Advertisements
Advertisements
प्रश्न
Solve the following pairs of equations:
`x/(3) + y/(4)` = 11
`(5x)/(6) - y/(3)` = -7
Advertisements
उत्तर
`x/(3) + y/(4)` = 11
⇒ 4x + 3y = 132 ....(i)
`(5x)/(6) - y/(3)` = -7
⇒ 5x - 2y = -42 ....(ii)
Multiplying eqn. (i) by 2 and eqn. (ii) by 3, we get
8x + 6y = 264 ....(iii)
15x - 6y = -126 ....(iv)
Adding eqns. (iii) and (iv), we get
23x = 138
⇒ x = 6
Substituting the value of x in eqn. (i), we get
4(6) + 3y = 132
⇒ 24 + 3y = 132
⇒ 3y = 108
⇒ y = 36
Thus, the solution set is (6,36).
APPEARS IN
संबंधित प्रश्न
For solving pair of equation, in this exercise use the method of elimination by equating coefficients:
y = 2x - 6; y = 0
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 )
Solve :
11(x - 5) + 10(y - 2) + 54 = 0
7(2x - 1) + 9(3y - 1) = 25
Solve :
`4x + [ x - y ]/8 = 17`
`2y + x - [ 5y + 2 ]/3 = 2`
Solve the following simultaneous equations :
2(3u - v) = 5uv
2(u + 3v) = 5uv
Solve the following simultaneous equations:
103a + 51b = 617
97a + 49b = 583
Solve the following pairs of equations:
`(2)/(x + 1) - (1)/(y - 1) = (1)/(2)`
`(1)/(x + 1) + (2)/(y - 1) = (5)/(2)`
Solve the following pairs of equations:
`(xy)/(x + y) = (6)/(5)`
`(xy)/(y - x)` = 6
Where x + y ≠ 0 and y - x ≠ 0
If 2 is added to the numerator and denominator it becomes `(9)/(10)` and if 3 is subtracted from the numerator and denominator it becomes `(4)/(5) `Find the fraction.
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.
