If 2 ^a + 3 ^b = 17 and 2 ^{a + 2} - 3 ^{b +1} = 5, then

Let the cost of one chair be ? x
and cost of one table be ? y.
By given condition,
10x + 6y = 6200 ..(i)
and 3x + 2y = 1900
? 9x + 6y = 5700 ...(ii)

On subtracting Eq. (ii), we get
x = ? 500

From Eq (i),
5000 + 6y = 6200
? 6y = 1200
? y = ? 200

The cost of 4 chair and 5 tables
= 4x + 5y
= 2000 + 1000
= ? 3000

ax + by = c and dx + ey = f
a₁/a₂ = a/d, b₁/b₂ = b/e, c₁/c₂ = c/f
? b₁/b₂ ? c₁/c₂
? b/e ? c/f
it represent a pair of parallel lines.
? a₁/a₂ ? b₁/b₂
? a/d ? b/e
Therefore, system has unique solutions and represents a pair of intersecting lines.

[?3 + x + ?3 - x] / [?3 + x - ?3 - x] = 2

Let ?3 + x = a and ?3 - x = b
Then, (a + b) / (a - b) = 2/1
? a + b = 2a - 2b
? a = 3b

? ?3 + x = 3?3 - x

On squaring bothv sides, we get
(?3 + x)² = (3?3 - x)²
? 3 + x = 9(3 - x)
? 3 + x = 27 - 9x
? 10x = 24
? x = 12/5

Given,
(x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11
? (x + y - 8)/2 = (x + 2y - 14)/3
? 3x + 3y - 24 = 2x + 4y - 28
? 3x + 3y - 2x - 4y = -28 + 24
x - y = -4 ...(i)

Again, (x + 2y - 14)/3 = (3x + y - 12)/11
? 11x + 22y - 154 = 9x + 3y - 36
? 2x + 19y = 118 ..(ii)


On multiplying Eq. (i) by 2 and subtracting from Eq., we get
2x - 2y = -8
2x + 19y = 118
-------------------
-21y = -126
? y = 6

On putting the value of y in Eq. (i), we get
x - 6 = -4
? x = 2
? x = 2 and y = 6

Let the number of gold coins initially be x and the number of non-gold coins be y.
According to the question,
3x = y

When 10 more gold coins, total number of gold coins becomes x + 10 and the number of non-gold coins remain the same at y.
Now, we have 2(x + 10) = y

Solving these two equations, we get
x = 20 and y = 60.
Total number of coins in the collection at the end is equal to
x + 10 + y = 20 + 10 + 60 = 90.

(p/x) + (q/y) = m ..(i)
(q/x) + (p/y) = n ...(ii)

On multiply Eq (i) by q and Eq. (ii) by p and subtracting, we get
(pq/x) + (q²)/y = mq
(pq/x) + (p²)/y = np
-----------------------------------------
(q²/y) - (p²/y) = mq - np
? (q² - p²) = y(mq - np)
? y = (q² - p²)/mp - np = (p² - q²)/np - mq

Again, on multiplying Eq. (i) by p and Eq. (ii) by q and subtracting, we get
(p²/x) + (pq/y) = mp
(q²/x) + (pq/y) = nq
-----------------------------------------
(p²/x) - (q²/x) = mp - nq
? (p² - q²) = x (mp - nq)
? x = (p² - q²)/(mp - nq)
? x = (p² - q²)/(mp - nq)
and y = (p² - q²) / (np - mq)

Let the capital of one is x and that of another is y .

According to the question.
x + 100 = 2(y - 100)
x + 100 = 2y - 200
or x - 2y = -300 ...(i)

Again, according to the question
y + 10 = 6(x - 10)
? y + 10 = 6x - 10
? 6x - y = 70 ..(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i) , we get
x - 2y = -300
12x - 2y = 140
----------------
-11x = -440
? x = 40

On putting the value of x in Eq. (i) , we get
40 - 2y = -300
? 2y = 340
? y = 170
So, their capital are ? 40 and ? 170.

Let the incomes of two persons be 8x and 5x and their expenditure be 2y and y , respectively.
? Saving = Income - Expenditure
? 1000 = 8x - 2y ...(i)
and 1000 = 5x - y ...(ii)


On multiplying Eq. (ii) by 2 and subtracting from Eq. . (i) , we get
8x - 2y = 1000
10x - 2y = 2000
----------------------
-2x = -1000
? x = 500

? Monthly incomes are
8x = ? 4000 and
5x = ? 2500

? Difference = ? 4000 - 2500 = ? 1500

Let the fraction be x/y,
According to the question,
(x + 5) / (y + 5) = 7/8
? 8x + 40 = 7x + 35
? 8x - 7y = -5 ..(i)

Again, according to the question,
(x + 3)/(y + 3) = 6/7
? 7x + 21 = 6y + 18
? 7x - 6y = -3 ..(ii)


On multiplying Eq. (i) by 6 and Eq. (ii) by 7 and subtracting, we get
48x - 42y = -30
49x - 42y = -21
---------------------
-x = -9
? x = 9
On putting te value of x in Eq. (i) , we get
72 - 7y = -5
? -7y = -5 - 72
? y = -5 - 72
? y = (-77) / (-7) = 11
? Required fraction = 9/11

Method 1 to solve the given question.
Lets assume the unit's digit is y and the ten's digit is x. then, the number is 10 x + y.
According to question, after interchanging the digits, the new number is 10y + x.
Then,
New Number - 18 = Original number
10y + x = 10x + y + 18
? 9y - 9x = 18
? y - x = 2 .....................(1)
Again according to question,
Sum of digits of Original Number = 8
x + y = 8 ..............................(2)
Add the equation (1) and (2) , we will get
y - x + x + y = 2 + 8
? 2y = 10
? y = 5
Put the value of Y in equation (2) , we will get
x + 5 = 8
? x = 8 - 5 = 3
Then , the original number = 10x + y
Put the value of x and y and get the original number.
The original number = 10x + y = 10 x 3 + 5 = 30 + 5 = 35

Method 2 to solve the given question.
Given that sum of digits of original number is 8.
Let us assume that unit digit is x , then ten digit will be 8 - x. (since sum of digits is 8.)
Now Original number = 10( 8 - x) + x
After interchanging the digits , the number = 10x + (8 - x) = 9x + 8
According to question.
New number = original number + 18
? 9x + 8 = 10(8 - x) + x + 18
? 9x + 8 = 80 - 10x + x + 18
? 9x + 8 = 98 - 9x
? 9x + 9x = 98 - 8
? 18x = 90
? x = 90/18
? x = 5
The original number = 10(8 - x) + x = 10(8 - 5) + 5 = 10(3) + 5 = 30 + 5 = 35