Resolve the equations 3x - 5y = -16 and 2x + 5y = 31 and calculate the x and y value.

3x - 5y = -16 ---------- (1) 2x + 5y = 31 ---------- (2)

x and y value

Let us calculate the value of x and y for the given linear equations using elimination method.

In the two equations coefficients of 'y' in both the equations are numerically equal. So, elimination of 'y' can be done easily. Now, add both the equation.

3x - 5y | = -16 ----- (1) | |

2x + 5y | = 31 ----- (2) | |

(1) + (2) | 5x | = 15 |

Therefore, x=15/5 = 3

Substitute the value x = 3 in any one of the equation. Let us substitute the x value in the 1st equation, 3x - 5y = -16.

3x - 5y | = -16 |

3(3)-5y | = -16 |

9-5y | = -16 |

-5y | = -16-9 |

-5y | = -25 |

y | = 25/5 |

y | = 5 |

Resolve the equations 6x + 4y = 6 and 7x - 8y = 10 and calculate the x and y value.

6x + 4y = 6 ---------- (1) 7x - 8y = 10 ---------- (2)

x and y value

Let us calculate the value of x and y for the given linear equations using elimination method.

In the given two equations Coefficients of 'x' and 'y' are numerically different. So, in-order to make the co-efficients equal, let us multiply the first equation by 2 If the first equation is multiply by 2 , then coefficient of y is equal numerically.

(1) * 2 | 12x +8y | = 12 ----- (3) |

7x - 8y | = 10 ----- (2) | |

(3) + (2) | 19x | = 22 |

Therefore, x = 22/19 = 1.158

Substitute the value x = 1.158 in (1) equation. 6x + 4y = 6 ---------- (1)

6(1.158) + 4y | = 6 |

6.948 + 4y | = 6 |

4y | = 6-6.948 |

4y | = -0.948 |

y | = -0.948/4 |

y | = -0.237 |

Therefore the solution is (x,y) = (1.158,-0.237)