Advertisements
Advertisements
प्रश्न
Solve the following pairs of equations:
`(3)/(5) x - (2)/(3) y + 1` = 0
`(1)/(3) y + (2)/(5) x ` = 4
Advertisements
उत्तर
`(3)/(5) x - (2)/(3) y + 1` = 0
⇒ 9x - 10y + 15 = 0
⇒ 9x - 10y = -15 ....(i)
`(1)/(3)y + (2)/(5)x` = 4
⇒ 5y + 6x = 60
⇒ 6x + 5y = 60 ....(ii)
Multiplying eqn. (ii) by 2, we get
12x + 10y = 120 ....(iii)
Adding eqns. (i) and (iii), we get
21x = 105
⇒ x = 5
Substituting the value of x in eqn. (ii), we get
6(5) + 5y = 60
⇒ 30 + 5y = 60
⇒ 5y = 30
⇒ y = 6
Thus, the solution set is (5,6).
APPEARS IN
संबंधित प्रश्न
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
3x - y = 23
`x/3 + y/4 = 4`
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :
`[ x - y ]/6 = 2( 4 - x )`
2x + y = 3( x - 4 )
If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.
Solve for x and y :
`[ y + 7 ]/5 = [ 2y - x ]/4 + 3x - 5`
`[ 7 - 5x ]/2 + [ 3 - 4y ]/6 = 5y - 18`
`(3)/x - (2)/y` = 0 and `(2)/x + (5)/y` = 19, Hence, find a if y = ax + 3.
The length of a rectangle is twice its width. If its perimeter is 30 units, find its dimensions.
If 1 is added to the denominator of a fraction, the fraction becomes `(1)/(2)`. If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.
Anil and Sunita have incomes in the ratio 3 : 5. If they spend in the ratio 1 : 3, each saves T 5000. Find the income of each.
Two mobiles S1 and S2 are sold for Rs. 10,490 making 4% profit on S1 and 6% on S2. If the two mobiles are sold for Rs.10,510, a profit of 6% is made on S1 and 4% on S2. Find the cost price of both the mobiles.
A and B can build a wall in `6(2)/(3)` days. If A's one day work is `1(1)/(4)` of one day work of B, find in 4 how many days A and B alone can build the wall.
