check the given equations have either unique solution or infinite solution or no solution. 5x-3y=8,10x-6y=19 About the author Sadie
No solution [tex]\\[/tex] Explanation: In the given pair of linear equations, For first Equation: a₁ = 5 b₁ = -3 c₁ = 8 [tex]\\[/tex] For the second Equation: a₂ = 10 b₂ = -6 c₂ = 19 [tex]\\[/tex] For finding whether the equations have unique solution, infinitely many solutions or no solution let’s compare the ratios. Here, [tex] \tt{\dfrac{5}{10} = \dfrac{ – 3}{ – 6} \neq \dfrac{8}{19}} \\ \\ [/tex] [tex] \implies \tt{ \dfrac{1}{2} = \dfrac{1}{2} \neq \dfrac{8}{19} } \\ \\ [/tex] [tex] \tt{ \dfrac{a _{1} }{a _{2}} = \dfrac{b_{1} }{b _{2} } \neq \dfrac{c _{1} }{c _{2} }} \\ \\ [/tex] So, the given pair of linear equations does not have any solution. KNOW MORE: If a1/a2 = b1/b2 = c1/c2, then the given pair of linear equations would have infinitely many solutions. If a1/a2 ≠ b1/b2 ≠ c1/c2, then the given pair of linear equations would have unique solution. Reply
No solution
[tex]\\[/tex]
Explanation:
In the given pair of linear equations,
For first Equation:
[tex]\\[/tex]
For the second Equation:
[tex]\\[/tex]
For finding whether the equations have unique solution, infinitely many solutions or no solution let’s compare the ratios.
Here,
[tex] \tt{\dfrac{5}{10} = \dfrac{ – 3}{ – 6} \neq \dfrac{8}{19}} \\ \\ [/tex]
[tex] \implies \tt{ \dfrac{1}{2} = \dfrac{1}{2} \neq \dfrac{8}{19} } \\ \\ [/tex]
[tex] \tt{ \dfrac{a _{1} }{a _{2}} = \dfrac{b_{1} }{b _{2} } \neq \dfrac{c _{1} }{c _{2} }} \\ \\ [/tex]
So, the given pair of linear equations does not have any solution.
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