Hey friends! Welcome to another descriptive session of Grade XI math, in previous article you guys have learned important terms about linear equations and inequalities of Algebra 1 unit of this syllabus but today in this article we are including two initial topics of Algebra ii that are related to linear inequalities. These topics are Solving Absolute Value Equations and Systems of linear equations/inequalities. Nothing much needs to tell about inequalities because that was very well explained in previous article. So directly starts with the main topics of today’s article.
Apart form starting anything we should now What is absolute value function? with Absolute value to solve equations and inequalities, let us first discuss the main definition of Absolute value. The term Absolute value is used when the definite numerical value is important rather than its sign. For example if you say that ‘x’ is 20 m far away from the original point then this causes the distance value as 20, when absolute value of ‘x’ is defined then it does not matter that in which direction it is being traveled.
In general if x = 20 or x = -20 then its absolute value is represented as │x│= 20. Absolute value inequalities are solved by using simple method that suggests to replace the inequality symbol with = sign. Once the replacement is done then solve the equation to find the critical numbers and then divide them into intervals and test to check the right one. Suppose an equation is │x - 2│< 3
So first replace it with = sign; │x - 2│=3
Now by evaluating the above equation student will have two values, x= 5 or x = -1 these two values cause two intervals, first interval includes values less than -1, second interval includes value between -1 and 5, and the third interval have values more than 5. Now test these interval values by putting them into inequality equation, it easily gets clear that only second interval value is satisfying the equation as:
When x lies in x interval then value of x can be 0, so by putting this in equation:
│x - 2│< 3│0 - 2│< 3
2<3 true
So the region of x is as -1 < x < 5
For solving less than sign inequalities, there is a short trick as:When Absolute value inequalities is in the following form
│Algebraic Expression│< K; here K is a positive real number. Then first rewrite the equation as:
-K <│Algebraic Expression│< K
The reason behind this short-trick preference is that whenever absolute value of quantity is less than a positive real number that time the Absolute quantity is situated in between negative and positive values of the real integer line. For example if │x│<3 then as an absolute value principle the value of x must be in 3 units but it can be of both directions so we can write this one as -3 < │x│<3.
Now if the Absolute value inequality includes a greater than symbol instead of less than to solve this one also with ease, students can follow the procedure defined below;
If │Algebraic Expression│> K; where k is real integer
Then it can be rewritten as: │Algebraic Expression│< -K
Now student have two equations ahead of just one, so solve them both and the solution will comprise the union of both Absolute inequality equation’s solution.
For example: evaluate │2x + 1 │≥ 3
First rewrite the equation as │2x + 1 │≥ 3 and │2x + 1 │≤ -3
Solve both of the inequality equation as:
as │2x + 1 │≥ 3
2x ≥ 2
x ≥ 2/2
x ≥ 1
Second equation: │2x + 1 │≤ -3
2x ≤ -4
X ≤ -4/2
X ≤ -2
As said before, solution is the union of these both
x ≥ 1 OR X ≤ -2
This is how any Absolute Value inequality is being resolved by using the short-tricks. Now let us move to the next session of this article and that is System of Linear Equations. Fraction of this topic is introduced in previous article but here we will do the total demonstration of it. While going through system of equations one thing for sure that student should learn the fundamentals of matrix operations because the standard applications that are used to evaluate system of equation are based on matrix analysis.
Let’s start with the actual meaning of system of linear equation: A linear system is a finite collection of linear equations that includes same number of unknown variables in it. As a standard form, a linear system of m equations and n variables is represented as:
aXIx1 + a12x2 + ……. + a1nxn = b1
a21x1 + a22x2 + ……. + a2nxn = b2
…. … …. .
…. … …. .
…. … …. .
am1x1 + am2x2 + ……+ amnxn = bm
Tuple set ( s1, s2,…..,sn), when these values are substituted in place of x variable then linear system is called Solution set of the system. Every existing linear system can be put into one of the following category:
No solution, Unique solution or infinite solution. Linear system is said to be consistent if its solution at-least comprise one solution and if there is not a single solution existing then the system is said to be inconsistent.
Let us take an example of solving a linear system,
If the system is as:
x1 - x2 + x3 - x4 = 2
x1 - x2 + x3 + x4 = 0
4x1 - 4x2 + 4x3 = 4
-2x1 + 2x2 - 2x3 + x4 = -3
Relevant matrices of the above given linear system is as:
1 -1 1 -1 2
1 -1 1 1 0
4 -4 4 0 4
-2 2 -2 1 -3
Now apply the row operations to convert the matrix into more linear form
We are applying following row eliminations to convert apply the row elimination algorithm on the matrices;
1. R2 - R1
2. R3 - 4R1
3. R4 + 2R1
After applying these eliminations the resultant matrices is as :
1 -1 1 -1 2
0 0 0 2 -2
0 0 0 4 -4
0 0 0 -1 1
Still the matrix is not in unit form so further row transformations are required,
1. (1/2)R2
2. R3 -2R2
3. R4 + (1/2)R2
The resultant matrix is as
1 -1 1 -1 2
0 0 0 1 -1
0 0 0 0 0
0 0 0 0 0
One more transformation is applied as R1 + R2
(1) [-1] [1] 0 1
0 0 0 (1) -1
0 0 0 0 0
0 0 0 0 0
Now the above matrix results the following equivalent linear system.
x1 = 1 + x2 - x3
x4 = -1
In above equation x2 and x3 are free variables so they can take arbitrary constants as x2= c1 and x3 = c2. Now the system is as:
x1 = 1 + c1 - c2
x2 = c1
x3 = c2
x4 = -1
Consider c1=0, c2 = 0 then obtained solution is as
x=-1
0
0
-1
Now for the homogeneous linear system AX= 0
x1 - x2 +x3 - x4 = 0
x1 - x2 +x3 + x4 = 0
4x1 - 4x2 + 4x3 = 0
-2x1 +2x2 - 2x3 +x4 = 0
Now because we have its equivalent analyzed matrix, so
1 -1 1 -1 0 (1) [-1] [1] 0 0
1 -1 1 1 0 equivalent to0 0 0 (1) 0
4 -4 4 0 0 0 0 0 0 0
-2 2 -2 1 0 0 0 0 0 0
So these above system provides two solution set for the given linear equation system, that’s why this system is said to be a consistent system.
This is how any linear system is sorted out but there are several more ways in which any system of equation can be resolved and for that you can go through the next article of grade XI Algebra ii. In next article we will continue the scenario of Algebra ii including the rest of system of equations solution procedures.
At any stage if student feels some difficulty in solving any math query then he can immediately access the online math tutoring service where expert online math tutors are available for 24 x7 hours and student can interact with them through remote connection. This online education service is the good platform to go with because due to Internet platform all the relevant data to study is easily available and you can compete your analytical skills with other students globally by attending scheduled online math tests. To have regular lesson sessions, students can open their account on these websites after which they can access the variety of tools of these websites in their study to add more ease factor, some of these tools are online math calculator, worksheets and video aids.
In upcoming posts we will discuss about Polynomials in Grade XI and translations. Visit our website for information on Maharashtra state board books
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