Advertisements
Advertisements
प्रश्न
If 7x – 15y = 4x + y, find the value of x : y. Hence, use componendo and dividend to find the values of:
`(3x^2 + 2y^2)/(3x^2 - 2y^2)`
उत्तर
7x – 15y = 4x + y
7x – 4x = y + 15y
3x = 16y
`x/y = 16/3`
`=> x^2/y^2 = 256/9`
`=> (3x^2)/(2y^2) = (768)/18 = 128/3` ...`("Multiplying both sides by" 3/2)`
`=> (3x^2 + 2y^2)/(3x^2 - 2y^2) = (128 + 3)/(128 - 3)` ...(Applying componendo and dividendo)
`=> (3x^2 + 2y^2)/(3x^2 - 2y^2) = 131/125`
APPEARS IN
संबंधित प्रश्न
If a : b = c : d, prove that: xa + yb : xc + yd = b : d.
If `(5x + 6y)/(5u + 6v) = (5x - 6y)/(5u - 6v)`; then prove that x : y = u : v.
If `(5x + 7y)/(5u + 7v) = (5x - 7y)/(5u - 7v)`, show that `x/y = u/v`
If x = `(4sqrt(6))/(sqrt(2) + sqrt(3)` find the value of `(x + 2sqrt(2))/(x - 2sqrt(2)) + (x + 2sqrt(3))/(x - 2sqrt(3)`
Solve x : `(sqrt(36x + 1) + 6sqrt(x))/(sqrt(36x + 1) -6sqrt(x))` = 9
Find x from the following equations : `(sqrt(12x + 1) + sqrt(2x - 3))/(sqrt(12x + 1) - sqrt(2x - 3)) = (3)/(2)`
Solve `(1 + x + x^2)/(1 - x + x^2) = (62(1 +x))/(63(1 + x)`
Given that `(a^3 + 3ab^2)/(b^3 + 3a^2b) = (63)/(62)`. Using componendo and dividendo find a: b.
If `(x + y)/(ax + by) = (y + z)/(ay + bz) = (z + x)/(az + bx)`, prove that each of these ratio is equal to `(2)/(a + b)` unless x + y + z = 0
If x = `"pab"/(a + b)`, provee that `(x + pa)/(x - pa) - (x + pb)/(x - pb) = (2(a^2 - b^2))/(ab)`