Advertisements
Advertisements
प्रश्न
A pair of linear equations which has a unique solution x = 2, y = –3 is ______.
पर्याय
x + y = –1, 2x – 3y = –5
2x + 5y = –11, 4x + 10y = –22
2x – y = 1, 3x + 2y = 0
x – 4y –14 = 0, 5x – y – 13 = 0
उत्तर
A pair of linear equations which has a unique solution x = 2, y = –3 is 2x + 5y = –11, 4x + 10y = –22.
Explanation:
L.H.S. = 2x + 5y
= 2(2) + 5(–3)
= 4 – 15
= –11
= R.H.S.
and L.H.S. = 4x + 10y
= 4(2) + 10(–3)
= 8 – 30
= –22
= R.H.S.
Since x = 2, y = –3 satisfy the equation.
∴ x = 2, y = –3 is a unique solution of these equations.
APPEARS IN
संबंधित प्रश्न
Find the value of k for which each of the following system of equations have no solution :
3x - 4y + 7 = 0
kx + 3y - 5 = 0
Find the value of k for which each of the following system of equations have no solution :
2x - ky + 3 = 0
3x + 2y - 1 = 0
Solve for x and y:
0.3x + 0.5y = 0.5, 0.5x + 0.7y = 0.74
Solve for x and y:
6(ax + by) = 3a + 2b,
6(bx – ay) = 3b – 2a
For what value of k, the system of equations
kx + 2y = 5,
3x - 4y = 10
has (i) a unique solution, (ii) no solution?
Find the value of k for which the system of equations
3x - y = 5, 6x - 2y = k
has no solution
Find the value of k for which the system of equations
kx + 3y + 3 - k = 0, 12x + ky - k = 0
has no solution.
The monthly incomes of A and B are in the ratio of 5:4 and their monthly expenditures are in the ratio of 7:5. If each saves Rs. 9000 per month, find the monthly income of each.
Find the value of k for which the system of equations 2x + 3y -5 = 0 and 4x + ky – 10 = 0 has infinite number of solutions.
The pair of equations x = a and y = b graphically represents lines which are ______.