Solve a Linear System by Graphing | jensenmath.ca | grade 10
Summary
TLDRThis video script introduces the concept of solving linear systems in a grade 10 math course. It defines a linear system as two or more linear equations considered together and explains the importance of finding the point of intersection. The script outlines three methods to solve these systems: graphing, substitution, and elimination, focusing on the graphing method in this lesson. It demonstrates how to rearrange equations into slope-intercept form, graph the lines, and find their intersection points. The video also discusses the possible outcomes of solving linear systems, including one solution, no solution, or infinite solutions, and emphasizes the importance of verifying solutions.
Takeaways
- 📚 The lesson is focused on solving linear systems, which are sets of two or more linear equations considered simultaneously.
- 📈 A linear system is solved by finding the point of intersection where two or more lines cross, graphically represented as the values of x and y that satisfy all equations.
- 📊 There are three main methods for solving linear systems: graphing, substitution, and elimination, with the first lesson covering graphing.
- 📉 To graph lines, equations are rearranged into the slope-intercept form \( y = mx + b \), where m is the slope and b is the y-intercept.
- 📐 The slope of a line is calculated using the formula \( \frac{change\ in\ y}{change\ in\ x} \) or \( \frac{y2 - y1}{x2 - x1} \).
- 🔍 When graphing, one can use the slope and y-intercept, x and y intercepts, or a table of values to plot the lines accurately.
- 🤔 There are three possible outcomes when solving a linear system: one solution (lines intersect), no solution (parallel and distinct lines), or infinitely many solutions (parallel and coincident lines).
- 📝 The process of solving by graphing involves plotting both lines, finding their point of intersection, and verifying the solution by substituting the intersection point into the original equations.
- 🔢 The script provides step-by-step examples of graphing lines and finding their points of intersection, emphasizing the importance of clear communication and verification of the solution.
- 📉 The importance of verifying the solution is highlighted to ensure the intersection point satisfies both equations in the system, confirming its accuracy.
- 🚫 The script also discusses scenarios where the lines do not intersect, such as when they are parallel and distinct, resulting in no solution, or when they are parallel and coincident, resulting in infinitely many solutions.
Q & A
What is the main focus of the first unit of the grade 10 math course?
-The main focus of the first unit of the grade 10 math course is solving linear systems.
What is a linear system in mathematics?
-A linear system is a set of two or more linear equations that are considered at the same time.
What is the definition of the point of intersection in the context of linear systems?
-The point of intersection is the point where two or more lines cross, which is the solution to the linear system when solved graphically.
What are the three main methods for solving a linear system mentioned in the script?
-The three main methods for solving a linear system are graphing, substitution, and elimination.
How does the script define 'solving' a linear system?
-Solving a linear system means finding the values of the variables that satisfy all of the equations in the system.
What is the equation in the form of y equals mx plus b used for?
-The equation y equals mx plus b is used to describe the relationship between the x and y coordinates of any point on a line segment in terms of the slope (m) and the y-intercept (b).
How can the slope of a line be calculated according to the script?
-The slope of a line can be calculated using the slope formula, which is the change in y over the change in x, written as (y2 - y1) / (x2 - x1).
What are the possible outcomes when solving a linear system graphically?
-The possible outcomes are one solution (if the lines intersect at one point), no solution (if the lines are parallel and distinct), or infinitely many solutions (if the lines are parallel and coincident).
Why is it important to verify the solution after finding the point of intersection?
-It is important to verify the solution by plugging the x and y values of the point of intersection into both original equations to ensure that the point satisfies both equations, confirming it is the correct solution to the linear system.
How does the script describe the process of graphing lines using the slope and y-intercept method?
-The script describes the process as rearranging the linear equations into the form y equals mx plus b, then using the slope (m) and y-intercept (b) to plot the y-intercept and additional points using the slope to fill the grid, and finally connecting the points with a straight line.
What is the significance of the y-intercept in the context of graphing lines?
-The y-intercept is the point where the line crosses the y-axis, and it is used in conjunction with the slope to accurately graph the line.
How can you determine if two lines are parallel?
-Two lines are parallel if they have the same slope but different y-intercepts.
What does it mean if two lines are coincident?
-If two lines are coincident, it means they have the same slope and y-intercept, and therefore, they lie exactly on top of each other, indicating an infinite number of solutions to the linear system.
How does the script suggest communicating the solution to a linear system?
-The script suggests communicating the solution clearly, either by stating the point of intersection as an (x, y) coordinate or by stating the values of x and y separately, and ensuring the solution is well-organized and easy for the teacher to identify.
Outlines
📚 Introduction to Solving Linear Systems
This paragraph introduces the first lesson of the grade 10 math course, focusing on solving linear systems. It defines a linear system as two or more linear equations considered simultaneously and explains the point of intersection where two or more lines cross. The objective is to find the values of variables that satisfy all equations in the system. The three main methods for solving linear systems—graphing, substitution, and elimination—are briefly introduced, with a focus on solving by graphing in this lesson.
📊 Graphing Lines Using Slope and Y-Intercept
This paragraph explains the method of graphing lines using the slope and y-intercept. It details how to rearrange linear equations into the form y = mx + b, where m is the slope and b is the y-intercept. The paragraph covers how to graph lines by plotting the y-intercept and using the slope to plot additional points. It also discusses the different scenarios when lines intersect: at one point (different slopes), no points (parallel and distinct), or infinitely many points (parallel and coincident).
📐 Example: Graphing Two Linear Equations
This paragraph walks through an example of graphing two linear equations: y = x + 4 and y = -x + 2. It explains the process of identifying the slope and y-intercept for each equation, plotting the y-intercepts, and using the slopes to plot additional points. The point of intersection for the two lines is identified and verified by substituting the intersection point into both equations to ensure it satisfies both. The importance of clearly communicating the final solution is emphasized.
📈 Verifying Solutions and Solving Another Example
The paragraph continues with verifying solutions by checking if the intersection point satisfies both equations. It then presents another example with two different lines: 2x + y = 5 and x - 2y = 10. The process of rearranging equations into slope-intercept form, plotting the y-intercepts, and using the slopes to graph the lines is detailed. The intersection point is identified and verified, demonstrating the method's accuracy.
📏 Estimating Intersection Points When Graphing
This paragraph discusses the challenges of estimating intersection points when graphing. It provides an example with two lines: 2x + 5y = -20 and 5x - 3y = -15. The equations are rearranged into slope-intercept form, and the lines are graphed. The intersection point is estimated due to the graph's scale limitations. The importance of learning algebraic methods in future lessons to avoid estimation errors is highlighted.
🔀 Exploring Parallel and Coincident Lines
This paragraph explores scenarios where lines are parallel and either distinct or coincident. It provides examples with equations that have the same slope but different y-intercepts (parallel and distinct, resulting in no solutions) and equations with the same slope and y-intercepts (parallel and coincident, resulting in infinitely many solutions). The graphical representation of these scenarios is explained to illustrate why these cases result in different numbers of solutions.
📏 Conclusion and Transition to Algebraic Methods
The final paragraph summarizes the lesson on solving linear systems by graphing, reiterating the different possible outcomes (one solution, no solutions, infinitely many solutions). It emphasizes the need for algebraic methods like substitution and elimination, which will be covered in upcoming lessons, to provide precise solutions without relying on graphical estimation. The lesson ends with a preview of the next topic: solving linear systems using the method of substitution.
Mindmap
Keywords
💡Linear System
💡Point of Intersection
💡Graphing
💡Slope
💡Y-Intercept
💡Substitution
💡Elimination
💡Parallel Lines
💡Coincident Lines
💡Verification
💡Graphical Solution
Highlights
Introduction to the concept of a linear system and its definition as two or more linear equations considered simultaneously.
Explanation of the point of intersection as the point where two or more lines cross.
Objective of solving a linear system is to find values of variables that satisfy all equations in the system.
Graphical representation of solving linear systems by finding the point where two lines intersect.
Introduction of three main methods for solving linear systems: graphing, substitution, and elimination.
Emphasis on the method of graphing lines using the slope and y-intercept for solving linear systems in this lesson.
Description of rearranging linear equations into the form y = mx + b for graphing purposes.
Explanation of the slope formula and its calculation using the change in y over change in x.
Different methods for graphing lines, including using x and y intercepts or creating a table of values.
Discussion of three possibilities when solving linear systems: one solution, no solutions, or infinitely many solutions.
Visual demonstration of the three possibilities for solving linear systems with different slopes.
Steps for solving linear systems by graphing, including rearranging equations and verifying the solution.
Example of solving a linear system with two lines, y = x + 4 and y = -x + 2, using graphing.
Verification process of a solution by plugging the point of intersection back into the original equations.
Demonstration of solving a second linear system with lines 2x + y = 5 and x - 2y = 10.
Illustration of estimating the point of intersection when exact points are not easily identifiable on the graph.
Examples of linear systems with no solutions due to parallel and distinct lines.
Examples of linear systems with infinitely many solutions due to parallel and coincident lines.
Conclusion and preview of the next lesson on the method of substitution for solving linear systems.
Transcripts
[Music]
let's get started with lesson one of the
first unit of the grade 10 math course
in this unit it's going to be focused on
solving linear systems so let's get that
definition out of the way first what is
a linear system a linear system is two
or more linear equations that are
considered at the same time
another definition we'll need is the
point of intersection
the point of intersection is the point
where two or more lines cross and like i
said in this unit we're focusing on
solving linear systems so what does
solving mean to solve a linear system
means to find the values of the
variables that satisfy all of the
equations in the system so we're going
to be finding the values of x and y
that satisfy both equations
graphically speaking that means we're
finding the point x y where the two
lines will intersect now there are three
main methods for solving a linear system
you can solve by graphing or there are
two algebraic ways called substitution
and elimination that we'll cover in
lessons two and three in this lesson
we're just going to be solving by
graphing and to solve by graphing we
basically just have to graph both lines
in the system and see where their point
of intersection is now there are lots of
ways to graph lines i'm going to be
using method one for graphing lines
which is using the slope and y-intercept
to graph the lines and to do that you'd
have to first rearrange the linear
equations into the form y equals mx plus
b then you can use m the slope and b the
y intercept to accurately graph the line
y equals mx plus b
is an equation that describes the
relationship between the x and y
coordinates of any point on a line
segment
it describes the relationship in terms
of m
the slope of the line which we can
calculate by doing the slope formula
change in y over change in x which we
could write as y2 minus y1 over x2 minus
x1 and b
the y-intercept of the line
you don't always have to use that method
you're welcome to graph it using its x
and y intercepts or creating a table of
values for the equation any method you
use should get you an accurate graph of
the line and allow you to find the point
of intersection let me go over the three
possibilities for what could happen when
you are trying to solve a linear system
before we fill out this chart let me
give you a visual demonstration of this
if the lines are not parallel meaning
they have different slopes the lines
will intersect at one point meaning we
will get one solution for x and y
if the lines are parallel meaning they
have the same slope they're either going
to be parallel and distinct meaning
they're going to have no solutions or
they'll be parallel and coincident
meaning that they have an infinite
number of solutions
so in this table let's fill out that
information so if the lines intersect at
one point we know that the slopes are
different
comparing the x and y intercepts of two
non-parallel lines isn't necessary to
determine if it has a point of
intersection or not we know if the lines
aren't parallel they have a point of
intersection
the y-intercepts and the x-intercepts
are usually different from each other
between the lines but i suppose they
could be the same if
on the x or y-axis is where the point of
intersection is
so i'll say that the intercepts are
usually different unless the lines
intersect on an axis how many solutions
do we get if we have different slopes
we get
one solution one point of intersection
now lines are parallel and distinct if
they have
the same slope
but
different
intercepts
if that happens we get no solutions to
the linear system
and you can see graphically there's
going to be no point where those lines
intersect
if the lines are parallel and coincident
it means the lines are right on top of
each other and that's because they have
the exact same slope and the exact same
intercepts
meaning we get infinitely many solutions
every point that's on one line is also
on the other line that's why there's
infinitely many solutions now we're
going to solve five linear systems
together using the method of graphing
here are the steps we're going to follow
for each of those systems we're going to
start by rearranging each linear
equation into the form y equals mx plus
b
i'll then graph both of the equations
and find the point of intersection
once i figure out where the point of
intersection is it's important that we
verify our answer plug the xy point of
the point of intersection into both of
the original equations to make sure that
xy point satisfies both equations that
verifies it's on both lines and proves
you have the correct answer and we
should clearly communicate our solution
making sure that your teacher is going
to be able to see what you think your
final answer is
let's try example one i have a linear
system i have two lines y equals x plus
four and y equals negative x plus two
i'm just going to show some rough work
for both lines and i'll do this for each
of the examples and i'll color code it
so i have
line one n is y equals one x plus four
now usually we don't write a coefficient
of one but i'm going to put it there
just so you can see that that is the
slope of this line right a line that's
in the form y equals mx plus b
its slope is equal to the m value which
in this case i see m is right here which
is 1. now when graphing lines it helps
to think of the slope as a fraction
right because slope is rise over run so
any whole number you could rewrite as
over 1.
so what i have here here's my m value
and my slope remember is rise
over run
and also i'm going to need to point out
what the y intercept is the y intercept
is the b value of the equation
the b value is 4
so i'm going to write the y intercept is
equal to the b value
which is 4. now to graph this line all i
have to do is first plot the y intercept
of 4
and then use my slope of 1 over 1 right
rise 1 run 1 to plot more points to fill
this grid so from my y-intercept i'm
going to rise one and run one remember a
positive rise means up and a positive
run means right so up one right one i'm
going to fill this grid
and now to plot points on the other side
of the y-intercept we do the exact
opposite instead of up one right one we
go down one left one
and notice these points also fall on the
same line
now i'm going to connect these points
with a straight line
and label this as line one
make sure when you're graphing lines you
put arrows on both ends of the line
showing that it does continue
next let me do the same process but for
line two line two is y equals negative
one x
plus 2. now notice this is also in the
format y equals mx plus b already
where
the slope
is equal to the m value which in this
case is negative 1 and like i said any
whole number you could rewrite as over 1
so that makes it easier to graph
and the y-intercept is the b value
which with this equation is 2.
here's our m-value it's the coefficient
of x here's our b value it's the
constant added after the x
and my slope i wrote it as a fraction
because slope is rise
over run so to graph this line we always
start by plotting the y intercept of 2
and then do rise over run
to plot points to fill the grid
so rise of negative one means down one
and run one means right one so i go down
one right one
and then to plot points on the other
side of the y-intercept do the exact
opposite instead of down one and right
one i go up one and left one and notice
oh there's the point of intersection
right there
i'll connect all the points with a
straight line making sure there's arrows
on both sides
i will label this one as line two
and i'm going to circle this point of
intersection right here
so our point of intersection
that point is the point negative one
three so i will say the point of
intersection and we can short form point
of intersection right that is poi the
point of intersection
is the point negative 1 3.
another way of writing your answer
instead of writing it as an x y point
you could say the solution is x equals
negative one comma y equals three those
are two different ways of communicating
the same solution now if you look at
this solution it's very well organized
and it's clear where the final answer is
when teachers are grading your work
they're required to grade your
communication of your solution as well
so make sure you're clearly
communicating the knowledge that you
have about how to solve the problem
one more thing that we should do with
each of these questions is we should
verify that the answer is correct
you can know if you got the correct
answer for these questions by checking
to make sure the x value of negative 1
and the y value of 3 satisfy both
equations let me show you how we can do
that check
to check the solution i need to check it
in both lines and i need to verify the
point is on both lines not just one
right verify it's on both lines prove
that that's where the lines intersect
making it the solution to the system so
let me check if the point negative one
three is on line 1. and how we do that
is i split the equation for line 1 into
its left side and right side
line 1 the left side is y the right side
is x plus 4.
so left side is y right side is x plus
4. if the point negative 1 3 is on this
line it'll make the left side equal to
the right side so let me just add up
here what solution i'm actually checking
i'm checking the solution x equals
negative 1 y equals 3.
so anytime i see a y i'm going to
replace it with three anytime i see an x
i'm going to replace it with negative
one and then simplify negative one plus
four is three i can see left side and
right side are the same value therefore
left side equals right side that point
is on that line
let's verify it's also on line
two
let me separate
the equation for line two into left side
right side left side is y
right side is negative x plus two
let's verify the solution change all the
y's to three change the x to negative
one so negative
negative one plus two
negative negative one is positive one
plus two is three hey that left side and
right side are equal again that means
the point is also on that line if the
point is on both of the lines that must
be where the lines intersect making it
the solution to the linear system let's
try another example we can do the next
view a little bit quicker so let's set
up our space for our work for
line
one
line one is two x plus y equals five
i want to rearrange this into the form y
equals mx plus b so i'm going to move
the 2x term to the other side of the
equation making it negative 2x and i'll
leave the positive 5 there
now you can see that the slope of this
line is equal to the m value which is
negative 2
or as a fraction we could think of it as
negative 2 over 1
and the y intercept
is equal to the b value which is
5. right here's the m
here's the b
and remember we write m as a fraction
because slope is rise
over run with that information i can
graph this line always start by plotting
the y intercept so plot five and my
slope is negative two over one rising
negative two means down two running one
means right one
down two right one and keep plotting
points until you fill the grid and to
plot points on the other side do the
opposite instead of down to right one go
up two left one
i'll connect it with a straight line and
label this line one
i will now go through the same process
for line two line two as x minus two y
equals ten
this needs to be rearranged into slope y
intercept form i'm actually going to
isolate y on the right this time
so i'm going to take this term negative
2 i move it to the right making it
positive 2y
and bring the constant 10 to the other
side making it negative 10. so i have x
minus 10 equals 2y
i want to isolate y currently it's being
multiplied by 2 so i need to divide both
sides of this equation by 2.
that means both terms on the left are be
going to be divided by 2. so i'll have a
half x
minus
10 over 2 is 5
equals 2y over 2 is y
so this is in the form mx plus b equals
y which is perfectly fine our m is a
half and our b is negative five
so that means my slope
is equal to my m value which is one over
two
and my y-intercept
is equal to my b value which is negative
five and remember slope is rise over run
so let's graph line two by first
plotting its y-intercept at negative
five
and then using the slope to plot more
points to fill the grid rise one means
up one run two means right two
and keep plotting points until we fill
the grid
and i can already see where the point of
intersection is going to be but let me
continue plotting points just to finish
off my line accurately
these two lines intersect at this point
right here that's the point four
negative three so we could write our
answer as a point of intersection
or we can write the value of x and value
of y separately
and like i said we should verify this is
the correct answer by making sure that
point 4 negative 3 is on both lines
so let me check the solution let me
verify it's on line 1 first so the
equation of line 1 was 2x plus y equals
5. so the left side is 2x plus y
and the right side is 5. and let me plug
in the x and y values that i got from my
solution x is 4
and y is negative 3.
so i have 8 plus negative 3 that means 8
minus 3 which is 5. left side equals
right side good that means the points on
that line
let's verify the point is also on the
other line and then that proves that's
the point of intersection
the equation of the other line was x
minus 2y equals 10. so the left side is
x minus 2y the right side
is 10.
let me plug in my point
4 for x negative three for y i have four
minus negative six that's four plus six
which is ten good left side equals right
side again which means the point four
negative three is on line two as well so
that's where the lines intersect okay so
the first two examples worked out very
nicely notice while we were graphing we
ended up plotting points right on top of
each other so it was very clear where
the point of intersection would be
when solving by graphing it doesn't
always work out that nicely let me show
you what i mean with part c line 1 is 2x
plus 5y equals negative 20. let me get
that into y equals mx plus b form i'll
start by moving
the 2x to the other side becomes
negative x
and then divide both sides by five
making sure to divide all terms by five
so negative two x over five is negative
two over five
times x
minus twenty divided by five is four
so there's the equation in y equals mx
plus b form where m is negative two over
five
and b is negative four so my slope is
negative two over five and my
y-intercept is negative four so to graph
this line we always start by plotting
the y intercept
and then use the slope rise negative two
run five to plot more points rise
negative two means down two run five
means right five
i won't be able to plot very many points
on this graph but i'll plot as many as i
can
to the other side of the y intercept do
the opposite instead of down two right
five go up two
and left five
and i've connected the points and
labeled the line so there's my line one
let's graph line two and see where they
intersect
line two is five x minus three y equals
negative fifteen
i'm going to rearrange this one to
isolate y on the right this time so i'll
have 5x
plus 15 equals 3y
and divide both sides by 3 i get 5 over
3x plus 15 over 3 which is 5 equals 3y
over 3 which is y
so there we have it slope y intercept
form
where m is 5 over 3
b is 5.
so my slope
is 5 over 3 and my y intercept is 5. let
me graph this line by starting by
plotting the y intercept and using the
slope of rise five
run three
so up five right three and the opposite
down five left three
there's my line 2 graphed
and now i'm going to try and find the
point of intersection
now the lines definitely cross
but my answer for this is going to have
to be an estimation they cross at this
point right here
what do i think the coordinates of that
point are
i'm not sure exactly but if i were to
estimate i'd say it looks like it's at
somewhere between negative 4 and
negative 5 as the x value
and somewhere between negative 2 and
negative 3 is the y value so i'm just
going to estimate
the point of intersection i think is
approximately
negative 4.3
comma negative 2.2
about
or another way of writing this the
solution
is once again i'll write the word
approximately to show that this is an
estimation
x equals negative 4.3
y equals negative 2.2
now i'm not going to try and verify this
solution because this has been an
estimation
this point negative 4.3 negative 2.2
might not be on either of the lines but
if i were to sub them back in and check
it should make left side and right side
of both equations like
approximately equal to each other but
there's no way to verify for this type
of solution if it's exactly correct or
not which is why in lesson two and three
we need to learn a better way of solving
linear systems so that we don't have to
do any estimation we'll learn algebraic
methods of substitution and elimination
that avoid us having to do any
estimation now remember when solving
linear systems you don't always get one
solution like we got in a b and c
part d and e are going to show you the
other scenarios
where you could get no solutions or
infinite solutions let's see which one
part d is so let's graph both lines i
mean i can tell right away that these
have the same slope but different
y-intercepts so i know they're parallel
and distinct and are going to have no
solutions but let's verify that
graphically
so line one is already in the form y
equals mx plus b
where our m value is our slope
and it's two which is two over one
and our y intercept is the b value which
is
3. so if i had to graph this line i
would plot the y-intercept of 3 and use
my slope of 2 over 1 to plot more points
now let's look at the properties of line
2.
once again line 2 is already in the
format y equals mx plus b
where our slope is equal to the m value
of
two which we could think of once again
as two over one
and our y intercept
equals the b value of the equation which
is negative four so if i were to graph
this one by plotting the y-intercept and
using the slope
notice i get a line that runs exactly
parallel to line one so it's never going
to intersect it therefore there are no
solutions to this linear system
so our final answer for this would be
the lines are parallel and distinct
therefore there are no solutions
or you could say there is no point of
intersection
last example let's work with line one
which is x plus y equals three let me
rearrange this into y equals mx plus b
form by moving that positive x to the
right becomes a negative x
now i can see that my slope
is equal to the m value the coefficient
of that x is negative one which as a
fraction is negative one over one
the y intercept
is equal to the b value which is
three
to graph this line plot the y intercept
and use the slope of rise negative one
run one that means down one right one
and then line two i have two x plus two
y equals six if i rearrange this one to
isolate y
i would have negative two x plus six on
the right
and then isolate the y by dividing both
sides by two so divide all terms by two
i'd get y equals negative x plus three
hey
notice exact same equation as line one
that tells me it's going to have the
same slope and the same y-intercept so
when i graph this one plot the
y-intercept
and then use the slope to plot more
points notice all the points are right
on top of each other all of these points
are solutions to the linear system so
from the equations i can see they have
the same slope and y intercept
meaning they're going to be parallel and
coincident meaning there are infinitely
many solutions to this system so let me
communicate that in my final answer oh
and i suppose i should somehow try and
show line two right on top of line one
so maybe
i'll put it right on top
and see if i can make it a little bit
transparent so we can see through to the
other one and i'll label that as line
one and line two so let me now write my
final answer the lines are parallel and
coincident
therefore there are infinitely many
solutions and that's it for this lesson
stay tuned for lesson two where we learn
the method of substitution to solve
linear systems
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