If (x + y - 8)/2 = (x + 2y - 14)/3 = (3x + y - 12)/11 then find the values of x and y, respectively.

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.

Given equations 2x + 4y = 6 and 4x + 8y = 6

then,
a₁/a₂ = 2/4 = 1/2;
b₁/b₂ = 4/8 = 1/2;
c₁/c₂ = 6/6 = 1
? a₁/b₂ = b₁/b₂ ? c₁/c₂

So there is no solution for these equations.

Given, 6x - 10y = 10 ..........(i)
and x/(x + y) = 5/7
? 7x = 5x + 5y
? 2x - 5y = 0 ...(ii)

On multiplying Eq. (ii) by 2 and subtracting from Ed.(i), we get
6x - 10y = 10
4x - 10y = 0
---------------------
2x = 10
? x = 5
Putting the value of x in Eq. (i), we get
30 - 10y = 10
? 10y = 20
? y = 2
? (x - y) = 5 - 2 = 3

Given equations of system
3x + y = 4 ...(i)
x + 2y = 8 ...(ii)
Here, a₁ = 3 , b₂ = 2 and c₂ = B
? a₁/a₂ = b₁/b₂ = c₁/c₂ = 1/2
So, the system of equations has infinite solutions, because it represents a parallel line.

Let cost of one saree and shirt be x and y, respectively.
2x + 4y = 16000 ....(i)
x + 6y = 16000 ....(ii)

On multiplying Eq. (ii) by 2 and subtracting from Eq. (i). we get
2x + 4y = 16000
2x + 12y = 32000
-------------------------
-8y = -16000
? y = 2000

Putting the value of y in Eq. (ii), we get
x +6 x 2000 = 16000
? x = 4000
? Cost of 12 shirts = 12y
= 12 x 2000 = ? 24000

[?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

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.

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

Given, 2^a + 3^b = 17
and 2^{a + 2} - 3^{b + 1} = 5
? 2² x 2² - 3^b x 3¹ = 5
? 4.2^a - 3.3^b = 5
Let 2^a = x amd 3^b = y
Then, x + y = 17 ...(i)
4x - 3y = 5 ...(ii)
On multiplying Eq. (i) by 3 and adding to Eq (ii), we get
3x + 3y = 51
4x - 3y = 5
------------------
7x = 56
? x = 8

On putting the value of x in Eq. (i), we get
8 + y = 17
? y = 9

Now, 2^a = x
? 2^a = 8 (2)³
? 3^b = y = 9
? 3^b = 3²
? b = 2
Hence, a = 3 and b = 2

(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)