Algebra For SBI PO : Set – 33

(b)

x² – 2x – 15 = 0

or,x² – 5x + 3x – 15 = 0

or, x(x – 5) + 3(x – 5) = 0

or,(x – 5) (x + 3) = 0

x = 5, -3

II. y 2 + 5y + 6 = 0

or, y 2 + 3y + 2y + 6 = 0

or, y(y + 3) + 2(y + 3) = 0

or,(y + 3)(y + 2) = 0

y = -3, -2

No relation

(e)

I . x²  – x – 12 = 0

or, x²  – 4x + 3x – 12 = 0

or, x(x – 4) + 3(x – 4) = 0

or, (x – 4) (x + 3) = 0

x = 4, -3

II . y 2 -3y + 2 = 0

or, y 2 – 2y – y + 2 = 0

or, y(y – 2) – 1 (y – 2) = 0

or, (y – 2)(y – 1) = 0

y = 2, 1

Hence, no relation can be established

(b)

x – √169 = 0

Or x = √169

x = 13

II.  y² – 169 = 0

or, y² = 169

y = ±13

Hence, x ≥ y

(c)

x² – 32 = 112

or, x² = 112 + 32 = 144

or, x = 144

x = ±12

II. y – 256 = 0

or, y = 256

y = 16

Hence, x < y

(e)

x² – 25 = 0

or, x² = 25

or, x = 25

x = ±5

II. y² – 9y + 20 = 0

or, y² – 5y – 4y + 20 = 0

or, y(y – 5) – 4(y – 5) = 0

or, (y – 5) (y – 4) = 0

y = 5, 4

Hence, no relation can be established.

(c)

3x + 5y = 69 … (i)

9x + 4y = 108 … (ii)

x + z = 12 … (iii)

Now, from (i) and (ii), we have

3x + 5y = 69 … (i) × 4

9x + 4y = 108 … (ii) × 5

12x + 20y = 276

45x + 20y = 540

– 33x = – 264

On subtracting, we get

or, 33x = 264

x=264/3=8

Putting the value of x in equation (i), we get 3 × 8

+ 5y = 69

or, 5y = 69 – 24 = 45

∴y=45/5=9

Again, putting the value of x in equation (iii),

we get

x + z = 12

or, z = 12 – 8 = 4

Hence, x < y > z

y = √(9^(3×1/3) ×9^(4×1/4) )= √(9 ×9 )……..(I)

2x + 5z = 54 .. (ii)

III. 6x + 4z = 74

or, 3x + 2z = 37 … (iii)

From equation (ii) × 2 – (iii) × 5, we get

4x + 10z = 108

15x + 10z = 185

– 11x = – 77

or, 11x = 77

x = 7

Putting the value of x in equation (ii), we get

2 × 7 + 5z = 54

or, 5z = 40

z = 8

Hence, x < y > z

(b) 2x + 3y + 4z = 66 … (i)

2x + y + 3z = 42 … (ii)

3x + 2y + 4z = 63 … (iii)

From (iii) and (i),

x – y = – 3 …(iv)

From equation (i) × 3 – equation (ii) × 4

6x + 9y + 12z = 198

8x + 4y + 12z = 168

2x + 5y = 30 … (v)

Solving equation (iv) and (v), we get

x = 5, y = 8

Now, on putting the value of x and y in equation

(i),

10 + 24 + 4z = 66

or, 4z = 32

Hence, x < y = z

(a)

(x + z)³ = 1728 = 12³

or, x + z = 12 …(i)

2x + 3y = 35 … (ii)

III. x – z = 2 …(iii)

Now, equation (i) and (ii),

x = 7, z = 5

Putting the value x in question (ii) we have,

2 × 7 + 3y = 35

or, 3y = 35 – 14 = 21

or, y = 25/5=5

Hence, x = y > z

(b)

4x + 5y = 37 … (i)

x + z = 8 … (ii)

7x + 3y = 36 … (iii)

From equation (i) and (iii),

4x + 5y = 37 … (i) × 3

7x + 3y = 36 … (ii) × 5

or, 12x + 15y = 111

35x + 15y = 180

-23x = – 69

x = 3

Putting the value of x in equation (i)

4 × 3 + 5y = 37

or, y = 25/5 =5

Now, putting the value of x in equation (ii)

z = 5. Hence, x < y = z