Advertisements
Advertisements
Question
Solve the following pairs of equations by reducing them to a pair of linear equations
`(7x-2y)/(xy) = 5`
`(8x + 7y)/(xy) = 15`
Solution
`(7x-2y)/(xy) = 5`
`⇒ (7x)/(xy) - (2y)/(xy) = 5`
`⇒ 7/y - 2/x = 5 ... (i)`
`(8x+7y)/(xy) = 15`
`⇒ (8x)/(xy) + (7y)/(xy) = 15`
`⇒ 8/y + 7/x = 15 ... (ii)`
Putting `1/x = p ` in (i) and (ii) we get,
7q - 2p = 5 ... (iii)
8q + 7p = 15 ... (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q - 14p = 35 ... (v)
16q + 14p = 30 ... (vi)
Now, adding equation (v) and (vi) we get,
49q - 14p + 16q + 14p = 35 + 30
⇒ 65q = 65
⇒ q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
⇒ 7p = 7
⇒ p = 1
Now,
p = 1/x = 1
⇒ 1/x = 1
⇒ x = 1
also, q = 1 = 1/y
⇒ 1/y = 1
⇒ y = 1
Hence, x =1 and y = 1 is the solution
APPEARS IN
RELATED QUESTIONS
Solve the following pairs of equations by reducing them to a pair of linear equations
`10/(x+y) + 2/(x-y) = 4`
`15/(x+y) - 5/(x-y) = -2`
Solve the following pairs of equations by reducing them to a pair of linear equations
`1/(3x+y) + 1/(3x-y) = 3/4`
`1/(2(3x-y)) - 1/(2(3x-y)) = (-1)/8`
Solve the following pair of linear equations
ax + by = c
bx + ay = 1 + c
Solve the following pair of linear equations.
(a − b) x + (a + b) y = a2− 2ab − b2
(a + b) (x + y) = a2 + b2
The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If 3 is added to the denominator and 2 is subtracted from the numerator, the fraction becomes `1/4`. Thus, we have
`(x-2)/(y+3)=1/4`
`⇒ 4(x-2)=y+3`
`⇒ 4x-8=y+3`
`⇒ 4x-y-11=0`
If 6 is added to the numerator and the denominator is multiplied by 3, the fraction becomes `2/3`. Thus, we have
`(x+6)/(3y)=2/3`
`⇒ 3(x+6)=6y`
`⇒ 3x +18 =6y`
`⇒ 3x-6y+18=0`
`⇒ 3(x-2y+6)=0`
`⇒ x-3y+6=0`
Here x and y are unknowns. We have to solve the above equations for x and y.
By using cross-multiplication, we have
`x/((-1)xx6-(-2)xx(-11))=(-y)/(4xx6-1xx(-11))=1/(4xx(-2)-1xx(-1))`
`⇒ x/(-6-22)=-y/(24+11)=1/(-8+1)`
`⇒ x/-28=-y/35=1/-7`
`⇒ x= 28/7,y=35/7`
`⇒ x= 4,y=5`
Hence, the fraction is`4/5`
The present age of a father is three years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.
Father's age is three times the sum of age of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
Two years ago, a father was five times as old as his son. Two year later, his age will be 8 more than three times the age of the son. Find the present ages of father and son.
Asha has only ₹1 and ₹2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹75, then the number of ₹1 and ₹2 coins are, respectively.
