# Verbal Problems

Chapter -10

Equations

Linear equations in two variables

A linear system of two equations with two variables can be written in the form.

ax+by =c

cx +dy = q

Where any of the constants can be zero with the exception that each equation must have at least one variable in it

The x and y –coordinates gives the solutions of pair of simultaneous equations in two variables.

Solving simultaneous linear equations in two variables

Substitution method

The equations are:

x + y= 6……………….1

4x – 3y =-4…………………2

Equation 1 can be written as

y= 6 - x……………….3

4x – 3y = - 4

4x – 3*(6-x) = -4

4x - 18 + 3x = -4

7x = 14

X= 2

Y= 4

Elimination method

The equations are:

x + y= 6……………….1

4x – 3y = -4…………………2

Multiplying equation 1 by 4, we get

[x + y= 6 ]*4

4x +4y = 24 ……………3

Subtracting 1 from 3

4x +4y = 24

4x – 3y = - 4

-      +            +

7y = 28

Y= 4

X= 2

The quadratic equation in its standard form is ax2 + bx + c = 0

The value of x can be calculated as

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

Here, a, b, c are the real numbers such that a ≠ 0

Nature of roots

i. if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$> 0 ,it has two distinct and real roots

ii.if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$< 0 ,it has no real roots, roots are imaginary.

iii. if$\sqrt {{{\rm{b}}^2} - {\rm{\: }}4{\rm{ac\: }}}$= 0 it has only one rootorrepeatedroots

Examples 1

5 years ago a man’s age was 5 times the age of his daughter’s. After 3 years twice his age will be equal to 6 times his daughter‘s age. What are their present ages?

Let the age of the man be x and the age of his daughter be y.

Or, (x – 5) = 5(y – 5)

Or, x – 5 = 5y – 25

Or, x – 5y = -20

So, x = 5y – 20 ….(i)

After 3 years hence,

Or, 2(x + 3) = 6(y + 3)

or, x + 3 = 3y + 9…(ii)

Putting value of x in (ii),

or, 5y – 20 + 3 = 3y + 9

or, 2y – 17 = 9

or, 2y = 26

So, y = 13 years.

So, x = 5 * 13 – 20

= 45years.

Examples 2

The area of the rectangular room is 45sq. cm if the length had been 3m less and breadth 1m , more it would have been a square . Find the length and breadth of room.

Or, l*b = 45

So, l = $\frac{{45}}{{\rm{b}}}$….(i)

Or, l – 3 = b + 1

Or, l – b = 4….(ii)

Or, $\frac{{45}}{{\rm{b}}} - {\rm{b}} = 4$

Or, $\frac{{45 - {{\rm{b}}^2}}}{{\rm{b}}} = 4$

Or, 45 – b2 = 4b

Or, b2 + 4b – 45 = 0

Or, b2 + 9b – 5b – 45 = 0

Or, b(b – 5) + 9(b – 5) = 0

Or, (b – 5)(b + 9) = 0

So, b = 5, -9.

So, b = 5cm

Now,

Or, l – b = 4

Or, l – 5 = 4

So, l = 9cm

Examples 3

If the ram gives one of the marble from what he possesses to Sita then they will have equal numbers of marbles. If Sita gives one of the marbles from what she possesses to Ram then Ram will have double of the marbles with she is left with. Find the number of marbles possessed by each initially.

Soln

Number of marbles possessed by ram = x

Number of marbles possessed by sita = y

By question,

x-1 = y +1

x-1 –y -1=0

x- y=2………..i

Again,

x +1= 2(y -1)

x-2y = -3 ………..ii

Solving i and ii

x = 7  and y= 5

Go Top