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प्रश्न
Solve the following simultaneous equation.
2x - y = 5 ; 3x + 2y = 11
उत्तर
2x - y = 5 ...(I)
3x + 2y = 11 ...(II)
Multiplying (I) with 2 we get
4x - 2y = 10 ...(III)
Adding (II) with (III)
3x + 2y = 11
+ 4x - 2y = 10
7x = 21
⇒ x = 3
Putting the value of x in (I) we get
∴ 2x - y = 5
⇒ 2 × 3 - y = 5
⇒ 6 - y = 5
⇒ y = 1
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संबंधित प्रश्न
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 the following pair of linear equation by the elimination method and the substitution method:
3x + 4y = 10 and 2x – 2y = 2
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.
The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Solve the following simultaneous equation.
`x/3 + 5y = 13 ; 2x + y/2 = 19`
By equating coefficients of variables, solve the following equation.
x − 2y = −10 ; 3x − 5y = −12
If 52x + 65y = 183 and 65x + 52y = 168, then find x + y = ?
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`
The ratio of two numbers is 2:3. If 5 is added in each numbers, then the ratio becomes 5:7 find the numbers.
The ratio of two numbers is 2:3.
So, let the first number be 2x and the second number be `square`.
From the given condition,
`((2x) + square)/(square + square) = square/square`
`square (2x + square) = square (square + square)`
`square + square = square + square`
`square - square = square - square`
`- square = - square`
x = `square`
So, The first number = `2 xx square = square`
and, Second number = `3 xx square = square`
Hence, the two numbers are `square` and `square`
Read the following passage:
Two schools 'P' and 'Q' decided to award prizes to their students for two games of Hockey ₹ x per student and Cricket ₹ y per student. School 'P' decided to award a total of ₹ 9,500 for the two games to 5 and 4 Students respectively; while school 'Q' decided to award ₹ 7,370 for the two games to 4 and 3 students respectively. |
Based on the above information, answer the following questions:
- Represent the following information algebraically (in terms of x and y).
- (a) What is the prize amount for hockey?
OR
(b) Prize amount on which game is more and by how much? - What will be the total prize amount if there are 2 students each from two games?
