20.Solving Systems of Linear Equation.pptx

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L.C. M8AL-Ii-j-1: Solve systems of linear equation by graphing, substitution and G r a d e 8
G R A D E 8
M A T H E M A T I C S
SYSTEMS OF LINEAR EQUATION

Content Standard:
The learner demonstrates understanding of
key concepts of linear equations
Performance Standard:
The learner is able to formulate real-life
problems involving linear equations

• How does the family’s power
consumption affect the amount of the
electric bill?

ELIMINATION
SUBSTITUTION
BY GRAPHING
EXIT
M
E
N
U

BY GRAPHING
A system of linear equations consists of two or
more linear equations.
This section focuses on only two equations at a
time.
The solution of a system of linear equations in two
variables is any ordered pair that solves both of
the linear equations.

BY GRAPHING
Example
Determine whether the given point is a solution of the following system.
point: (– 3, 1)
system: x – y = – 4 and 2x + 10y = 4
•Plug the values into the equations.
First equation: – 3 – 1 = – 4 true
Second equation: 2(– 3) + 10(1) = – 6 + 10 = 4 true
•Since the point (– 3, 1) produces a true statement in both equations, it
is a solution.

BY GRAPHING
Example
Determine whether the given point is a solution of the following system
point: (4, 2)
system: 2x – 5y = – 2 and 3x + 4y = 4
Plug the values into the equations
First equation: 2(4) – 5(2) = 8 – 10 = – 2 true
Second equation: 3(4) + 4(2) = 12 + 8 = 20  4 false
Since the point (4, 2) produces a true statement in only one equation, it
is NOT a solution.

BY GRAPHING
•Since our chances of guessing the right coordinates to try for a
solution are not that high, we’ll be more successful if we try a
different technique.
•Since a solution of a system of equations is a solution common
to both equations, it would also be a point common to the
graphs of both equations.
•So to find the solution of a system of 2 linear equations, graph
the equations and see where the lines intersect.

BY GRAPHING
Example 1
Solve the following
system of equations
by graphing.
2x – y = 6 and
x + 3y = 10
First, graph 2x – y = 6.
Second, graph x + 3y = 10.
The lines APPEAR to intersect at (4, 2).
x
y
(-5, 5)
(-5, 5)
(-2, 4)
(1, 3)
(0, -6)
(3, 0)
(6, 6)
(4, 2)

BY GRAPHING
Example 2
Solve the following
system of equations
by graphing.
– x + 3y = 6 and
3x – 9y = 9
First, graph – x + 3y = 6.
Second, graph 3x – 9y = 9.
The lines APPEAR to be parallel.
x
y
(-6, 0)
(0, 2)
(6, 4)
(0, -1)
(3, 0)
(6, 1)

BY GRAPHING
Example 3
Solve the following
system of equations
by graphing.
x = 3y – 1 and
2x – 6y = –2
x
y
First, graph x = 3y – 1.
(7, -2)
(-1, 0)
(5, 2)
Second, graph 2x – 6y = –2.
(-4, -1)
(2, 1)
The lines APPEAR to be identical.

BY GRAPHING
•There are three possible outcomes when graphing two linear
equations in a plane.
•One point of intersection, so one solution
•Parallel lines, so no solution
•Coincident lines, so infinite # of solutions
•If there is at least one solution, the system is considered to
be consistent.
•If the system defines distinct lines, the equations are
independent.

BY GRAPHING
Activity 1
Solve the following system of
equations by graphing.
x + y = 6 and
x = y + 2
x
y
GRAPH

Let’s Summarize
How will you
solve system of
equations by
graphing?
BY GRAPHING
1. Graph both equations on
a Cartesian plane.
2. Find the point of
intersection of the graphs,
if it exists.

Let’s Summarize
BY GRAPHING
The solution
to a system of linear equations corresponds
to the coordinates of the points of
intersection of the graphs of the equations.

Work Alone:
Let’s
Check!!!
BY GRAPHING
Solve the following system of
equations by graphing.
x + y = 5 and
x - y = 1
x
y

Reference:
Grade 8 Teachers’ Guide, pp. 286-290
Grade 8 Learner’s Manual, pp. 270-273
BY GRAPHING

MAIN
MENU
EXIT
Click “Main Menu” to
choose another method in
Finding the Equation of a
Line
Click “EXIT” to
end
presentation

BY SUBSTITUTION
Another method (beside getting
lucky with trial and error or
graphing the equations) that can
be used to solve systems of
equations is called the
substitution method.

BY SUBSTITUTION
TRY THESE!
•Solve for the indicated variable in terms of the other
variable.
•1. 4x + y = 11 y =
•2. 5x – y = 9 y =
•3. 4x + y = 12 x =
•4. 2x + 3y = 6 y =
– 4x + 11
5x - 9
-y/4 +3
-2x/3 + 2

BY SUBSTITUTION
Steps
• Solve the value of one variable in terms of the other. You can use
either of the equations.
• Substitute the expression in step 1 into the other equation. This will
result in an equation in one variable. Solve the equation.
• Back – substitute the value obtained from step 2 in either of the
original equations to find the value of the remaining variable.
• Check the proposed solutions in both of the equations in the system.

BY SUBSTITUTION
Example 1
Solve the following system using the substitution
method.
3x – y = 6 and – 4x + 2y = –8
1. Solving the first equation for y,
3x – y = 6
–y = –3x + 6 (subtract 3x from both sides)
y = 3x – 6 (multiply both sides by – 1)
continued

BY SUBSTITUTION
Example 1 continued
2. Substitute this value for y in the second equation.
–4x + 2y = –8
–4x + 2(3x – 6) = –8 (replace y with result from first equation)
–4x + 6x – 12 = –8 (use the distributive property)
2x – 12 = –8 (simplify the left side)
2x = 4 (add 12 to both sides)
x = 2 (divide both sides by 2)
continued

BY SUBSTITUTION
3. (Back) Substitute x = 2 into the first equation, then solve for
y.
y = 3x – 6
y = 3(2) – 6
y = 6 – 6
y = 0
Our computations have produced the point (2, 0).
Example 1 continued
continued

BY SUBSTITUTION
4. Check the point in the original equations.
First equation,
3x – y = 6
3(2) – 0 = 6 true
Second equation,
–4x + 2y = –8
–4(2) + 2(0) = –8 true
The solution of the system is (2, 0).
Example 1 continued

BY SUBSTITUTION
Example 2
method.
x + y =6 and x = y + 2
Since the second equation gives the value of x, skip
step 1.
continued

BY SUBSTITUTION
Example 2 continued
2. Substitute.
x + y = 6
y+2 + y = 6 (replace x with result from first equation)
2y + 2 = 6 (combine similar terms)
2y + 2-2 = 6-2 (subtract 2 to both sides of the equation)
2y = 4 (divide both sides by 2)
y = 2
continued

BY SUBSTITUTION
3. (Back) Substitute y = 2 into the second equation, then
solve for x.
x = y + 2
x = 2 + 2 (substitute y with 2)
x = 4 (simplify the right side)
Our computations have the point (4, 2).
Example 2 continued
continued

BY SUBSTITUTION
4. Check the point in the original equations.
First equation,
x + y =6
4 + 2 = 6 true
Second equation,
x = y + 2
4 = 2 + 2 true
The solution of the system is (4, 2).
Example 2 continued

TRY THESE!
BY SUBSTITUTION
method.
1. x + y =6
x = y + 2
2. y = 2x – 5
3y – x = 5
LET’S
CHECK!

Let’s Summarize
How to solve
system of
equations by
substitution?
Steps:
1. Solve one variable in terms of
the other variable.
3. Back.
4. Check.
BY SUBSTITUTION
2. Substitute.

Work Alone
Let’s
Check!
BY SUBSTITUTION
method.
1. x = y - 1
2x – 4 = 8
2. x + y = 11
x = y + 5

Reference:
Grade 8 Learner’s Manual pp. 274-275
BY SUBSTITUTION

BY ELIMINATION
Another method that can be used to solve
systems of equations is called the addition
or elimination method.

TRY THESE!
Find the value of the variable that would make the equation
true.
1. 5x = 15
2. -3x = 21
3. 9x = -27
4. x + 7 = 10
5. 3y – 5 = 4
BY ELIMINATION
x = -7
x = 3
x = 17
x = -3
x = 3

BY ELIMINATION
You multiply both equations by numbers that will
allow you to combine the two equations and
eliminate one of the variables.

ACTIVITY 1
BY ELIMINATION
Solve
x + y = 5
x − y = 1
by using the elimination method.
1. Are there terms in the system which are additive inverses? If
yes, what are they?
x + y = 5
x − y = 1
(Add the two equations)
2x = 6 x = 3 (Divide both sides by 2)

ACTIVITY 1 Continued
BY ELIMINATION
2. Solve for the value of y by substituting the value of x in either
of the two equations.
x + y = 5
3 + y = 5 (Substitute x with 3)
3 – 3 + y = 5 - 3 (Subtract 3 to both sides of the equation)
y = 2 (Simplify)
The solution is: (3,2)

BY ELIMINATION
3. Check the proposed solutions in both equations.
(3,2)
First equation: x +y = 5 3 + 2 = 5 5 = 5 TRUE
Second Equation: x – y = 1 3 – 2 = 1 1 = 1 TRUE
The solution is: (3,2)

ACTIVITY 2
BY ELIMINATION
Solve the following system of equations using the elimination method.
6x – 3y = –3 and 4x + 5y = –9
Multiply both sides of the first equation by 5 and the second by 3.
First equation,
5(6x – 3y) = 5(–3)
30x – 15y = –15 (use the distributive property)
Second equation,
3(4x + 5y) = 3(–9)
12x + 15y = –27 (use the distributive property)

BY ELIMINATION
Combine the two resulting equations
(eliminating the variable y).
30x – 15y = –15
12x + 15y = –27
42x = –42
x = –1 (divide both sides by 42)

BY ELIMINATION
Substitute the value for x into one of the original
equations.
6x – 3y = –3
6(–1) – 3y = –3 (replace the x value in the first equation)
–6 – 3y = –3 (simplify the left side)
–3y = –3 + 6 = 3 (add 6 to both sides and simplify)
y = –1 (divide both sides by –3)
Our computations have produced the point (–1, –1).

BY ELIMINATION
Check the point in the original equations.
First equation,
6x – 3y = –3
6(–1) – 3(–1) = –3 true
Second equation,
4x + 5y = –9
4(–1) + 5(–1) = –9 true
The solution of the system is (–1, –1).

LET’S PRACTICE!
Let’s
Check!
BY ELIMINATION
Solve the following system of equations using the elimination
method.
1. x – y = –3
3x + y = 19
2. 2x + y = 7
-2x + 3y = 5

Let’s Summarize
How to solve
system of
equations by
elimination?
BY ELIMINATION
1) Rewrite each equation in standard form,
eliminating fraction coefficients.
2) If necessary, multiply one or both equations
by a number so that the coefficients of a
chosen variable are opposites.
3) Add the equations.
4) Find the value of one variable by solving
equation from step 3.
5) Find the value of the second variable by
substituting the value found in step 4 into
either original equation.
6) Check the proposed solution in the original
equations

WORK ALONE
Let’s
Check!
BY ELIMINATION
Solve the following system of equations using the elimination
method.
1. x + y = 10
x - y = 8
2. 3x - y = 8
4x + y = 6

Reference:
Grade 8 Learner’s Manual pp.275-276
BY ELIMINATION

ACTIVITY 1
BACK
BY GRAPHING

Work Alone:
BACK
BY GRAPHING

TRY THESE!
1. x = 4
y = 2
(4,2)
2. x = 4
y = 3
(4,3)
BACK
BY SUBSTITUTION

WORK ALONE
1. x = 6
y = 7
The solution of the system is (6,7)
2. x = 8
y = 3
BACK
BY SUBSTITUTION

LET’S PRACTICE!
BACK
BY ELIMINATION
1. x = 4
y = 7
2. y = 3
x = 2

WORK ALONE
BACK
BY ELIMINATION
1. x = 9
y = 1
2. y = -2
x = 2
The solution of the system is (2,-2)

HAPPY TO LEARN!!!
I MATH!!!

20.Solving Systems of Linear Equation.pptx

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