For which value(s) of k will the pair of equations k x + 3y = k – 3 12x + ky = k have No solution?
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For which value(s) of k will the pair of equations k x + 3y = k – 3 12x + ky = k have No solution?
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For the pair of equations having No solution, k = -6
Given,
The pair of linear equations:
kx + 3y = k – 3 …(1)
12x + ky = k …(2)
To Find,
Value of k, for pair of equations having No solution.
Solution,
We can solve this problem by the following process:
On comparing the equations (1) and (2) with standard form of linear equation ax + by = c = 0 We get,
a₁ = k, b₁ = 3, c₁ = -(k – 3)
a₂ = 12, b₂ = k, c₂ = – k
For no solution of the pair of linear equations, the condition is
[tex]\frac{a{1} }{a{2}} = \frac{b{1} }{b{2}} \neq \frac{c{1} }{c{2}}[/tex]
[tex]\frac{k}{12} = \frac{3 }{k} \neq \frac{-(k-3) }{-k}[/tex] ...(3)
By dividing the equation(3) into two parts we obtain,
[tex]\frac{k}{12} = \frac{3 }{k}[/tex] ....(4) and [tex]\frac{3 }{k} \neq \frac{-(k-3) }{-k}[/tex] ....(5)
On solving equation(4) we get,
⇒ [tex]\frac{k}{12} = \frac{3 }{k}[/tex]
⇒ [tex]k^{2} = 12* 3[/tex]
⇒ [tex]k = \sqrt{36}[/tex]
⇒ [tex]k[/tex] = ± 6
On solving equation(5) we get,
⇒ [tex]\frac{3 }{k} \neq \frac{-(k-3) }{-k}[/tex]
⇒ [tex]3k \neq k^{2} - 3k[/tex]
⇒ [tex]k^{2} - 6k \neq 0[/tex]
⇒ [tex]k \neq 0[/tex] and [tex]k \neq 6[/tex]
From the above we obtain [tex]k = -6[/tex]
Hence, For the pair of equations having No solution, k = -6
Answer:
The value of k for which the pair of equations will have no solution is - 6.
Step-by-step-explanation:
The given linear equations are
kx + 3y = k - 3
kx + 3y - ( k - 3 ) = 0 - - - ( 1 )
12x + ky = k
12x + ky - k = 0 - - - ( 2 )
Comparing equation ( 1 ) with a₁x + b₁y + c₁ = 0, we get,
Comparing equation ( 2 ) with a₂x + b₂y + c₂ = 0, we get,
For system of linear equations to have no solution,
a₁ / a₂ = b₁ / b₂ ≠ c₁ / c₂
⇒ k / 12 = 3 / k ≠ - ( k - 3 ) / - k
Now,
k / 12 = 3 / k
⇒ k * k = 3 * 12
⇒ k² = 36
⇒ k = ± √36
⇒ k = ± √( 6 * 6 )
⇒ k = ± 6
∴ k = 6 OR k = - 6
Now,
a₁ / a₂ ≠ c₁ / c₂
⇒ k / 12 ≠ - ( k - 3 ) / - k
⇒ k / 12 ≠ ( k - 3 ) / k
⇒ k * k ≠ 12 ( k - 3 )
⇒ k² ≠ 12k - 36
⇒ k² - 12k + 36 ≠ 0
⇒ k² - 6k - 6k + 36 ≠ 0
⇒ k ( k - 6 ) - 6 ( k - 6 ) ≠ 0
⇒ ( k - 6 ) ( k - 6 ) ≠ 0
⇒ ( k - 6 ) ≠ 0
⇒ k - 6 ≠ 0
⇒ k ≠ 6
∴ The value of k for which the pair of equations will have no solution is - 6.