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Vladimir [108]
2 years ago
11

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

Mathematics
1 answer:
Irina18 [472]2 years ago
3 0

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  =  \dfrac{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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