Question

1 point
Q.20: The value of x and y of pair of
equations : 3x - y = 3 and 9x - 3y = 9
are :-
0 Infinity
0 1,1
0 -3,3
O 0,0​

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow 9x-3y=9  -------eq^1$

$\sf\dashrightarrow 3x - y = 3 ----------eq^2$

METHOD IN USE,

$\sf\large\dashrightarrow\underline ELIMINATION \:BY\:SUBSTITUTING,$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore 9x-3y=9  --eq^1$

$\sf\therefore 3x - y = 3 --eq^2$

TAKING EQUATION 2 ,

$\sf\implies 3x - y = 3$

$\sf\implies 3x = 3 + y$

$\sf\implies x =  \dfrac{3 + y}{3}---eq^3$

SUBSTITUTING VALUE OF X (EQUATION 3) IN EQUATION 1,

WE GET,

$\sf\implies 9 \times \dfrac{3 + y}{3} - 3y = 9$

$\sf\implies \dfrac{27+9y}{3} -3y =9$

$\sf\implies \dfrac{27-9y-9y}{3}=9$

$\sf\implies \dfrac{\cancel{9}(3-1y-1y)}{\cancel{3}}=9$

$\sf\implies 3(3-2y)=9$

$\sf\implies 3-2y= \dfrac{9}{3}$

$\sf\implies 3-2y= \cancel\dfrac{9}{3}$

$\sf\implies3-2y= 3$

 $\sf\implies 2y=0$

$\sf\implies y=0$

$\rm{\boxed{\bf{ y=0}}}$

now,

SUBSTITUTING THE VALUE OF Y IN EQUATION 3,

$\sf\implies x =  \dfrac{3 + y}{3}$

$\sf\implies x= \dfrac{3 + 0}{3}$

$\sf\implies x= \dfrac{3}{3}$

$\sf\implies x= \cancel \dfrac{3}{3}$

$\sf\implies x=1$

$\rm{\boxed{\bf{ x=1}}}$

$\large\underline\bold{hence, x=1 \:and\:y=0}$

_________________________

Answer 2

$\huge\underline{\underline{\bold{\green{AnSwEr}}}}$

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow 9x-3y=9  -------eq^1$

$\sf\dashrightarrow 3x - y = 3 ----------eq^2$

METHOD IN USE,

$\sf\large\dashrightarrow\underline ELIMINATION \:BY\:SUBSTITUTING,$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore 9x-3y=9  --eq^1$

$\sf\therefore 3x - y = 3 --eq^2$

TAKING EQUATION 2 ,

$\sf\implies 3x - y = 3$

$\sf\implies 3x = 3 + y$

$\sf\implies x =  \dfrac{3 + y}{3}---eq^3$

SUBSTITUTING VALUE OF X (EQUATION 3) IN EQUATION 1,

WE GET,

$\sf\implies 9 \times \dfrac{3 + y}{3} - 3y = 9$

$\sf\implies \dfrac{27+9y}{3} -3y =9$

$\sf\implies \dfrac{27-9y-9y}{3}=9$

$\sf\implies \dfrac{\cancel{9}(3-1y-1y)}{\cancel{3}}=9$

$\sf\implies 3(3-2y)=9$

$\sf\implies 3-2y= \dfrac{9}{3}$

$\sf\implies 3-2y= \cancel\dfrac{9}{3}$

$\sf\implies3-2y= 3$

 $\sf\implies 2y=0$

$\sf\implies y=0$

$\rm{\boxed{\bf{ y=0}}}$

now,

SUBSTITUTING THE VALUE OF Y IN EQUATION 3,

$\sf\implies x =  \dfrac{3 + y}{3}$

$\sf\implies x= \dfrac{3 + 0}{3}$

$\sf\implies x= \dfrac{3}{3}$

$\sf\implies x= \cancel \dfrac{3}{3}$

$\sf\implies x=1$

$\rm{\boxed{\bf{ x=1}}}$

$\large\underline\bold{hence, x=1 \:and\:y=0}$

_________________________