Advertisements
Advertisements
प्रश्न
Solve for x and y:
`x/a + y/b = 2, ax – by = (a^2 – b^2)`
उत्तर
The given equations are:
`x/a + y/b = 2`
`⇒(bx+ay)/(ab)` = 2 [Taking LCM]
⇒bx + ay = 2ab …….(i)
Again, ax – by =` (a^2 – b^2)` …..(ii)
On multiplying (i) by b and (ii) by a, we get:
`b^2x + bay = 2ab^2` ……..(iii)
`a^2x – bay = a(a^2 – b^2)` …….(iv)
On adding (iii) from (iv), we get:
`(b^2 + a^2)x = 2a^2b + a(a^2 – b^2)`
`⇒(b^2 + a^2)x = 2ab^2 + a^3 – ab^2`
`⇒(b^2 + a^2)x = ab^2 + a^3`
`⇒(b^2 + a^2)x = a(b^2 + a^2)`
`⇒x =( a(b^2+ a^2))/((b^2+ a^2)) = a`
On substituting x = a in (i), we get:
ba + ay = 2ab
⇒ ay = ab
⇒y = b
Hence, the solution is x = a and y = b.
APPEARS IN
संबंधित प्रश्न
In the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:
x − 3y = 3
3x − 9y = 2
Find the value of k for which each of the following system of equations has infinitely many solutions :
\[kx + 3y = 2k + 1\]
\[2\left( k + 1 \right)x + 9y = 7k + 1\]
Find the value of k for which each of the following system of equations has infinitely many solutions :
2x +3y = k
(k - 1)x + (k + 2)y = 3k
Solve for x and y:
3x - 5y - 19 = 0, -7x + 3y + 1 = 0
Solve for x and y:
`(x + y - 8)/2 = (x + 2y -14)/3 = (3x + y - 12)/11`
Find the value of k for which the system of equations has a unique solution:
4x - 5y = k,
2x - 3y = 12.
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 linear equations has a unique solution:
(k – 3) x + 3y – k, kx + ky - 12 = 0.
Find the values of k for which the system of equations 3x + ky = 0,
2x – y = 0 has a unique solution.
Find the value of k for which the system of equations x + 2y – 3 = 0 and 5x + ky + 7 = 0 is inconsistent.