Linear Equation, Inconsistence

Inconsistance

Consider the two equations.

X + Y = 4
2X + 2Y = 7 or X + Y = 3.5

The second equation establishes a different relationship between X and Y as compared to the first equation. So they are said to be Inconsistance. Such a set of two equations do not have any solution.
So to get a unique solution, the number of equations should be equal to or more than no. of variables, these equations should be independent to each other and should be consistent with each other.
Let us take an example of set of equations that follow all the above restrictions.
x + 2y = 4 and 2x + 3y = 6
If you solve these two equations, we get the values of x = 0 and y = 2.
Graphically,

Fro graph also, the intersection points of two lines is, (0, 2) which is the unique solution.

Corollary: (For two variable linear equations)
For Unique Solution, the two lines would intersect each other at one point. If the two lines are parallel to each other we have no solution and if the two lines coincide, we shall have infinite solutions.

Number of solutions:
We can find the possible number of solutions without solving the algebraic equations or drawing the graphs. If the general representation of linear equation is,
ax + by = M
and cx + dy = N
where a and c are coefficient of x, b and d are coefficients of y and M and N are constants, then for Unique solution, a/c  b/d
for No solution, a/c = b/d M/N
for infinite solutions, a/c = b/d = M/N

Methods to solve the Linear Equations

1. Substitution Method

In this method, we find from one of the equations the value of one variable in terms of the other and substitute it in the second equation and thus find the value of the other variable.

Example: Solve x + y = 5… (i.)
3x + 2y = 12… (ii.)

Sol: We have x+ y = 5 or x = 5-y
Substituting in 2nd equation
3(5-y) + 2y = 12
15-y = 12
Or y = 3
x = 5-y = 5-3 = 2
Thus required solution is x = 2, y = 3

2. Elimination Method

The principle of this method consists of multiplying the coefficients of the equations by suitable numbers such that the coefficients of the one variable may become the some in both the equations. Then, by adding or subtracting, we get the value of one of the variables.
Then by substituting this value of this variable in either of the equation, we find the value of the other variable or unknown quantity.

Example: Solve 2x + 3y = 12
5x + 7y = 29
Sol: We have 2x + 3y = 12….... (i)
5x + 7y = 29….... (ii)
Here we see that if we multiply (i.) by 5 and (ii.) by 2 the coefficient of x in both equations would become the same
5(2x + 3y = 12)
Or 10x + 15y = 60.....… (iii)
2(5x + 7y = 29)
Or 10x + 14y = 58…..... (iv)
Subtracting (iv) from (iii)
15y – 14y = 60 – 58
Or y = 2
Substituting value of y in (1)
2x + 3 x 2 = 12
Or 2x = 12 – 6
Or 2x = 6
Or x = 3
Hence solution is x = 3, y = 2

Example: What is the solution of the following simultaneous equations?
x + y + z = 6, x + 2y + 3z = 14 and x + 3y + z = 10
Sol: x + y + z = 6 … (i)
x + 2y + 3z = 14 … (ii)
x + 3y + z = 10 … (iii)
From (i), we get z = 6 – x – y
Substitute it in (ii) and (iii).
We have from (ii)
x + 2y + 18 – 3x –3y = 14 or 2x + y = 4
Similarly, from (iii)
x + 3y + 6 – x – y = 10 or 2y = 4 or y = 2
Solving, we get y = 2, x = 1, z = 3.

Example: Find for what value of K would there be a unique solution for the given set of equations. 2x – 3y = 1 and Kx + 5y = 7
Sol: If two equations ax + by = M and cx + dy = N have a unique solution, then a/c  b/d. So in the above problem,
2/K-3/5
Solving, K –10/3.

Example: Find the value of K for which there is no solution for the given set of equations.
2x – Ky = -3 and
3x + 2y = 1
Sol: If two equations ax + by = M and cx + dy = N have no solution, then
a/c = b/d M/N
So,
2/3=-k/2-3
Solving k = -(4/3) or K-3
So, k = -(4/3)

Example: Find the value of K for which there are infinite solutions for the given set of equations. 5x + 2y = K and 10x + 4y = 3
Sol: If two equations ax + by = M and cx + dy =N have infinite
solutions, then a/c = b/d = M/N
So, for the two sets of equation to have infinite solutions, we have
5/10 = 2/4 = k/3
Hence, k = 3/2

Example: If the numerator and the denominator of a certain fraction are both increased by 1 each, then the fraction will be equal to ½. Instead, if the numerator and the denominator are increased by 4, fraction equal to 2/3. Find the fraction.

Sol: Let the fraction be x/y where x is the numerator and y is the denominator. When both numerator and denominator are increased by 1 each, we have
(x + 1)/(y + 1) = ½
2(x + 1) = y + 1 => 2x – y = -1 ……(1)
When both numerator and denominator are increased by 4 each, we have
(x + 4)/(y + 4) = 2/3
3(x + 4) = 2(y +4) => 3x – 2y = -4 …….(2)
Solving the two equations (1) and (2), we get x = 2 and y = 5. Hence the fraction is 2/5

Example: Find the values of x and y following equations: 6/(x + y) + 5/(x – y) = 7, 12/(x + y) – 3/(x – y) = 1

Sol: Substitute p = 1/(x + y) and q = 1(x – y) in the given equations. Then we get
6p + 5q = 7 -------- (1)
12 p – 3q =1 --------- (2)
Now these are two equations in two unknowns that we are familiar with.
Solving these two equations, we get p = 1/3 and q = 1
Since p = 1/(x + y) and q = 1/(x – y) we get 1/(x + y) = 1/3 and 1/(x – y) = 1
x + y = 3 and x – y = 1. By solving these two equations,
we get x = 2 and y = 1

Example: The sum of the two digits of a two-digit number is 8 and if the digits are reversed, the resulting number is 36 lower than the original number. Find the original
number.

Sol: Let the two digits of the given number be x and y such that then given number is 10x + y.
Since the sum of the digits is 8, we have x + y = 8 ------ (1)
When the digits are reversed, y becomes the tens digit and x, the units digit.
The number then becomes (10y + x)
So (10x+y) - (10y + x) = 36
or, 9x – 9y = 36 or x – y = 4
Solving equations (1) and (2) we get x = 6 and y = 2 and thus the number is 62.

Example: Four years from now, father’s age will be four times son’s age. Nine years from now, father’s age will be three times son’s age. When are their present ages?

Sol: Let f be father’s present age and s be the son’s present age.
Four years from now, father will be f + 4 and son will be s + 4 years old.
f + 4 = 4(s + 4)
f – 4s = 12 ------(1)
Nine years from now, father will be f + 9 and son will be s + 9 years old.
f + 9 = 3(s + 9) => f – 3s = 18 ----- (2)
We now have two equations in two unknowns. Solving these two equations, we get f = 36 and s = 6
Father’s age is 36 years and son’s age is 6 years.

Inequalities

We just learnt about equations. Now instead of equating variables with a constant, if we use < or > sign to have a range of solutions, we are dealing with inequalities.
So, If x = 4, then x can take only one value.
But if x > 4, then x can take infinite vales all of which are greater than 4.
All these conditions that deal with > or < situations come under the category of
Inequalities.
Graphs are used extensively to solve questions on inequalities.

Basic rules of inequalities

1. If X > Y, then –X < - Y.
2. If X> Y and A > B, then X + A > Y + B, but it is not necessary that X – A > Y – B.
3. If x/a > y/b, then bx > ay, only if a, b > 0. This is an important rule and you may often have to use it in solving data sufficiency problems.

Example: Solve the following in-equations:
(i) 0 < -x/3 < 1 (ii) 6≤ -3(2x – 4) < 12
(iii) -3 ≤ 4 – 7x < 18 (iv) -2 < 1 -3x < 7
(v) -7 < 2x – 3 < 7 (vi) -12 < 3x – 5 ≤ - 4.
Sol: (i) We have, 0 < -x/3 < 1 => 0 > x > -3
-3 < x < 0
(Sign of inequality reverses on multiplying with negative number).
Hence, the interval (-3, 0) is the solution set of the given system of inequations.

(ii) We have, 6≤ -3(2x – 4) < 12
6/-3 ≥ -3(2x – 4)/-3 > 12/ -3
-2 ≥ 2x – 4 > - 4
-2 + 4 ≥ 2x – 4 + 4 > - 4 + 4
2 ≥ 2x > 0
1 ≥ x > 0
0 < x ≤ 1
Hence, the interval (0, 1) is the solution set of the given system of inequations.

(iii) We have, -3 ≤ 4 – 7x < 18
-3 -4 ≤ 4 – 7x – 4 < 18 – 4
-7 ≤ - 7x < 14
-7/-7 ≥ -7x/-7 > 14/-7 => 1 ≥ x > -2
-2 < x ≤ 1
Hence, the interval (-2, 1) is the solution set of the given system of inequations.

(iv) We have, -2 < 1 – 3x < 7
-2 -1 < 1 – 3x – 1 , 7 -1
-3 < - 3x < 6
-3/-3 > -3x/-3 > 6/-3
1 > x > -2
-2 < x < 1
Hence, the interval (-2, 1) is the solution set of the given system of inequations.

(v) We have, -7 < 2x – 3< 7
-7 + 3 < 2x – 3 + 3 < 7 + 3
-4 < x < 10
-4/2 < 2x/2 < 10/2
-2 < x < 5
Hence, the interval (-2, 5) is the solution set of the given system of inequations.

(vi) We have, -12 < 3x – 5 ≤ - 4
-12 + 5 < 3x – 5 + 5 ≤ -4 + 5
-7 < 3x ≤ 1
-7/3 < x ≤ 1/3
Hence, the interval (-7/3, 1/3) is the solution set of the given system of inequations.

Example: Draw the graph of the solution set of the inequations 2x + 3y  6,
x + 4y 4, x 0 and y 0.
Sol: First we draw the graph of 2x + 3y 6.
Consider the line 2x + 3y = 6.
Now, 2x + 3y = 6 => x/3 + y/2 = 1.
This line meets the axes at A (3,0) and B(0,2). Plot these points on a
graph paper and join them by hick line AB. Consider the point (0,0).
Clearly, it does not lie on 2x + 3y = 6.
Clearly, (0,0) does not satisfy 2x + 3y =6.
Therefore, the solution set of 2x + 3y =6 is represented by the line
AB and the part of the plane separated by AB, not containing (0,0).
Next, we draw the graph of x + 4y =4.
Consider the line x + 4y = 4.
Now, x + 4y = 4 => x/4 + y/1 = 1.
This line meets the axes at C (4,0) and D(0,1).
Plot these points on the same graph paper as above and join them by
thick line CD.
Consider the point (0,0). It does not lie on the line x + 4y = 4. Clearly, (0,0) satisfies the inequation x + 4y  4.
Therefore, the solution set of x + 4y =4 is line CD and that part of
the plane separated by CD which contains (0,0).
Clearly, x = 0 consists of y-axis and the plane on the right side of y-axis. And, y = 0 consists of x-axis and the plane above x-axis.

Hence, the shaded region represents the solution set of the given system of inequations.
Note: Just a small note. The above question can never be asked in CAT. Nobody can expect you to draw a graph and solve. So why did we mentioned it. Just for your understanding in general. So understand what is equality and that is it.