Module 1
Lesson 1
Review on Linear Equations
1.1 Linear Equations in One Variable
A linear equation in one variable, say x, is in the form
ax + b = 0
where a and b are constants and a ≠ 0.
Since a and b are referred to as constants, the variable x is the unknown.
Examples:
Find the solution set of each of the following:
1) 5x + 2 = 3(x + 2)
5x + 2 = 3x + 6 by multiplication
5x – 3x = 6 – 2
by transportation
2x
2
=
4
2
dividing both
sides by 2
Check:
If x = 2, substitute to the given equation
5x + 2 = 3(x + 2)
5(2) + 2 = 3(2 + 2)
10 + 2 = 3(4)
12 = 12
Therefore, the solution set S.S = 2 satisfies the equation.
2𝑦
2)
3
2𝑦
𝑦
+ =7
2
𝑦
6 ( 3 + 2 = 7)
4𝑦 + 3𝑦 = 42
7𝑦 = 42
𝑦=6
Check:
Substitute y = 6 in the equation.
2𝑦
𝑦
+ =7
3
2(6)
3
2
6
+2=7
7
12
3
+3= 7
4+3=7
7=7
Therefore, the solution set S.S = 6 satisfies the equation.
1.2 Linear Equations in Two Variables
An equation reducible to the form
ax + by + c = 0
where a, b, and c are constants, a, b ≠ 0 is called a linear equation in x and y. this is also
known as the general form of an equation. The graph of a linear equation is a straight line
that is, the set of points satisfying the equation always lie on that line. It also follows that
a linear equation in two variables has no unique solution.
Graph of a Linear Equation in Two Variables.
To find the set of points that will satisfy a given linear equation in two variables, it is best
to use the graphical presentation. There are two ways in which equations can be solved
by the graphical method, namely:
1. By assigning values to x and solving for y or vice-versa
2. By finding the x-intercept and y-intercept of the given equation. (This is done by
setting one of the variables to zero and solving for the other variable.)
Examples:
Graph the following linear equations:
1) x + 2y = 6
Solution:
By assigning values to x and solving for y
A
B
C
D
x
0
1
2
6
y
3
2.5
2
0
4
3
2
1
0
0
2
4
6
8
Alternative Solution:
By solving for the x- and y-intercepts
Let x = 0
x + 2y = 6
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0 + 2y = 6
y=3
Therefore, y-intercept is the ordered pair (0,3).
Let y = 0
x + 2y = 6
x + 2(0) = 6
x =6
Therefore, x-intercept is the ordered pair (6,0).
Then plotting the x- and y-intercepts on a rectangular coordinate plane, we have
4
3
2
1
0
0
2
4
6
8
2) 2x + 3y = 12
x-intercept (6,0)
y-intercept (0,4)
5
4
3
2
1
0
0
2
4
6
8
1.3 System of Linear Equations
Many problems in business and economics lead to what is called a system of linear
equations. A system of linear equations in two variables x and y consists of two equations
of the type
a1x + b1y = c1
a2x + b2y = c2
where a1, b1, c1, a2, b2, and c2 are six given constants. The solution of the system defined by
the two equations given is the set of values of x and y that satisfies both equations.
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There are three different ways in finding the common solution for the system of equations
in two variables:
1. Elimination by addition or subtraction
2. Substitution
3. Graphical method
A. Elimination by Addition or Subtraction
Example 1:
Find the common solution of
x+y=6
eq. 1
x–y=4
eq. 2
using addition:
x+y= 6
x–y= 4
2x
2
x
=
=
10
2
5
dividing both
sides by 2
Substitute x = 5 to either eq. 1 or eq. 2.
Using eq. 1
x+y=6
5+y=6
y=6–5
y=1
Therefore, the solution set that will satisfy both equations is the ordered pair (5,1). To
check if this is correct, substitute to both equations.
x+y=6
x–y=4
5+1=6
5–1=4
6=6
4=4
Using subtraction:
3𝑥 + 𝑦 = 9
5𝑥 + 4𝑦 = 22
10
12𝑥 + 4𝑦 = 36
−5𝑥 − 4𝑦 = −22
7𝑥 = 14
𝑥=2
Then, substitute x = 2 to either eq. 1 or eq. 2
Using eq. 2
5𝑥 + 4𝑦 = 22
5(2) + 4𝑦 = 22
10 + 4𝑦 = 22
4𝑦 = 22 − 10
4𝑦 = 12
𝑦=3
Therefore, the solution set is the ordered pair (2, 3). To check if this is correct, substitute
to both equations.
3𝑥 + 𝑦 = 9
3(2) + 3 = 9
6+3=9
9=9
B. By Substitution
Example 2:
Find the common solution of
2x + y = 6
eq. 1
2x + 3y = 12
eq. 2
Using Substitution:
Solve for x in terms of y using eq. 1 or eq. 2
Using eq. 1
11
eq. 2
2𝑥 + 3𝑦 = 12
2(
6−𝑦
2
) + 3𝑦 = 12
6 − 𝑦 + 3𝑦 = 12
−𝑦 + 3𝑦 = 12 − 6
2𝑦 = 6
𝑦=3
if y = 3, substitute to either eq. 1 or eq. 2
Using eq. 1
2𝑥 + 𝑦 = 6
2𝑥 + 3 = 6
2𝑥 = 6 − 3
2𝑥 = 3
𝑥=
3
2
Therefore, the solution set is the ordered pair (3/2, 3)
Checking:
Eq.1
Eq. 2
2x + y = 6
2x + 3y = 12
2(3/2) + 3 = 6
2(3/2) + 3 (3) = 12
3+3=6
3 + 9 = 12
6=6
12 = 12
C. By Graphical Method
Example 3:
Solve the following system of linear equations using graphical method:
x–y=3
2x + y = 6
L1
L2
1st solution:
x–y=3
x
y
L1
A
1
-2
B
3
0
C
5
2
12
2x + y = 6
L2
E
2
2
x
y
F
1
4
G
3
0
Then, plot these points on the rectangular coordinate plane.
5
4
3
2
1
0
-1 0
-2
-3
2
4
6
Therefore, the common point of intersection is at pt. B (3,0), which is the solution to the
system of linear equations.
S.S = (3,0)
2nd Solution:
By finding the x-intercept and the y-intercept
eq. 1
eq. 2
x–y=3
2x + y = 6
x-intercept
(3,0)
(3,0)
y-intercept
(0,-3)
(0,6)
Solving for the x-intercept in eq. 1
If
y=0
x–y=3
x–0=3
x=3
therefore, the x-intercept is the ordered pair (3,0)
Solving for the y-intercept in eq. 1
If
x=0
x–y=3
0–y=3
y = -3
Therefore, the y-intercept is the ordered pair (0, -3).
Thus, in solving for the x-intercept or the y-intercept on any given linear equation simply
assign zero to one of the given variables then, solve for the other variable.
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After finding the values of the x- and y-intercepts of the given equation, the next step is
to plot them on the rectangular coordinate plane.
Thus we have
8
6
4
2
0
-2
0
1
2
3
4
-4
The point of intersection of the two lines is at (3,0), therefore, the solution set to the system
S.S. = (3,0).
Note:
If the solution set cannot be visualized clearly using the graphical method, check using
any of the other two methods.
----------------------------------------------------Exercise 1.1
A. Find the solution set for each of the following:
1) 3(x – 4) = 2x + 5
2)
6+
y
3
=
2(1 + y)
3) 2y + 2 = 4(2 – y)
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B. Graph the solution set for each of the following:
1) 3x + 4y = 24
2) 2x + 5y = 10
3) 3x (2) = 4y – 12
Exercise 1.2
1) 3x + y = 6
x–y=6
Use Elimination by Addition
2) 2x + y = -7
x–y=4
Use Elimination by Subtraction
3) 2x – 2y = 24
x– y=2
Use Substitution
4) 3x + y = 5
5x + 3y = 55
Use the Graphical Method
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