Question

4(х+у+1)= х+3у-2 и 5х+у-3=2(2х-у)+3

Answer 1

$\left \{ {{4(x+y+1)=x+3y-2} \atop {5x+y-3=2(2x-y)+3}} \right.$

4(x+y+1)=x+3y-2 
3x+y+4+2=0 

y=-3x-6

 

5x+y-3=2(2x-y)+3

5x+(-3x-6)-3=2(2x-(-3x-6))+3

2x-6-3-2*(2x-(-3x-6))-3=0

2x-9-2*(5x+6)-3=0

2x-9-10x-12-3=0

-8x-21-3=0

-8x-24=0

x=-24/8
x=-3

4(x+y+1)=x+3y-2 
4((-3)+y+1)=(-3)+3y-2 
-8+4y+3-3y+2=0
-5+y+2=0
-3+y=0

y=3

Ответ: x=-3; y=3.

Answer 2

To solve the system of equations:

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

Let's simplify and solve these equations step by step:

1. Distribute the 4 on the left side: 4x + 4y + 4 = x + 3y - 2

Rearrange the terms: 4x - x + 4y - 3y = -2 - 4

Combine like terms: 3x + y = -6 ----(Equation 1)

2. Distribute the 2 on the right side: 5x + y - 3 = 4x - 2y + 3

Rearrange the terms: 5x - 4x + y + 2y = 3 + 3

Combine like terms: x + 3y = 6 ----(Equation 2)

Now, we have a system of two equations: Equation 1: 3x + y = -6 Equation 2: x + 3y = 6

We can solve this system by either substitution or elimination method. Let's use the elimination method.

Multiply Equation 1 by 3 and Equation 2 by 1, to make the coefficients of x in both equations the same:

3 * (3x + y) = 3 * (-6) ---> 9x + 3y = -18 ----(Equation 3) 1 * (x + 3y) = 1 * 6 ---> x + 3y = 6 ----(Equation 4)

Now subtract Equation 4 from Equation 3:

(9x + 3y) - (x + 3y) = -18 - 6 8x = -24

Divide both sides by 8: x = -24/8 x = -3

Substitute the value of x into Equation 4:

(-3) + 3y = 6

Combine like terms: 3y = 9

Divide both sides by 3: y = 3

So, the solution to the system of equations is x = -3 and y = 3.