to simplify this on
the right hand side, we could
divide both sides of the
equation by 3-- we can divide
everything by 3-- and we're
left with 0 is equal to x
squared plus x minus 2.
We could do the quadratic
equation, or complete the
square, or all sorts of crazy
things, but this is actually
very factorable-- 0 is equal
to two numbers, which are 2
and negative 1.
When you multiply them, you get
negative 2, and when you
add them, you get positive 1.
So this is x plus 2
times x minus 1.
That tells us that x plus 2 is
equal to 0, or x minus 1 could
be equal to 0.
Subtract 2 from both
sides of this.
You could get x is equal
to negative 2.
Or add 1 to both sides of
this, x is equal to 1.
These are our two solutions-- x
is equal to negative 2 or x
is equal to 1.
Let's verify it.
When I put x is equal to
negative 2-- let's do it over
here-- what do I get?
Negative 2 squared
is positive 4.
And then you put a negative
there, so it's negative 4--
the point here is negative
2, and then negative 4.
That's what happens when you
put negative 2 there.
And then what happens
when I put 1?
1 squared is 1, and then if you
put a negative there, it's
negative 1.
So it's 1, negative 1.
And so these are both points on
this equation right here,
on this function.
If you look at this one over
here, if you put negative 2
over here-- 2 times negative
2 squared.
Negative 2 squared
is positive 4.
2 times positive 4 is 8.
3 times negative 2 is negative
6, so 8 minus 6.
8 minus 6 is 2, and 2 minus
6 is equal to negative 4.
The point negative 2, negative
4 is on this function.
They both share that
point in common, so
they intersect there.
If I put 1 into this, I
get 2 plus 3 minus 6.
2 plus 3 is 5, minus 6, when x
is 1, which is negative 1.
The point 1, negative 1 is
also on this top graph.
We can plot these points:
negative 2 comma negative 4.
Negative 2-- 1, 2, 3,
4-- is right there.
That's going to be a point
of intersection.
Then we have the point
1, negative 1.
1, negative 1 is also going
to be a point of
intersection there.
Let's graph this second one and
just verify-- they say,
check your solution by graphing
the equation.
Let's graph the first one.
This is pretty easy to
graph-- y is equal
to negative x squared.
It's going to intersect
the point 0, 0.
It's going to be a
downward slope,
downward opening parabola.
When x is positive or negative
1, y is going to be negative
1, because you square them and
then you take the negative.
When x is positive or negative
2, y is going
to be negative 4.
So the graph will look
something like this--
4, 5, 6, 7, 8, 9.
The graph will look something
like that.
That is that equation
right here.
This is going to be an upward
opening parabola, and a quick
way-- we could do all of that
completing the square, but if
you think about it, let's just
apply the quadratic equation
right here.
I want to show you something,
a quick way of where the
vertex formula comes
from when you apply
the quadratic formula.
If you take this, and you wanted
to find its zeros, you
would say x is equal to negative
b-- which is negative
3-- plus or minus the square
root of b squared-- which is
9-- minus 4ac.
Minus 4 times 2 times
negative 6.
All of that over 2 times
a, or 2 times 2.
Now, we can evaluate this and
figure out the zeros of this,
but the real point why I wanted
to show you this is
because we always have
two solutions.
Let me actually write
down the quadratic
formula here for you.
The quadratic formula is
negative b plus or minus the
square root of b squared
minus 4ac over 2a.
You always end up-- or if this
is a positive number-- with
two solutions, and they're
equal distant.
They're this far away, this far
over 2a away from negative
b over 2a-- we could write this
as negative b over 2a
plus or minus the square
root of b squared
minus 4ac over 2a.
You have at most two solutions
that are equal distant from
this x value right here.
We've seen in multiple videos:
what is that point that is
equidistant from the
two solutions,
that is right in between?
That's going to be the line of
symmetry, or the x value of
the vertex.
This is where the x value of the
vertex formula comes from:
x val of vertex is negative
b over 2a.
If we wanted to find the vertex,
the x value of this
guy right here, we just take
negative b-- which is negative
3-- over 2 times a, 2
times 2, which is 4.
So, x is equal to negative
3/4 is the
vertex of this parabola.
And when x is equal to negative
3/4, what is y?
We could actually-- that's a
little bit more complicated
right here, but I'll
just go through it.
It's going to be 2 times
9 over 16 minus 9
over 4 minus 6.
Let me actually get the
calculator out for this one.
It's going to be 18 over 60.
Let me just get the calculator
out, because it will be
simpler, and I don't want
to waste your time doing
arithmetic.
It's going to be 18 divided by
16 minus 9 divided by 4 minus
6 is equal to negative 7.125.
This is equal to
negative 7.125.
So the vertex occurs when x is
negative 3/4-- when x is right
there-- and y is negative 7.
This is essentially 7 1/8, so
1, 2, 3, 4, 5, 6, 7.125.
So it's a little over 7.
The vertex of this top graph
is right over there.
It's symmetric around
the vertex.
That is the line of symmetry,
and so this top graph is going
to look something like this.
Just like that and we're done.
We've found our two points of
intersection, right over there
and right over there, and
when you graph it it
looks pretty good.
by substitution.
Check your solution by graphing
the equation.
Let's do it by substitution.
We know that y is equal to this
thing over here, and we
also know that y is equal
to this thing over here.
So they're going to intersect
when they both equal the same
y, or when this thing is equal
to this thing, or when
negative x squared is equal to
2x squared plus 3x minus 6.
The x value, which these two are
equal, is going to be the
x values where these y's are
equal, so that's going to be
their point of intersection.
That will be an x and y
pair that satisfies
both of these equations.
Let's solve for x here.
A good starting point-- let's
just add x squared to both
sides, and we end up with
0 is equal to 3x squared
plus 3x minus 6.
Then, just to simplify this on
the right hand side, we could
divide both sides of the
equation by 3-- we can divide
everything by 3-- and we're
left with 0 is equal to x
squared plus x minus 2.
We could do the quadratic
equation, or complete the
square, or all sorts of crazy
things, but this is actually
very factorable-- 0 is equal
to two numbers, which are 2
and negative 1.
When you multiply them, you get
negative 2, and when you
add them, you get positive 1.
So this is x plus 2
times x minus 1.
That tells us that x plus 2 is
equal to 0, or x minus 1 could
be equal to 0.
Subtract 2 from both
sides of this.
You could get x is equal
to negative 2.
Or add 1 to both sides of
this, x is equal to 1.
These are our two solutions-- x
is equal to negative 2 or x
is equal to 1.
Let's verify it.
When I put x is equal to
negative 2-- let's do it over
here-- what do I get?
Negative 2 squared
is positive 4.
And then you put a negative
there, so it's negative 4--
the point here is negative
2, and then negative 4.
That's what happens when you
put negative 2 there.
And then what happens
when I put 1?
1 squared is 1, and then if you
put a negative there, it's
negative 1.
So it's 1, negative 1.
And so these are both points on
this equation right here,
on this function.
If you look at this one over
here, if you put negative 2
over here-- 2 times negative
2 squared.
Negative 2 squared
is positive 4.
2 times positive 4 is 8.
3 times negative 2 is negative
6, so 8 minus 6.
8 minus 6 is 2, and 2 minus
6 is equal to negative 4.
The point negative 2, negative
4 is on this function.
They both share that
point in common, so
they intersect there.
If I put 1 into this, I
get 2 plus 3 minus 6.
2 plus 3 is 5, minus 6, when x
is 1, which is negative 1.
The point 1, negative 1 is
also on this top graph.
We can plot these points:
negative 2 comma negative 4.
Negative 2-- 1, 2, 3,
4-- is right there.
That's going to be a point
of intersection.
Then we have the point
1, negative 1.
1, negative 1 is also going
to be a point of
intersection there.
Let's graph this second one and
just verify-- they say,
check your solution by graphing
the equation.
Let's graph the first one.
This is pretty easy to
graph-- y is equal
to negative x squared.
It's going to intersect
the point 0, 0.
It's going to be a
downward slope,
downward opening parabola.
When x is positive or negative
1, y is going to be negative
1, because you square them and
then you take the negative.
When x is positive or negative
2, y is going
to be negative 4.
So the graph will look
something like this--
4, 5, 6, 7, 8, 9.
The graph will look something
like that.
That is that equation
right here.
This is going to be an upward
opening parabola, and a quick
way-- we could do all of that
completing the square, but if
you think about it, let's just
apply the quadratic equation
right here.
I want to show you something,
a quick way of where the
vertex formula comes
from when you apply
the quadratic formula.
If you take this, and you wanted
to find its zeros, you
would say x is equal to negative
b-- which is negative
3-- plus or minus the square
root of b squared-- which is
9-- minus 4ac.
Minus 4 times 2 times
negative 6.
All of that over 2 times
a, or 2 times 2.
Now, we can evaluate this and
figure out the zeros of this,
but the real point why I wanted
to show you this is
because we always have
two solutions.
Let me actually write
down the quadratic
formula here for you.
The quadratic formula is
negative b plus or minus the
square root of b squared
minus 4ac over 2a.
You always end up-- or if this
is a positive number-- with
two solutions, and they're
equal distant.
They're this far away, this far
over 2a away from negative
b over 2a-- we could write this
as negative b over 2a
plus or minus the square
root of b squared
minus 4ac over 2a.
You have at most two solutions
that are equal distant from
this x value right here.
We've seen in multiple videos:
what is that point that is
equidistant from the
two solutions,
that is right in between?
That's going to be the line of
symmetry, or the x value of
the vertex.
This is where the x value of the
vertex formula comes from:
x val of vertex is negative
b over 2a.
If we wanted to find the vertex,
the x value of this
guy right here, we just take
negative b-- which is negative
3-- over 2 times a, 2
times 2, which is 4.
So, x is equal to negative
3/4 is the
vertex of this parabola.
And when x is equal to negative
3/4, what is y?
We could actually-- that's a
little bit more complicated
right here, but I'll
just go through it.
It's going to be 2 times
9 over 16 minus 9
over 4 minus 6.
Let me actually get the
calculator out for this one.
It's going to be 18 over 60.
Let me just get the calculator
out, because it will be
simpler, and I don't want
to waste your time doing
arithmetic.
It's going to be 18 divided by
16 minus 9 divided by 4 minus
6 is equal to negative 7.125.
This is equal to
negative 7.125.
So the vertex occurs when x is
negative 3/4-- when x is right
there-- and y is negative 7.
This is essentially 7 1/8, so
1, 2, 3, 4, 5, 6, 7.125.
So it's a little over 7.
The vertex of this top graph
is right over there.
It's symmetric around
the vertex.
That is the line of symmetry,
and so this top graph is going
to look something like this.
Just like that and we're done.
We've found our two points of
intersection, right over there
and right over there, and
when you graph it it
looks pretty good.
About this Video
|
4.94/5 Rating |
|
Nov 22, 2024 Last updated |
Video Description: Non-linear systems of equations 2 for SAT 2024 is part of Mathematics for Digital SAT preparation.
The notes and questions for Non-linear systems of equations 2 have been prepared according to the SAT exam syllabus.
Information about Non-linear systems of equations 2 covers all important topics for SAT 2024 Exam.
Find important definitions, questions, notes, meanings, examples, exercises and tests below for Non-linear systems of equations 2.
Introduction of Non-linear systems of equations 2 in English is available as part of our Mathematics for Digital SAT for SAT & Non-linear systems of equations 2 in
Hindi for Mathematics for Digital SAT course. Download more important topics related with notes, lectures and mock test series for
SAT Exam by signing up for free.
Description
Video Lecture & Questions for Non-linear systems of equations 2 Video Lecture | Mathematics for Digital SAT - SAT full syllabus preparation | Free video for SAT exam to prepare for Mathematics for Digital SAT.
Information about Non-linear systems of equations 2
Here you can find the meaning of Non-linear systems of equations 2 defined & explained in the simplest way possible.
Besides explaining types of Non-linear systems of equations 2 theory, EduRev gives you an ample number of questions to practice Non-linear systems of equations 2 tests,
examples and also practice SAT tests.
|
185 videos|124 docs|75 tests
|
|
Explore Courses for SAT exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
study material
,ppt
,shortcuts and tricks
,mock tests for examination
,past year papers
,practice quizzes
,Free
,Objective type Questions
,Non-linear systems of equations 2 Video Lecture | Mathematics for Digital SAT
,Non-linear systems of equations 2 Video Lecture | Mathematics for Digital SAT
,MCQs
,Viva Questions
,Exam
,Non-linear systems of equations 2 Video Lecture | Mathematics for Digital SAT
,video lectures
,Extra Questions
,Sample Paper
,Important questions
,Summary
,Semester Notes
,Previous Year Questions with Solutions
;
Study Non-linear systems of equations 2 on the App
Students of SAT can study Non-linear systems of equations 2 alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Non-linear systems of equations 2,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Non-linear systems of equations 2 is prepared as per the latest SAT syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup