Question

(a+3)x+4y=3a+5,ax+(a-1)y=2​

Answer 1

ax+3x+4y=3a+5

4y=3a+5-ax-3x  Делим правую часть на 4

y= $\frac{3}{4}$a+$\frac{5}{4}$-$\frac{1}{4}$ax-$\frac{3}{4}$x

ax+(a-1)y=2

ax-ay-y=2

Дальше никак не решить

Answer 2

To solve this system of linear equations for the variables a, x, and y, you can use the method of substitution or elimination. In this case, let's use the elimination method.

Given equations:

1. (a + 3)x + 4y = 3a + 5
2. ax + (a - 1)y = 2

First, let's eliminate one of the variables. We can do this by multiplying both sides of Equation (2) by (a + 3) to make the coefficients of x in both equations equal. This will allow us to eliminate x when we subtract Equation (2) from Equation (1).

Multiply Equation (2) by (a + 3): (a + 3)[ax + (a - 1)y] = (a + 3)(2)

This gives us: a(ax) + (a - 1)(a + 3)y = 2(a + 3)

Now, let's expand both sides: a^2x + (a^2 + 2a - a - 3)y = 2(a + 3)

Simplify the equation: a^2x + (a^2 + a - 3)y = 2(a + 3)

Now, we have two equations:

1. (a + 3)x + 4y = 3a + 5
2. a^2x + (a^2 + a - 3)y = 2(a + 3)

Next, we'll eliminate x by subtracting Equation (2) from Equation (1):

[(a + 3)x + 4y] - [a^2x + (a^2 + a - 3)y] = (3a + 5) - [2(a + 3)]

Now, distribute the negative sign on the left side: (a + 3)x - a^2x + (a^2 + a - 3)y - 4y = 3a + 5 - 2(a + 3)

Combine like terms: (a + 3 - a^2)x + (a^2 + a - 3 - 4)y = 3a + 5 - 2a - 6

Simplify further: (-a^2 + a + 3)x + (a^2 + a - 3 - 4)y = a - 1

Now, let's simplify the equation:

1. (-a^2 + a + 3)x + (a^2 + a - 7)y = a - 1

Now, you have a system of two linear equations in two variables (x and y):

1. (-a^2 + a + 3)x + (a^2 + a - 7)y = a - 1
2. ax + (a - 1)y = 2

You can solve this system of equations for the values of x and y by using any appropriate method, such as substitution or elimination. The solution will depend on the specific values of 'a' that you have. If you provide a specific value for 'a', I can help you find the solution.