Advertisements
Advertisements
प्रश्न
Show that the following system of equations has a unique solution:
2x - 3y = 17,
4x + y = 13.
Also, find the solution of the given system of equations.
उत्तर
The given system of equations is:
2x - 3y - 17 = 0 ….(i)
4x + y - 13 = 0 …..(ii)
The given equations are of the form
`a_1x+b_1y+c_1 = 0 and a_2x+b_2y+c_2 = 0`
where, `a_1 = 2, b_1= -3, c_1= -17 and a_2 = 4, b_2 = 1, c_2= -13`
Now,
`(a_1)/(a_2) = 2/4 = 1/2 and (b_1)/(b_2) = (−3)/1 = -3`
Since, `(a_1)/(a_2) ≠ (b_1)/(b_2)`, therefore the system of equations has unique solution.
Using cross multiplication method, we have
`x/(b_1c_2− b_2c_1) = y/(c_1a_2− c_2a_1) = 1/(a_1b_2− a_2b_1)`
`⇒ x/(−3(−13)−1×(−17)) = y/(−17 ×4−(−13)×2) = 1/(2 ×1−4×(−3))`
`⇒ x/(39+17) = y/(−68+26) = 1/(2+12)`
`⇒ x/56 = y/(−42) = 1/14`
`⇒ x = 56/14, y = (−42)/14`
⇒ x = 4, y = -3
Hence, x = 4 and y = -3.
APPEARS IN
संबंधित प्रश्न
The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
For what value of , the following system of equations will be inconsistent?
4x + 6y - 11 = 0
2x + ky - 7 = 0
Determine the values of a and b so that the following system of linear equations have infinitely many solutions:
(2a - 1)x + 3y - 5 = 0
3x + (b - 1)y - 2 = 0
Solve for x and y:
3x - 5y - 19 = 0, -7x + 3y + 1 = 0
Solve for x and y:
`(bx)/a + (ay)/b = a^2 + b^2, x + y = 2ab`
Show that the system of equations
6x + 5y = 11,
9x + 152 y = 21
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 sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3. They are in the ratio of 2: 3. Determine the fraction.
Write the number of solutions of the following pair of linear equations:
x + 2y -8=0,
2x + 4y = 16
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.