Question

6p-3q=1 , 5p+q=1 by substitution method ​

Answer 1

Answer:

p = 1/3, q = 1/3

Step-by-step explanation:

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Answer 2

Answer:

The solution of the given simultaneous equations is

$\displaystyle{\boxed{\red{\sf\:(\:p\,,q\:)\:=\:\left(\:\dfrac{4}{21}\:,\:\dfrac{1}{21}\:\right)\:}}}$

Step-by-step-explanation:

The given simultaneous equations are

6p - 3q = 1 - - - ( 1 ) &

5p + q = 1 - - - ( 2 )

Now,

5p + q = 1 - - - ( 2 )

⇒ q = 1 - 5p

By substituting this value in equation ( 1 ), we get,

6p - 3q = 1 - - - ( 1 )

⇒ 6p - 3 * ( 1 - 5p ) = 1

⇒ 6p - 3 + 15p = 1

⇒ 6p + 15p = 1 + 3

⇒ 21p = 4

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:p\:=\:\dfrac{4}{21}}}}}$

By using the value of p,

q = 1 - 5p

$\displaystyle{\implies\sf\:q\:=\:1\:-\:5\:\times\:\dfrac{4}{21}}$

$\displaystyle{\implies\sf\:q\:=\:1\:-\:\dfrac{20}{21}}$

$\displaystyle{\implies\sf\:q\:=\:\dfrac{21\:-\:20}{21}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:q\:=\:\dfrac{1}{21}}}}}$

∴ The solution of the given simultaneous equations is

$\displaystyle{\boxed{\red{\sf\:(\:p\,,q\:)\:=\:\left(\:\dfrac{4}{21}\:,\:\dfrac{1}{21}\:\right)\:}}}$