Question

20х-15у=51 , 4х-3у=10,2 система уравнений?​

Answer 1

Ответ:

решение смотри на фотографии

Объяснение:

Answer 2

Yes, the equations you provided form a system of two linear equations with two variables. Here's how you can solve it using the method of substitution or elimination:

Method 1: Substitution

1. Solve one equation for one variable in terms of the other variable. Let's solve the second equation for x: 4x - 3y = 10 4x = 10 + 3y x = (10 + 3y) / 4

2. Substitute the expression for x in terms of y into the other equation: 20x - 15y = 51 20((10 + 3y) / 4) - 15y = 51

   Simplify the equation: (20/4)(10 + 3y) - 15y = 51 5(10 + 3y) - 15y = 51 50 + 15y - 15y = 51 50 = 51

   This equation is inconsistent, meaning there is no solution that satisfies both equations simultaneously. Therefore, the system of equations is inconsistent, and there are no values of x and y that satisfy both equations.

Method 2: Elimination

1. Multiply the first equation by 4 and the second equation by 20 to eliminate the x variable: Equation 1: 20x - 15y = 51 (multiply by 4) -> 80x - 60y = 204 Equation 2: 4x - 3y = 10 (multiply by 20) -> 80x - 60y = 200

2. Subtract equation 2 from equation 1 to eliminate the x variable: (80x - 60y) - (80x - 60y) = 204 - 200 -60y + 60y = 4 0 = 4

   This equation is inconsistent, which means there is no solution that satisfies both equations simultaneously. Therefore, the system of equations is inconsistent, and there are no values of x and y that satisfy both equations.

In conclusion, the system of equations you provided does not have a solution.