Algebra - Aptitude Questions and Answers

9^x * 3^y = 2187
(3²)^x . 3^y = 2187
3^{2x + y} = 3⁷
2x + y = 7 …...(i)

Again,
2^3x 2^2y - 4^xy = 0
2^3x * 2^2y = 4^xy
2^{3x +2y} = (2²)^xy
3x + 2y = 2xy …...(ii)
From Eqs. (i) and (ii)
3x + 2(7 - 2x ) = 2x(7 - 2x )
3x + 14 - 4x = 14x -4x²
4x² - 15x + 14 = 0
(x - 2)(4x - 7) = 0

Thus x = 2 or 7/4
y = 3 or 7/2

x + y = 5 or 21/4

x = √(a+b) - √(a-b) / √(a+b) + √(a-b)
x = (√(a+b) - √(a-b))*(√(a+b) - √(a-b)) / (√(a+b) + √(a-b)) * (√(a+b) - √(a-b))
x = (√(a+b) - √(a-b))² / (a+b) - (a-b)
x = (a+b) + (a-b) -2*√(a+b)*√(a-b) / 2b
x = 2*a -2*√(a² - b²) / 2b
x = a - √(a² - b²) / b
a - b*x= √(a² - b²)
After squaring both the sides
(a - bx )² = a² - b²
a² + b²x² - 2abx = a² - b²
b²x² - 2abx + b² = 0
bx² - 2ax + b = 0

Now, y = t ^{t / t-1} = ( t ^{1 / t-1})^t = x^t …..(i)
y / x = t ^{t / t-1} / t ^{1 / t-1}
= t ^{t / t-1 - 1 / t-1}
= t ^{t / t-1 - 1 / t-1}
= t ^{t - 1 / t - 1}
y / x = t …..(ii)
From Eqs. (i) and (ii), we get y = x ^y/x
=> y^x = x^y

x = 2 + 2^1/3 + 2^2/3 …..(i)
x - 2 = 2^{1/3 _*} (2^1/3 + 1)
After cubing both the sides
(x - 2)³ = 2* (2^1/3 + 1)³
x³ - 3*x² *2 + 3*x*2² - 2³ = 2*( (2^1/3)³ + 3*(2^1/3)²*1 + 3*2^1/3*1² + 1³)
x³ - 6x² + 12x - 8 = 2*(2 + 3*2^2/3 + 3*2^1/3 + 1)
x³ - 6x² + 12x - 8 = 2*(3 + 3*2^2/3 + 3*2^1/3)
x³ - 6x² + 12x - 8 = 6*(1+ 2^2/3 + 2^1/3)
From Eqs (i)
x³ - 6x² + 12x - 8 = 6*(1+ x - 2)
x³ - 6x² + 12x - 8 = 6(x - 1)
x³ - 6x² + 12x - 8 = 6x - 6
x³ - 6x² + 6x = 2

Let z=√(x/y) , then, z = 24/5 + 1/z
z = 24z + 5 / 5z
5z² - 24z - 5 = 0
5z² - 25z + z - 5 = 0
5z( z - 5) + 1( z - 5) = 0
( z - 5)(5z + 1) = 0
z = 5 or -1/5 that is √(x/y)= 5 or -1/5
Let consider
√(x/y)= 5
x/y= 25
x = 25y ....(i)

As is given x + y = 26
then 25y + y = 26 From Eqs (i)
26y = 26
y = 1
Hence x = 25
and xy = 25 * 1 = 25

As α αnd β αre the roots of the equqtion x2 + px + q = 0
therefore α + β = - p αnd αβ = q
Now,
α² + β² = (α + β)² - 2αβ
= (- p)² - 2q
= p² - 2q

As is given, a³ = 335 + b³ and a = 5 + b , thus
a³ - b³ = 335 …(i)
a - b = 5 …(ii)

As, (a - b)³ = a³ - b³ - 3ab(a - b)
From Eqs (i) and (ii)
5³ = 335 - 3ab(5)
125 = 335 - 15ab
ab = 14
Also, (a + b)² = (a - b)² + 4ab
= 5² + 4*14
= 25 + 56
= 81

Hence a + b = 9

linear equations a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
Has unique solution if a1/a2 ≠ b1/b2

for the euqation,
kx + 3y + 1 = 0
2x + y + 3 = 0
the unique solution,
k/2 ≠ 3/1
k ≠ 6

5^{1 + x} + 5^{1 - x} = 26
(5*5^x) + (5*5^{- x}) = 26
(5*5^x) + (5/5^x) = 26
5(5^x + 1/5^x) = 26

Let 5^x = y , then
5y² - 26y +5 = 0
5y² - 25y - y + 5 = 0
5y( y - 5) - 1( y - 5) = 0
( y - 5)(5y - 1) = 0
y = 5 , 1/5
that is , 5^x = 5 or 5^-1
therefore x = 1 , -1

a³ + b³ = (a + b)³ - 3ab(a + b)
= (5)³ - 3 * 6 * 5 = 125 - 90 = 35

(7 - 12x) - (3x - 7) = 14
7 - 12x - 3x + 7 = 14
- 15x + 14 = 14
- 15x = 0
x = 0

6x2 - 11x - 35 = 0
6x2 - 21x + 10x - 35 = 0
3x(2x - 7) + 5(2x - 7) = 0
(3x + 5)(2x - 7) = 0
x = -5/3, x = 7/2

the distance between the points, c² = (x_A − x_B)² + (y_A − y_B)²

(K - 4)² + (0 + 8)² = (10)²
K² + 16 - 8K + 64 = 100
K² - 8K - 20 = 0
K² - 10K + 2K - 20 = 0
K(K - 10) + 2(K - 10) = 0
(K + 2)(K - 10) = 0
K = - 2, K = 10

Let (2, 3) is (x1, y1) and (-4, 1) is (x2, y2)

The equation of a line passing through two points (x1, y1) and (x2, y2) is given by
y - y1 = m(x - x1) , m is the slope

there, m = (y2 - y1)/(x2 - x1) = (1-3)/(-4-2) = -2/-6 = 1/3

Then the equation is:
y - 3 = (1/3)*(x-2)
3y - 9 = x - 2
3y - x = 7
x - 3y = - 7

Let quadratic equation be
ax² + bx + c = 0
If a and b are roots, then
α + β = -b/a
and αβ = c/a

Since Aman made a mistake in writing down the constant term.
α + β = 4 + 3 = 7
and Alok made a mistake in writing down the coefficient of x.
αβ = 3 * 2 = 6
So, equation will be
x² - (α + β)x + αβ = 0
x² - 7x + 6 = 0
(x - 6)(x - 1) = 0
x = 6, 1

We have, 2x + 4y = 6 and 4x + 8y = 8

a1 = 2, b1 = 4, c1 = -6
and
a2 = 4, b2 = 8, c2 = -8
Now,
a1/a2 =2/4 =1/2
b1/b2 =4/8 =1/2
c1/c2 =-6/-8 =3/4

When there is no solution, the equations are called inconsistent. This happens, when the lines are parallel.

Here, a1/a2 = b1/b2 = 1/2 ≠ c1/c2

Hence, system of equation is inconsistent.

Given, 3x+2y=12 , xy=6

(3x+2y)² =12²
(3x)² + (2y)² + 2 (3x) (2y)=144
9x² +4y² +12xy=144
9x² +4y² +(12×6)=144
9x² +4y² +72=144
9x² +4y² =144 - 72
9x² +4y² =72