Advertisements
Advertisements
Question
Solve the following simultaneous equation.
`x/3 + 5y = 13 ; 2x + y/2 = 19`
Solution
`x/3 + 5y = 13` ...(I)
x + 15y = 13 × 3
∴ x + 15y = 39 ...(II)
`2x + y/2 = 19` ...(III)
`(4x+y)/2 = 19`
4x + y = 19 × 2
∴ 4x + y = 38 ...(IV)
Multiplying (II) with 4 we get,
4x + 60y = 156 ...(V)
Subtracting (IV) from (V),
4x + 60y = 156
4x + y = 38
- - -
59y = 118
⇒ y = 2
Putting the value of y in (IV) we get,
∴ 4x + y = 38
⇒ 4x + 2 = 38
⇒ 4x = 36
⇒ x = 9
APPEARS IN
RELATED QUESTIONS
Solve the following system of equations by using the method of elimination by equating the co-efficients.
`\frac { x }{ y } + \frac { 2y }{ 5 } + 2 = 10; \frac { 2x }{ 7 } – \frac { 5 }{ 2 } + 1 = 9`
Solve (a – b) x + (a + b) y = `a^2 – 2ab – b^2 (a + b) (x + y) = a^2 + b^2`
Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:
The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Form the pair of linear equation in the following problem, and find its solution (if they exist) by the elimination method:
Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.
In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than twice number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.
Solve the following simultaneous equation.
x - 2y = -1 ; 2x - y = 7
Solve the following simultaneous equation.
x − 2y = −2 ; x + 2y = 10
By equating coefficients of variables, solve the following equation.
5x + 7y = 17 ; 3x - 2y = 4
The sum of the two-digit number and the number obtained by interchanging the digits is 132. The digit in the ten’s place is 2 more than the digit in the unit’s place. Complete the activity to find the original number.
Activity: Let the digit in the unit’s place be y and the digit in the ten’s place be x.
∴ The number = 10x + y
∴ The number obtained by interchanging the digits = `square`
∴ The sum of the number and the number obtained by interchanging the digits = 132
∴ 10x + y + 10y + x = `square`
∴ x + y = `square` .....(i)
By second condition,
Digit in the ten’s place = digit in the unit’s place + 2
∴ x – y = 2 ......(ii)
Solving equations (i) and (ii)
∴ x = `square`, y = `square`
Ans: The original number = `square`
Solve: 99x + 101y = 499, 101x + 99y = 501
