Solve the system y= 4x + 1 and 3x+2y =13

Question 1

Question 2

Given :

${\qquad \sf \dashrightarrow y= 4x + 1 \: –––– \sf \: (i)}$

${\qquad \sf \dashrightarrow 3x + 2y = 13 \: –––– \sf \: (ii)}$

Now, Substituing the value of y in equation (ii) :

${\qquad \sf \dashrightarrow \: 3x + 2y = 13 }$

${\qquad \sf \dashrightarrow \: 3x + 2(4x + 1) = 13 }$

${\qquad \sf \dashrightarrow \: 3x + 8x + 2 = 13 }$

${\qquad \sf \dashrightarrow \: 11x + 2 = 13 }$

${\qquad \sf \dashrightarrow \: 11x = 13 - 2 }$

${\qquad \sf \dashrightarrow \: 11x = 11 }$

${\qquad \sf \dashrightarrow \: x = (11)/(11) }$

${\qquad \sf \dashrightarrow \bf \: x = 1 }$

Now, substituting the value of x in equation (i) :

${\qquad \sf \dashrightarrow y= 4x + 1}$

${\qquad \sf \dashrightarrow y= 4(1) + 1}$

${\qquad \sf \dashrightarrow y= 4 + 1}$

${\qquad \bf \dashrightarrow y= 5}$

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