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Question
Show that the following system of equations has a unique solution:
`x/3 + y/2 = 3, x – 2y = 2.`
Also, find the solution of the given system of equations.
Solution
The given system of equations is:
`x/3 + y/2 = 3`
`⇒ (2x+3y)/6 = 3`
2x + 3y = 18
⇒2x + 3y – 18 = 0 ….(i)
and
x – 2y = 2
x – 2y – 2 = 0 …..(ii)
These equations are of the forms:
`a_1x+b_1y+c_1 = 0 and a_2x+b_2y+c_2 = 0`
where, `a_1 = 2, b_1= 3, c_1= -18 and a_2 = 1, b_2= -2, c_2= -2`
For a unique solution, we must have:
`(a_1)/(a^2) ≠ (b_1)/(b_2), i.e., 2/1 ≠ 3/(−2)`
Hence, the given system of equations has a unique solution.
Again, the given equations are:
2x + 3y – 18 = 0 …..(iii)
x – 2y – 2 = 0 …..(iv)
On multiplying (i) by 2 and (ii) by 3, we get:
4x + 6y – 36 = 0 …….(v)
3x - 6y – 6 = 0 ……(vi)
On adding (v) from (vi), we get:
7x = 42
⇒x = 6
On substituting x = 6 in (iii), we get:
2(6) + 3y = 18
⇒3y = (18 - 12) = 6
⇒y = 2
Hence, x = 6 and y = 2 is the required solution
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