Question

Y - 5x = 3, 5x-y=2 y - 3x = 4, y+3x=4 2y + 4x = 2, 6x+3y=3

Answer 1

Ответ:

Заранее спасибо тем кто поможет!

Объяснение:

Answer 2

To solve these systems of equations, we can use the method of substitution or elimination. Let's solve each system one by one:

System 1: Equation 1: Y - 5x = 3 Equation 2: 5x - y = 2

We can solve this system using the method of substitution: From Equation 1, we can isolate Y: Y = 5x + 3

Substitute this value of Y into Equation 2: 5x - (5x + 3) = 2 5x - 5x - 3 = 2 -3 = 2 (inconsistent equation)

Therefore, System 1 does not have a solution.

System 2: Equation 1: y - 3x = 4 Equation 2: y + 3x = 4

We can solve this system using the method of elimination: Add the two equations together: (y - 3x) + (y + 3x) = 4 + 4 2y = 8 y = 4

Substitute the value of y back into either equation: y - 3x = 4 4 - 3x = 4 -3x = 0 x = 0

Therefore, the solution to System 2 is x = 0 and y = 4.

System 3: Equation 1: 2y + 4x = 2 Equation 2: 6x + 3y = 3

We can solve this system using the method of elimination: Multiply Equation 1 by 3 and Equation 2 by 2 to make the coefficients of y in both equations the same: 6y + 12x = 6 6x + 3y = 3

Now, subtract the two equations: (6y + 12x) - (6x + 3y) = 6 - 3 6y - 3y + 12x - 6x = 3 3y + 6x = 3

Divide the entire equation by 3: y + 2x = 1

We now have a new equation: y + 2x = 1

This equation is not independent of the previous two equations. It is a linear combination of the original equations. Therefore, System 3 does not have a unique solution.

To summarize:

System 1 does not have a solution. System 2 has a solution: x = 0 and y = 4. System 3 does not have a unique solution.