# Formula - Algebraic Expression & Inequalities

VARIABLE

The unknown quantities used in any equation are known as variables. Generally, they are denoted by the last English alphabet x, y, z etc.
An equation is a statement of equality of two algebraic expressions, which involve one or more unknown quantities, called the variables.

LINEAR EQUATION

An equation in which the highest power of variables is one, is called a linear equation. These equations are called linear because the graph of such equations on the x–y cartesian plane is a straight line.

Linear Equation in one variable : A linear equation which contains only one variable is called linear equation in one variable.The general form of such equations is ax + b =c, where a, band care constants and a ≠ 0.
All the values of x which satisfy this equation are called its solution(s).

NOTE
An equation satisfied by all values of the variable is called an identity. For example : 2x + x = 3x.
Example 1

1. Solve 2x – 5 = 1
Sol.
2x – 5 = 1
=> 2x = 1 + 5
=> 2x = 6
=> x = 3.

Example 2

2. Solve 7x – 5 = 4x + 11
Sol.
7x – 5 = 4x + 11
=> 7x – 4x = 11 + 5 (Bringing like terms together)
=> 3x = 16 => x = 16/3 = 5(1/3).

Example 3

3. Solve $\mathbf{\frac{4}{x} - \frac{3}{2x}}$ = 5
Sol.
$\frac{4}{x} - \frac{3}{2x}$ = 5
=> 5/2x = 5 => 10x = 5
=> x = 5/10 = 1/2.

APPLICATIONS OF LINEAR EQUATIONS WITH ONE VARIABLES
Example 4

4. The sum of the digits of a two digit number is 16. If the number formed by reversing the digits is less than the original number by 18. Find the original number.

Sol. Let unit digit be x.
Then tens digit= 16 – x
∴ Original number= 10 × (16 – x) + x
= 160 – 9x.
On reversing the digits, we have x at the tens place and(16 – x) at the unit place.
∴ New number = 10x + (16 – x) = 9x + 16
Original number – New number = 18
(160 – 9x) – (9x + 16) = 18
160 – 18x – 16 = 18
– 18x + 144 = 18
– 18x = 18 – 144
=> 18x = 126
=> x =7
∴ In the original number, we have unit digit = 7
Tens digit =(16 –7) = 9
Thus, original number = 97

Example 5

5. The denominator of a rational number is greater than its numerator by 4. If 4 is subtracted from the numerator and 2 is added to its denominator,the new number becomes 1/6. Find the original number.

$\therefore \frac{x - 4}{x + 4 + 2} = \frac{1}{6}$

=> $\frac{x - 4}{x + 6} = \frac{1}{6}$

=> 6(x – 4) = x + 6
=> 6x – 24 = x + 6 => 5x = 30
∴ x = 6
Thus, Numerator = 6, Denominator = 6+ 4 = 10.
Hence the original number = 6/10.

Example 6

6. A man covers a distance of 33 km in 1(3/2) hours; partly on foot at the rate of 4 km/hr and partly on bicycle at the rate of 10 km/hr. Find the distance covered on foot.

Sol. Let the distance covered on foot be x km.

∴ Distance covered on bicycle = (33 – x) km

∴ Time taken on foot = $\frac{Distance}{Speed} = \frac{x}{4} \; hr$.

∴ Time taken on bicycle = $\frac{30 - x}{10} \; hr$

The total time taken = 7/2 hr.

$\frac{x}{4} + \frac{33 - x}{10} = \frac{7}{2}$

=> $\frac{5x + 66 - 2x}{20} = \frac{7}{2}$

6x +132 = 140

6x = 140 – 132

6x = 8

x = 1.33 km.
The distance covered on foot is 1.33 km.

Linear equation in two variables

General equation of a linear equation in two variables is ax + by + c= 0, where a, b ≠ 0 and c is a constant, and x and y are the two variables.
The sets of values of x and y satisfying any equation are called its solution(s).
Consider the equation 2x + y = 4. Now, if we substitute x = –2 in the equation, we obtain 2.(–2) + y = 4 or – 4 + y = 4 or y = 8. Hence (–2, 8) is a solution. If we substitute x = 3 in the equation, we obtain 2.3 + y = 4 or 6 + y = 4 or y = –2 Hence (3, –2) is a solution. The following table lists six possible values for x and the corresponding values for y,i.e.six solutions of the equation.

SYSTEMS OF LINEAR EQUATION

Consistent System :
A system (of 2 or 3 or more equations taken together) of linear equations is said to be consistent,if it has atleast one solution.

Inconsistent System:
A system of simultaneous linear equations is said to be inconsistent,if it has no solutions at all.
e.g. X + Y = 9; 3X + 3Y = 8
Clearly there are no values of X & Y which simultaneously satisfy the given equations. So the system is inconsistent.

REMEMBER
The system a1x + b1y = c1 and a2x + b2y = c2 has :
=> a unique solution, if $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.
=> Infinitely many solutions, if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$.
=> No solution, if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$.
=> The homogeneous system a1x + b1y = 0 and a2x + b2y = 0 has the only solution x = y = 0 when $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$.
=> The homogeneous system a1x + b1y = 0 and a2x + b2y = 0 has a non-zero solution only when $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$, and in this case, the system has an infinite number of solutions.
Example 7

7. Find k for which the system 3x– y = 4,kx + y = 3 has a infinitely many solution.
Sol. The given system will have infinite solution,
if $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$ i.e.$\frac{3}{k} = \frac{-1}{1}$ or k = 3.

Example 8

8. Find k for which the system 6x– 2y = 3,kx – y = 2 has a unique solution.
Sol. The given system will have a unique solution,
$if \; \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ i.e. $\frac{6}{k} \neq \frac{-2}{-1} \; or \; k \neq 3$.

Example 9

9. What is the value of k for which the system x + 2y = 3, 5x + ky = –7 is inconsistent?
Sol. The given system will be inconsistent if $\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ i.e if $\frac{1}{5} = \frac{2}{k} \neq \frac{3}{-7}$ or k = 10.

Example 10

10. Find k such that the system 3x + 5y = 0,kx + 10y = 0 has a non-zero solution.
Sol. The given system has a non zero solution,
if $\frac{3}{k} = \frac{5}{10}$ or k = 6

An equation of the degree two of one variable is called quadratic equation.
General form : ax2 + bx + c = 0...........(1) where a, b and c are all real number and a ≠ 0.
For Example :
2x2 – 5x + 3 = 0; 2x2 – 5 = 0; x2 + 3x = 0
A quadratic equation gives two and only two values of the unknown variable and both these values are called the roots of the equation.The roots of the quadratic equation (1) can be evaluated using the following formula.

$x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

The above formula provides both the roots of the quadratic equation, which are generally denoted by α and β,

say $\alpha = \frac{-b + \sqrt{b^{2} - 4ac}}{2a} \; and \; \beta = \frac{-b - \sqrt{b^{2} - 4ac}}{2a}$

The expression inside the square root b2 - 4ac is called the DISCRIMINANT of the quadratic equation and denoted by D. Thus, Discriminant (D) = b2 – 4ac.

Example 11

11. Which of the following is a quadratic equation?

(a) x1/2 + 2x + 3 = 0

(b) (x - 1)(x + 4) = x2 + 1

(c) x4 - 3x + 5 = 0

(d) (2x + 1)(3x - 4) = 2x2 + 3

Sol.(d) Equations in options (a) and (c) are not quadratic equations as in (a) max. power of x is fractional and in (c), it is not 2 in any of the terms.

For option (b), (x – 1)(x +4) = x2 + 1

or x2 + 4x - x + 4 = x2 + 1

or 3x - 5 = 0

which is not a quadratic equations but a linear.

For option (d), (2x + 1)(3x – 4) = 2x2 + 3

or 6x2 - 8x + 3x - 4 = 2x2 + 3

or 4x2 - 5x - 7 = 0
which is clearly a quadratic equation.

Example 12

12. Solve 2x2 + 6 = 7x

Sol. 2x2 + 6 = 7x

=> 2x2 – 7x + 6 =0

=> 2x2 – 4x – 3x + 6 = 0

=> 2x(x – 2) – 3(x –2) = 0

=> (2x –3)(x – 2) =0

Either 2x – 3 = 0 or x – 2 = 0

=> 2x = 3 or x =2

=> x = 3/2 or x = 2

∴ Solutions or roots of given quadratic equation

2x2 + 6 = 7x are 3/2 and 2.

Example 13

13. Solve $\mathbf{x - \frac{1}{x} = 1\tfrac{1}{2}}$

Sol. $x - \frac{1}{x} = 1\tfrac{1}{2} \Rightarrow \frac{x^{2} - 1}{x} = \frac{3}{2}$

=> 2(x2 - 1) = 3x

=> 2x2 - 2 = 3x

=> 2x2 - 3x - 2 = 0

=> 2x2 - 4x + x - 2 = 0

=> 2x(x - 2) + 1(x - 2) = 0

=> (2x + 1)(x - 2) = 0

Either 2x + 1 = 0 or x - 2 = 0

=> 2x = -1 or x = 2

=> x = -1/2 or x = 2

∴ x = -1/2 , 2 are solutions.

Nature of Roots

The nature of roots of the equation depends upon the nature of its discriminant D.
1. If D < 0, then the roots are non-real complex, Such roots areal ways conjugate to one another. That is, if one root is p + iq then other is p – iq, q ≠ 0. 2. If D = 0, then the roots are real and equal. Each root of the equation becomes -b/2a Equal roots are referred as repeated roots or double roots also. 3. If D > 0 then the roots are real and unequal.

4. In particular,if a, b, c are rational number,D > 0 and D is a perfect square, then the roots of the equation are rational number and unequal.

5. If a, b, c, are rational number, D > 0 but D is not a perfect square, then the roots of the equation are irrational (surd).Surd roots are always conjugate to one another, that is if one root is p + √q then the other is p - √q , q>0.

6. If a = 1, b and c are integers, D > 0 and perfect square, then the roots of the equation are integers.

Sign of Roots

Let α,β are real roots of the quadratic equation

ax2 + bx + c = 0 that is D = b2 - 4ac ≥ 0. Then

1.Both the roots are positive if a and c have the same sign and the sign of b is opposite.

2.Both the roots are negative if a, b and c all have the same sign.

3.The Roots have opposite sign if sign of a and c are opposite.

4.The Roots are equal in magnitude and opposite in sign if b = 0 [that is its roots α and –α]

5.The roots are reciprocal if a = c. [that is the root are α and 1/α]

6.If c =0. then one root is zero.

7.If b = c = 0, then both the roots are zero.

8.If a = 0, then one root is infinite.

9.If a = b = 0, then both the roots are infinite.

10.If a = b = c = 0, then the equation becomes an identity

11.If a + b + c = 0 then one root is always unity and the other root is c/a, hence the root are rational provided a, b, c,are rational.

Example 14

14. The solutions of the equation $\mathbf{\sqrt{25 - x^{2}} = x - 1 \; are:}$

(a) x = 3 and x = 4

(b) x = 5 and x = 1

(c) x = – 3 and x = 4

(d) x = 4 and x = – 3

Sol. (d) $\sqrt{25 - x^{2}} = x - 1 \; are:$

or 25 - x2 = (x - 1)2 or 25 - x2 = x2 + 1 - 2x

or 2x2 - 2x - 24 = 0 or x2 - x - 12 = 0

or (x - 4)(x + 3) = 0 or x = 4,x = -3

Example 15

15. Which of the following equations has real roots ?

(a) 3x2 + 4x + 5 = 0

(b) x2 + x + 4 = 0

(c) (x - 1)(2x - 5) = 0

(d) 2x2 - 3x + 4 = 0

Sol.(c) Roots of a quadratic equation

ax2 + bx + c = 0 are real if b2 - 4ac ≥ 0

Let us work with options as follows.

Option (a) : 3x2 + 4x + 5 = 0

b2 - 4ac = 42 - 4(3)(5) = -44 < 0.

Thus, roots are not real.

(b) : x2 + x + 4 = 0

b2 - 4ac = 12 - 4(1)(4) = 1 - 16 = -15 < 0. Thus, roots are not real. (c) :(x –1) (2x – 5) = 0 => 22 - 7x + 5 = 0

b2 - 4ac = (-7)2 - 4(2)(5) = 49 - 40 = 9 > 0.

Thus roots are real.

or x = 1 and x = 5/2 > 0; Thus, equation has real roots.

(d) : 2x2 - 3x + 4 = 0

b2 - 4ac = (-3)2 - 4(2)(4) = 9 - 32 = -23 < 0.

Thus, roots are not real. Hence, option (c) is correct.

Example 16

16. If 2x2 - 7xy + 3y2 = 0, then the value of x:y is :

(a)3 : 2

(b)2 : 3

(c)3 : 1 or 1 :2

(d)5 : 6

Sol.(c) 2x2 - 7xy + 3y2 = 0

$2\left ( \frac{x}{y}\right )^{2} - 7\left ( \frac{x}{y} \right ) + 3 = 0$

$\frac{x}{y} = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{7 \pm \sqrt{49 - 24}}{2 \times 2} = \frac{7 \pm 5}{4} = 3,\frac{1}{2}$

=> $\frac{x}{y} = \frac{3}{1} \; or \; \frac{x}{y} = \frac{1}{2}$

Example 17

17. If a + b + c = 0 and a,b,c, are rational numbers then the roots of the equation
(b + c – a)x2 + (c + a – b) x + (a + b – c) = 0 are

(a) rational
(b) irrational
(c) non real
(d) none of these.

Sol.(a) The sum of coefficients
= (b + c - a) + (c + a - b) + (a + b - c) = a + b + c = 0 (given)
∴ x = 1 is a root of the equation
∴ The other root is $\frac{a + b - c}{b + c - a}$ , which is rational as a,b, c, are rational
Hence, both the roots are rational.

ALTERNATIVE :

D = (c + a - b)2 - 4(b + c - a)(a + b - c)

= (-2b)2 - 4(-2a)(-2c) = 4b2 - 16ac

= 4(a + c)2 - 16ac = 4[(a + c)2 - 4ac] = 2[(a - c)2]

D is a perfect square. Hence, the roots of the equation are rational.

Example 18

18. Both the roots of the equation

(x - b)(x - c) + (x - b)(x - a) + (x - a)(x - b) = 0

(a) dependent on a, b, c
(b) always non real
(c) always real
(d) rational

Sol.(c) The equation is

3x2 - 2(a + b + c)x + (bc + ca + ab) = 0

The discriminant

D = 4(a + b + c)2 - 4.3.(bc + ca + ab)

= 4[a2 + b2 + c2 - ab - bc - ca]

= 2[(a2 - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2)]

= 2[(a - b)2 + (b - c)2 + (c - a)2] ≥ 0

∴ Roots are always real.

Symmetric Functions of Roots

An expression in α, β is called a symmetric function of α, β if the function is not affected by interchanging α and β. If α , β are the roots of the quadratic equation ax2 + bx + c = 0 , a ≠ 0 then,

Sum of roots : α + β = -b/a = $-\frac{coefficient \; of \; x}{coefficient \; of \; x^{2}}$

and Product of roots : αβ = c/a = $\frac{constant \; term}{coefficient \; of \; x^{2}}$

Note
1.Above relations hold for any quadratic equation whether the coefficients are real or non-real complex.

2.With the help of above relations many other symmetric functions of α and β can be expressed in terms of the coefficients a, b and c.

3.Recurrence Relation
αn + βn = (α + β)(αn - 1 + βn - 1) - αβ(αn - 2 + βn - 2)

4.Some symmetric functions of roots are

(i) α2 + β2 = (α + β)2 - 2αβ

(ii) α - β = $\pm \sqrt{\left ( \alpha + \beta \right )^{2} - 4\alpha \beta }$

(iii) α2 - β2 = ±(α + β)(α - β) = ±(α + β)$\sqrt{\left ( \alpha + \beta \right )^{2} - 4\alpha \beta }$

(iv) α3 + β3 = (α + β)3 - 3αβ(α + β)

(v) α3 - β3 = (α - β)3 + 3αβ(α - β) &
α - β = ±$\sqrt{\left ( \alpha + \beta \right )^{2} - 4\alpha \beta }$

(vi) α4 + β4 = (α2 - β2)2 - 2α2β2
= [(α + β)2 - 2αβ]2 - 2(αβ)2

(vii) α4 - β4 = (α2 - β2)(α2 + β2)
= [(α + β)2] - 2αβ [$\pm \sqrt{\left ( \alpha + \beta \right )^{2} - 4\alpha \beta }$

FORMATION OF QUADRATIC EQUATION WITH GIVEN ROOTS

=> An equation whose roots are α and β can be written as (x – α) (x – β) = 0 or x2 – (α + β) x + αβ = 0 or x2 – (sum of the roots) x + product of the roots = 0.

=> Further if α and β are the roots of a quadratic equation ax2 + bx + c = 0, then
ax2 + bx + c = a(x – α) (x – β) is an identity.
A number of relations between the roots can be derived using this identity by substituting suitable values of x real or imaginary.

Condition of a Common Root between two quadratic equations

a1x2 + b1x + c1 = 0 ......(i)

and a2x2 + b2x + c2 = 0 ......(ii)

Let α be a common root of the two equations

Then, a1α2 + b1α + c1 = 0 and a2α2 + b2α + c2 = 0
On solving we get

Which gives the common root as well as the condition for common root.

Condition that two quadratic equations have both the Roots Common

Suppose that the equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have both the roots common.

then $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

If the coefficients of two quadratic equations are rational (real) and they have one irrational (imaginary) root common then they must have both the roots common as such roots occur in conjugate pair.

Example 19

19. Of the following quadratic equations, which is the one whose roots are 2 and – 15 ?

(a) x2 - 2x + 15 = 0
(b) x2 + 15x - 2 = 0
(c) x2 + 13x - 30 = 0
(d) x2 - 30 = 0

Sol.
(c) Sum of roots = 2 – 15 = – 13
Product of roots = 2 × (– 15) = – 30
Required equation
= x2 - x(sum of roots) + product of roots = 0

=> x2 + 13x - 30 = 0

Example 20

20. If a and b are the roots of the equation x2 - 6x + 6 = 0, then the value of a2 + b2 is:

(a) 36
(b) 24
(c) 17
(d) 6

Sol.(b) The sum of roots = a+ b = 6
Product of roots = ab = 6
Now, a2 + b2 = (a + b)2 - 2ab = 36 - 12 = 24

Example 21

21. If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then the value of $\mathbf{\frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha }}$ is:

(a) $\frac{3bc - a^{3}}{b^{2}c}$

(b) $\frac{3bc - b^{3}}{a^{2}c}$

(c) $\frac{3abc - b^{2}}{a^{3}c}$

(d) $\frac{ab - b^{2}c}{2b^{2}c}$

Sol.(b) Here, α + β = -b/a and αβ = c/a

Thus, $\frac{\alpha ^{2}}{\beta } + \frac{\beta ^{2}}{\alpha} = \frac{\alpha ^{3} + \beta ^{3}}{\alpha \beta }$

= $\frac{(\alpha + \beta )(\alpha ^{2} - \alpha \beta + \beta ^{2})}{\alpha \beta }$ ....(i)

Now, (α2 + β2 - αβ) = [(α + β)2 - 2αβ - αβ]

= [(α + β)2 - 3αβ]

Hence (i) becomes

$\Rightarrow \frac{\left ( \alpha + \beta \right )[\left ( \alpha + \beta \right )^{2} - 3\alpha \beta ]}{\alpha \beta } = \frac{\frac{-b}{a}[\frac{b^{2}}{a^{2}} - \frac{3c}{a}]}{\frac{c}{a}}$

= $\frac{-b}{c}\left [ \frac{b^{2} - 3ac}{a^{2}} \right ] = \frac{3abc - b^{3}}{a^{2}c}$

Example 22

22. If a, b are the two roots of a quadratic equation such that a + b = 24 and a – b = 8, then the quadratic equation having a and b as its roots is :

(a) x2 + 2x + 8 = 0
(b) x2 - 4x + 8 = 0
(c) x2 - 24x + 128 = 0
(d) 2x2 + 8x + 9 = 0

Sol.(c) a + b = 24 and a – b = 8

=> a = 16 and b = 8 => ab = 16 × 8 = 128

A quadratic equation with roots a and b is

x2 - (a + b)x + ab = 0 or x2 - 24x + 128 = 0

Inequations

A statement or equation which states that one thing is not equal to another, is called an inequation.

Symbols

‘<’ means “is less than” ‘>’ means “is greater than”
‘≤’ means “is less than or equal to”
‘≥’ means “is greater than or equal to”
For example :
(a) x < 3 means x is less than 3.
(b) y ≥ 9 means y is greater than or equal to 9.

PROPERTIES

1. Adding the same number to each side of an equation does not effect the sign of inequality, it remains same, i.e. if x > y then, x + a > y + a.

2. Subtracting the same number to each side of an inequation does not effect the sign of inequaltiy, i.e., if x < y then,x – a < y – a. 3. Multiplying each side of an inequality with same number does not effect the sign of inequality,i.e., if x ≤ y then ax ≤ ay (where, a> 0).

4. Multiplying each side of an inequality with a negative number effects the sign of inequality or sign of inequality reverses, i.e., if x < y then ax > ay (where a < 0). 5. Dividing each side of an inequation by a positive number does not effect the sign of inequality,i.e., if x ≤ y then $\frac{x}{a} \leq \frac{y}{a}$ (where a > 0)

6. Dividing each side of an inequation by a negative number reverses the sign of inequality,i.e., if x > y then $\frac{x}{a} < \frac{y}{a}$ (where a < 0)

REMEMBER
If a > b and a, b, n are positive, then an > bn but a-n < b-n.
For example 5 > 4; then 53 > 43 or 125 > 64, but

5-3 < 4-3 or 1/125 < 1/64. => If a > b and c > d, then (a + c) > (b + d).

=> If a > b > 0 and c > d > 0, then ac > bd.

=> If the signs of all the terms of an inequality are changed,then the sign of the inequality will also be reversed.

MODULUS :

1. If a is positive real number,x and y be the fixed real numbers,then
(i) |x – y | < a <=> y – a < x < y + a

(ii) | x – y| ≤ a <=> y – a ≤ x ≤ y + a

(iii) |x –y | > a <=> x > y + a or x < y – a

(iv) | x – y| ≥ a <=> x ≥ y + a or x ≤ y – a

2.Triangle inequality :

(i) | x + y| ≤ |x| + |y|, ∀ x , y ∈ R

(ii) | x - y| ≥ |x| - |y|, ∀ x , y ∈ R

Example 23

23. If a – 8 = b, then determine the value of | a - b| - | b - a|.

(a) 16
(b) 0
(c) 4
(d) 2

Sol.(b) | a - b| = |8| = 8
=> | b - a| = |-8| = 8
=> | a - b| - | b - a| = 8 - 8 = 0.

Example 24

24. Solve : 3x + 4 ≤ 19, x ∈ N

Sol. 3x + 4 ≤ 19

3x + 4 - 4 ≤ 19 - 4 [Subtracting 4 from both sides]

3x ≤ 15

3x/3 ≤ 15/3

x ≤ 5; x ∈ N

∴ x = {1, 2, 3, 4, 5}

Example 25

25. Solve 5 ≤ 2x - 1 ≤ 11

Sol. 5 ≤ 2x - 1 ≤ 11

5 + 1 ≤ 2x - 1 + 1 ≤ 11 + 1 [Adding 1 to each sides]

6 ≤ 2x ≤ 12

$\frac{6}{2} \leq \frac{2x}{2} \leq \frac{12}{2}$ [divided each side by 2]

3 ≤ x ≤ 6

=> x = {3, 4, 5, 6}.

Example 26

26. The solution set of
x - 2y ≥ 0; 2x - y ≤ 2; x ≥ 0; y ≥ 0 is :

(a) Empty
(b) Bounded
(c) Neither empty nor bounded
(d) None of these

Sol.(a) Plotting the given inequations, we get the following graph :

There is no common region. Hence, the solution set is empty.

Example 27

27. If x ≥ 0, y ≥ 0 and (x + y) ≤ 1, then the maximum value of (2x + 3y) is

(a) 2
(b) 3
(c) 4
(d) 5

Sol.(b) It is given that x ≥ 0, y ≥ 0 and (x + y) ≤ 1.

x + y ≤ 1 => 2(x + y) ≤ 2 => 2x + 2y ≤ 2.

=> 2x + 2y + y ≤ 2 + y

=> 2x + 3y ≤ 2 + 1 = 3. (since y ≤ 1).

Example 28

28. If $\mathbf{x^{2} + x + \frac{1}{x^{2}} + \frac{1}{x} < 0}$ then which of the following is true?

(a) x + 1/x > -2
(b) x + 1/x < -2
(c) x + 1/x < 1
(d) Both (a) and (c)

Sol.(d) Given that $\left ( x^{2} + \frac{1}{x^{2}} \right ) + \left ( x + \frac{1}{x} \right ) < 0$ $\left ( x + \frac{1}{x} \right )^{2} + \left ( x + \frac{1}{x} \right ) -2 < 0$

Substituting x + 1/x = y, we get

y2 + y - 2 < 0 => (y – 1) (y +2) < 0

∴ either y – 1 < 0; y + 2 >0

or y + 2 < 0; y –1 > 0

i.e.,y < 1, y > – 2 or y < –2; [ y > 1 ] <= this is not possible

Therefore, –2 < y < 1

i.e. -2 < |x + 1/x| < 1.

APPLICATIONS

Formulation of Equations/Expressions : A formula is an equation, which represents the relations between two or more quantities.
For example :
Area of parallelogram (A) is equal to the product of its base (b) and height (h), which is given by
A = b × h
or A = bh.
Perimeter of triangle (P),
P = a + b + c, where a, b and c are three sides.

Example 29

29. Form the expression for each of the following:
(a) 5 less than a number is 7.

(b) Monika’s salary is 1500 less than thrice the salary of Surbhi.

Sol. (a) Expression is given by
x – 5 = 7, where x is any number

(b) Let the salary of Surbhi be Rs. x and salary of Monika be Rs. y.
Now, according to the question y = 3x – 1500

More Applications of Equations : Problems on Ages can be solved by linear equations in one variable, linear equations in two variables,and quadratic equations.

Example 30

30. Kareem is three times as old as his son. After ten years, the sum of their ages will be 76 years. Find their present ages.

Sol. Let the present age of Kareem’s son be x years.

Then, Kareem’s age = 3x years

After 10 years, Kareem’s age = 3x + 10 years

and Kareem’s son’s age = x + 10 years
∴ (3x + 10) + (x + 10) = 76

=> 4x = 56 => x = 14

∴ Kareem’s present age = 3x = 3 × 14 = 42 years

Kareem’s son’s age = x = 14 years.

Example 31

31. The present ages of Vikas and Vishal are in the ratio 15 : 8. After ten years, their ages will be in the ratio 5 : 3. Find their present ages.

Sol. Let the present ages of Vikas and Vishal be 15x years and 8x years.
After 10 years,
Vikas’s age = 15x + 10 and Vishal’s age = 8x + 10

$\therefore \frac{15x + 10}{8x + 10} = \frac{5}{3}$

=> 3(15x + 10) = 5(8x + 10)

=> 45x + 30 = 40x + 50

=> 5x = 20 => x = 20/5 = 4

∴ Present age of Vikas = 15x = 15 × 4 = 60 years
Present age of Vishal = 8x = 8 × 4 = 32 years.

Example 32

32. Father’s age is 4 less than five times the age of his son and the product of their ages is 288. Find the father’sage.

Sol. Let the son’s age be x years.

So, father’s age = 5x – 4 years.

∴ x(5x – 4) = 288

=> 5x2 – 4x – 288 = 0 => 5x2 – 40x + 36x – 288 = 0

=> 5x (x– 8) + 36 (x– 8) = 0

=> (5x + 36)(x – 8) = 0

Either x – 8 = 0 or 5x + 36 = 0 => x = 8 or x = -36/5

x cannot be negative; therefore, x = 8 is the solution.

∴ Son’s age = 8 years and Father’s age = 5x – 4 = 36 years.

If present age of the father is F times the age of his son. Tyears hence, the father’s age become Z times the age of son then present age of his son is given by $\frac{(Z - 1)T}{\left ( F - Z \right )}$.

Example 33

33. Present age of the father is 9 times the age of his son. One year later, father’s age become 7 times the age of his son. What are the present ages of the father and his son.

Sol. By the formula

Son’s age = $\frac{(7 - 1)}{\left ( 9 - 7 \right )} \times 1 = \frac{6}{2} \times 1 = 3 \; years$

So, father’s age = 9 × son’s age = 9 × 3 = 27 years.

If T1 years earlier the age of the father was n times the age of his son, T2 years hence, the age of the father becomes m times the age of his son then his son’s age is given by

Son’s age = $\frac{T_{2}(n - 1) + T_{1}(m - 1)}{n - m}$

Example 34

34. 10 years ago, Shakti’s mother was 4 times older than her. After 10 years, the mother will be twice older than the daughter. What is the present age of Shakti?

Sol. By using formula,

Shakti’s age = $\frac{10(4 - 1) + 10(2 - 1)}{4 - 2} = 20 \; years$

Present age of Father : Son = a : b

After /Before T years = m : n

Then son’s age = b × $\frac{T(m - n)}{an - bm}$

and Father’s age = a × $\frac{T(m - n)}{an - bm}$

Example 35

35. The ratio of the ages of the father and the son at present is 3 : 1. Four years earliar, the ratio was 4: 1. What are the present ages of the son and the father?

Sol. Ratio of present age of Father and Son = 3 : 1

4 years before = 4 : 1

Son’s age = $1 \times \frac{4(4 - 1)}{4 \times 1 - 3 \times 1} = 12 \; years.$

Father’s age = $3 \times \frac{4(4 - 1)}{4 \times 1 - 3 \times 1} = 36 \; years.$