Question

FOR EQUATION 3X + 3Y =21 COMPLETE THE FOLLOWING TABLE GIVEN BELOW [FORMAT Y=---,X=----,(X,Y)=------,(X,Y)=----.]

Answer 1

ANSWER✔

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

$\dashrightarrow x_1= 2$

$\dashrightarrow y_2=7$

$\dashrightarrow equation= 3x+3y=21$

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

$\sf\dashrightarrow x_1= 2$

$\therefore finding\:value\:of\:y_1$

$\dashrightarrow 3x+3y= 21$

$\implies 3(2)+3(y)=21$

$\implies 6+3y=21$

$\implies 3y=21-6$

$\implies y= \dfrac{15}{3}$

$\implies y= \cancel \dfrac{15}{3}$

$\implies y_1=5$

$\large{\boxed{\bf{ \star\:\: y_1=5\:\: \star}}}$

$\sf\dashrightarrow y_2= 7$

$\therefore finding\:the\:value\:of\:x_2$

$\implies 3x+3(7)=21$

$\implies 3x+21=21$

$\implies x=0$

$\large{\boxed{\bf{ \star\:\:x_2=0 \:\: \star}}}$

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Answer 2

$\large\underline\bold{SOLUTION \red{\huge {\checkmark}}}$

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

$\dashrightarrow x_1= 2$

$\dashrightarrow y_2=7$

$\dashrightarrow equation= 3x+3y=21$

$\sf\dashrightarrow x_1= 2$

$\therefore finding\:value\:of\:y_1$

$\dashrightarrow 3x+3y= 21$

$\implies 3(2)+3(y)=21$

$\implies 6+3y=21$

$\implies 3y=21-6$

$\implies y= \dfrac{15}{3}$

$\implies y= \cancel \dfrac{15}{3}$

$\implies y_1=5$

$\large{\boxed{\bf{ \star\:\: y_1=5\:\: \star}}}$

$\sf\dashrightarrow y_2= 7$

$\therefore finding\:the\:value\:of\:x_2$

$\implies 3x+3(7)=21$

$\implies 3x+21=21$

$\implies 3x=21-21$

$\implies x= \dfrac{0}{3}$

$\implies x=0$

$\large{\boxed{\bf{ \star\:\:x_2=0 \:\: \star}}}$

$\sf\large\dashrightarrow (x_1,y_1)=(2,5)\\ \dashrightarrow (x_2,y_2)= ( 0,7)$

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