Advertisements
Advertisements
प्रश्न
Solve the following simultaneous equations:
41x + 53y = 135
53x + 41y = 147
Advertisements
उत्तर
The given equations are
41x + 53y = 135 ....(i)
53x + 41y = 147 ....(ii)
Multiplying eqn. (i) by 53 and eqn. (ii) by 41, we get
2173x + 2809y = 7155 ....(iii)
2173x + 1681y = 6027 ....(iv)
Subtracting eqn. (iv) from eq. (iii), we get
1128y = 1128
⇒ y = 1
Substituting the value of y in eqn. (i), we get
41x + 53(1) = 135
⇒ 41x + 53 = 135
⇒ 41x = 135 - 53
⇒ 41x = 82
⇒ x = 2
Thus, the solution set is (2, 1).
APPEARS IN
संबंधित प्रश्न
If 10y = 7x - 4 and 12x + 18y = 1; find the values of 4x + 6y and 8y - x.
Solve the following simultaneous equations :
3(2u + v) = 7uv
3(u + 3v) = 11uv
Solve the following simultaneous equations:
13a - 11b = 70
11a - 13b = 74
Solve the following pairs of equations:
`(3)/(2x) + (2)/(3y)` = 5
`(5)/x - (3)/y` = 1
Solve the following pairs of equations:
`(3)/x - (1)/y` = -9
`(2)/x + (3)/y` = 5
If 10y = 7x - 4 and 12x + 18y = 1 ; find the value of 4x + 6y and 8y - x.
The length of a rectangle is twice its width. If its perimeter is 30 units, find its dimensions.
The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes `(1)/(2)`. Find the fraction.
A solution containing 12% alcohol is to be mixed with a solution containing 4% alcohol to make 20 gallons of solution containing 9% alcohol. How much of each solution should be used?
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.
