Advertisements
Advertisements
प्रश्न
Solve the following pairs of equations by reducing them to a pair of linear equations
`(7x-2y)/(xy) = 5`
`(8x + 7y)/(xy) = 15`
उत्तर
`(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
संबंधित प्रश्न
Solve the following pairs of equations by reducing them to a pair of linear equations
`4/x + 3y = 14`
`3/x - 4y = 23`
Formulate the following problems as a pair of equations, and hence find their solutions:
2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Formulate the following problems as a pair of equations, and hence find their solutions:
Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solve the following pair of linear equations.
152x − 378y = − 74
− 378x + 152y = − 604
Solve the following for x:
`1/(2a+b+2x)=1/(2a)+1/b+1/(2x)`
Find the value of following determinant.
`|(-1,7),(2,4)|`
A two-digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Let the numerator and denominator of the fraction be x and y respectively. Then the fraction is `x/y`
If the numerator is multiplied by 2 and the denominator is reduced by 5, the fraction becomes `6/5`. Thus, we have
`(2x)/(y-5)=6/5`
`⇒ 10x=6(y-5)`
`⇒ 10x=6y-30`
`⇒ 10x-6y+30 =0`
`⇒ 2(5x-3y+15)=0`
`⇒ 5x - 3y+15=0`
If the denominator is doubled and the numerator is increased by 8, the fraction becomes `2/5`. Thus, we have
`(x+8)/(2y)=2/5`
`⇒ 5(x+8)=4y`
`⇒ 5x+40=4y`
`⇒ 5x-4y+40=0`
So, we have two equations
`5x-3y+15=0`
`5x-4y+40=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/((-3)xx40-(-4)xx15)=-y/(5xx40-5xx15)=1/(5xx(-4)-5xx(-3))`
`⇒ x/(-120+60)=(-y)/(200-75)=1/(-20+15)`
`⇒x/(-60)=-y/125``=1/-5`
`⇒ x= 60/5,y=125/5`
`⇒ x=12,y=25`
Hence, the fraction is `12/25`
The sum of a numerator and denominator of a fraction is 18. If the denominator is increased by 2, the fraction reduces to 1/3. Find the fraction.
Ten years ago, a father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be then. Find their present ages.